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CBSE maths XI STD

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Irrational Numbers You might have learnt about representing various types of numbers such as Natural numbers, Whole numbers, Integers, Rational numbers on the number line. Rational Numbers: Numbers that can be written in the form of , where p, q and . The collection of Rational numbers is denoted by Q. Between any two rational numbers there exists infinitely many rational numbers. Irrational numbers:Numbers which cannot be expressed in the form of , where p, q and . The collection of irrational numbers is denoted by . Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Using this theorem we can represent the irrational numbers on the number line. Real Numbers We already know about various types of numbers such as Natural numbers, Whole numbers, Integers, Rational and Irrational numbers. We already learnt how to represent these numbers on the number line. A number that can be expressed as , where , is known as Rational number. Collection of rational numbers is denoted by Q.
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Page 1: CBSE maths XI STD

Irrational Numbers

You might have learnt about representing various types of numbers such as Natural numbers, Whole numbers, Integers, Rational numbers on the number line.

Rational Numbers: Numbers that can be written in the form of , where p, q and .

The collection of Rational numbers is denoted by Q. Between any two rational numbers there exists infinitely many rational numbers.

Irrational numbers:Numbers which cannot be expressed in the form of , where p, q and.

The collection of irrational numbers is denoted by .

Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Using this theorem we can represent the irrational numbers on the number line.

Real Numbers

We already know about various types of numbers such as Natural numbers, Whole numbers, Integers, Rational and Irrational numbers. We already learnt how to represent these numbers on the number line.

A number that can be expressed as , where , is known as Rational number. Collection of rational numbers is denoted by Q.

Page 2: CBSE maths XI STD

A number that cannot be expressed as , where , is called an irrational number. Collection of irrational numbers is denoted by . The square root of any prime number is irrational. There are infinitely many irrational numbers between two rational numbers.The collection of real numbers is the collection of all rational numbers and irrational numbers together. It is represented by R. A real number is either rational or irrational.A rational number can be expressed as its decimal expansion.In case of a division:

• If remainder becomes zero after certain stage, then the decimal expansion is terminating.

• If the remainder never becomes zero but repeats after certain stage, then the decimal expansion is non-terminating recurring.

The decimal expansion of a rational number is either terminating or non-terminating recurring. The decimal expansion of an irrational number is non-terminating non-recurring. Every real number can be represented on a number line uniquely.This process of visualisation of representing a decimal expansion on the number line is known as the process of successive magnification.

Page 3: CBSE maths XI STD

Operations on Real Numbers

We already learnt that the collection of rational numbers and irrational numbers together is known as Real Numbers. This collection is denoted by R.The sum, difference and the product of two rational numbers is always a rational number. The quotient of a division of one rational number by a non-zero rational number is a rational number. Rational numbers satisfy the closure property under addition, subtraction, multiplication and division.The sum, difference, multiplication and division of irrational numbers are not always irrational. Irrational numbers do not satisfy the closure property under addition, subtraction, multiplication and division.Real numbers satisfy the commutative, associative and distributive laws. These can be stated as :

• Commutative Law of Addition:

• Commutative Law of Multiplication: • Associative Law of Addition: • Associative Law of Multiplication: • Distributive Law: or

We can represent real numbers on the number line. The square root of any positive real number exists and that also can be represented on number lineThe sum or difference of a rational number and an irrational number is an irrational number.The product or division of a rational number with an irrational number is an irrational number.Some of the basic identities involving square roots are:

a, b, c and d are positive real numbers. The process of converting the denominator into a rational number is called rationalising the denominator.

Page 4: CBSE maths XI STD

Laws of Exponents

An exponent is a mathematical notation that represents how many times a base is multiplied by itself. Other terms used to define exponents are ‘power’ or ‘index’. An exponential term is a term that can be expressed as a base raised to an exponent. For example, in an exponential expression ,

is the base and ‘ is the exponent.The exponent can be a number or a constant; they can also be a variable. They are generally positive real numbers, but they can also be negative numbers.Laws of exponents: If a and b are any real numbers then

Where m and n are rational numbers.

Polynomials in One Variable

An expression of the form where

, , are real numbers and is called a polynomial in . Degree of a polynomial: The highest power of the variable in a polynomial is called degree of the polynomial.Zero polynomial: The polynomial "0", which has no term at all, is called Zero polynomial.

Page 5: CBSE maths XI STD

Constant Polynomial: It is a polynomial of degree 0. The value of constant function is constant irrespective of values of "x". Linear polynomial: A polynomial of degree one is called a first-degree or linear polynomial. The

general form of such a polynomial is , where . In a linear polynomial the maximum number of terms is two.Quadratic polynomial: A polynomial of degree two is called a second degree or quadratic

polynomial. Its general form is , where . In a Quadratic polynomial the maximum number of terms is three.Cubic polynomial: A polynomial of degree three is called a third-degree or cubic polynomial and

is represented as , where .Such a polynomial can have a maximum of four terms.Zeros of a Polynomial: Roots of a polynomial: It is a solution to the polynomial equation p(x)=0 i.e. a number "a" is said to be a zero of a polynomial if p(a) = 0.If we draw the graph of p(x) =0, the values where the curve cuts the X-axis are called Zeros of the polynomial. Methods to Determine Zero of a Polynomial Trial and error Method:

Equating polynomial to zero: In this method, you can find the zero of the polynomial by taking as the subject.

, ,

,

p(a) = 0

Remainder Theorem

Remainder= Dividend - (Divisor × Quotient)

Remainder Theorem: if is a polynomial in , and is divided by then the

remainder is i.e, . Here

Based on the remainder theorem:

If a polynomial is divided by , then the remainder is .

If a polynomial is divided by , then the remainder is .

If a polynomial is divided by , then the remainder is .

Remainder= Dividend - (Divisor × Quotient)

Page 6: CBSE maths XI STD

p(a) or or or

Factor Theorem

A is an consisting of multiple terms of

These terms are combined by using mathematical operators - addition, subtraction and multiplication.

The of the must be .

The of the in a is called

A term can be a , a or a product of the two.

A real number that precedes the variable is called the .

The general form of a is where

, .... , are real numbers and is called a polynomial in of degree nA polynomial whose degree is equal to two is called a Quadratic Polynomial.A polynomial whose degree is equal to three is called a cubic Polynomial.

Factor Theorem: is a polynomial and is a real number, if is divided by a linear polynomial , then the remainder is .

Factorization of Polynomials Using Algebraic Identities

We know that the algebraic identities of second degree are

These identities can be used to factorise quadratic polynomials.A polynomial is said to be cubic polynomial if its degree is threeThe algebraic identities used in factorising a third degree polynomial are:

Cubic polynomials can be factorised using Factor theorem

Page 7: CBSE maths XI STD

Factor theorem: is a polynomial and is a real number, if , then is a

factor of

Cartesian Plane

A Cartesian plane, named after the mathematician Rene Descartes, is a plane with a rectangular

coordinate system that associates each point in the plane, with an ordered pair . The horizontal line XOX′ is the and the vertical line YOY′ is the . The line OX is called positive x-axis, is called negative X-axis, OY is called positive y-axis and is called negative y-axis.The lines and divides the plane into four equal parts. Each part is called a quadrant. The quadrants , are called first quadrant, second quadrant, third quadrant and fourth quadrant respectively. If P ( ) be any point in the Cartesian plane, then is called -coordinate or Abscissa and is called coordinate or Ordinate,

The equation of x- axis is y=0 and equation of y-axis is x=0.Equation of any line parallel to x-axis is y=k and equation of any line parallel to y-axis is x=k. The general form of equation of a line is ax+by+c=0, Where, a, b and c are real numbers and a 0 or b 0.

Page 8: CBSE maths XI STD

ax+by+c=0y=0y=kx=0x=k

Introduction to Linear Equation in Two Variables

An equation which can be put in the form is called a linear equation in two

variables and where a, b and c are real numbers and a and b are not both zero.

Here a and b are called coefficients of x and y respectively and c is called constant term.

The equation is called linear because the equation is of the first degree.

A solution is an ordered pair of real numbers which satisfies the equation. In general, any linear equation in two variables will have infinitely many solutions.

Graphical Representation of Linear Equations in Two Variables

An in the form of where a, b and c are real numbers and a and b are not both zero is called a linear equation in two variables x and y.A linear equation in two variables has infinitely many solutions.The solutions of a linear equation can be obtained by substituting different values for x in the equation to find the corresponding values of y. The values of x and y are represented as an order pair.To plot the graph of this equation, we’ll first find its solutions algebraically and then plot the points on the graph.

Any linear equation of the form represents a straight line on the graph. Every point that satisfies the linear equation lies on the line.Every point that lies on the line is a solution of the linear equation.

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A point that does not lie on the line is not solution of the linear equation.

The equation of X axis is Any line parallel to x-axis is of the form y=k

The equation of y-axis is Any line parallel to y-axis is of the form x=k

ax+by+c=0

y=0x=0y=kx=k

Introduction to Euclid's Geometry

The word 'Geometry' is derived from the Greek words 'Geo' means 'Earth' and 'Metron' means to 'Measure'. Around 325 BC Euclid, a teacher of mathematics at Alexandria in Egypt, collected all the known work and arranged it in his famous treatise, called 'Elements'. He divided the 'Elements' into thirteen chapters, each called a book. These books influenced the whole world's understanding of geometry for generations to come. Euclid's Definitions:1: A point is that which has no part.2: A line is breadthless length.3: The ends of a line are points. 4: A straight line is a line which lies evenly with the points on itself. 5: A surface is that which has length and breadth only.6: The edges of a surface are lines.7: A plane surface is a surface which lies evenly with the straight lines on itself.

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Euclid's Postulates:1: A line can be drawn from any point to any point.2: A terminated line can be produced indefinitely. 3: It is possible to describe a circle with any centre any distance.4: All right angles are equal to one another.5: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.

Euclid's Axioms:1: Things which are equal to the same things are also equal to one another.2: If equals are added to equals, then the wholes are equal.3: If equals are subtracted from equals, then the remainders are equal.4: Things which coincide with one another are equal to one another.5: The whole is greater than the part.6 : Things which are double of the same things are equal to one another.7 : Things which are halves of the same things are equal to one another.

A line is a collection of points along a straight path. A line has no endpoints.A line segment is a part of a line with two end points. a part of line with one end point is called a ray.All the points that lie on the same line are collinear points.Consider the two rays AB and AC originating from the same point A.The union of two rays AB and AC is called an angle.The rays that form the angle are called the arms of the angle.The intersection point is called the vertex of the angle.The size of an angle is measured in degrees

An angle that measures less than 90 but more than 0 is called an acute angle.A right angle is an angle measuring ninety degrees, formed by the intersection of two perpendicular lines.Angles greater than 90 degrees but less than 180 degrees are known as obtuse angles.An angle that is equal to 180 degrees is called a straight angle.A reflex angle is greater than 180 degrees but less than 360 degrees.Two angles are said to be adjacent if they have a common arm and a common vertex.

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Linear pair of angles: Two adjacent angles are said to form a linear pair if their sum is 1800.Vertically opposite angles: When two lines intersect four angles are formed. The angles that are opposite to each other are called vertically opposite angles. Two angles are said to be complementary, if their sum is 90 degrees.Two angles are supplementary if their sum is 180 degrees.They may or may not be adjacent angles.Intersecting lines can be defined as two or more lines that meet at one point.Parallel lines can be defined as lines on the same plane that never intersect.

Theorem: when two lines intersect each other, then the vertically opposite angles are equal.

Parallel Lines

Two lines having only one point in common are called intersecting lines

Two lines in the same plane are parallel if they are equidistant.

A line that intersects two or more lines, at a different points is called a Transversal

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When a transversal intersects two lines, eight angles are formed, four angles at each point, P and Q respectively.We identify these angles by their positions.

1, 2, 7 and 8 are called ∠ ∠ ∠ ∠ exterior angles 3, 4, 5 and 6 are called ∠ ∠ ∠ ∠ Interior angles 1 and 5 , 2∠ ∠ ∠ and 6, 4 and 8 , 3 and 7 are Pair of ∠ ∠ ∠ ∠ ∠ corresponding angles 1 and 7 ,∠ ∠ 2∠ and 8 are pair of ∠ alternate exterior angles 4 and 6 , 3 and∠ ∠ ∠ 5 are Pair of ∠ alternate interior angles 4 and∠ 5∠ , 3 and∠ 6 are ∠ Consecutive Interior angles on the same side of the transversal

When a transversal intersects two linesPair of corresponding angles is equalPair of alternate exterior angles is equal.Pair of alternate interior angles is equal.Consecutive interior angles on the same side of the transversal are supplementary.

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Theorem: If a transversal intersects two parallel lines, then each pair of corresponding angles are equal

Theorem: If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other.

Theorem: If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Theorem: If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.

Theorem: If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Theorem: If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.Lines which are parallel to the same line are parallel to each other.

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Theorem: The sum of three angles of a triangle is 1800.Theorem: In a triangle if a side is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Page 15: CBSE maths XI STD

Congruence of Triangles

In general we come across many figures having same shape and same size. Such objects are known as as congruent figures. Two triangles are said to be congruent if all the sides and the angles of one triangle are respectively equal to corresponding sides and angles of other triangle.Consider triangles ABC and DEF, If AB = DE, BC = EF, AC = DF and

then DABC @ DDEF. Two triangles can be said to be congruent by using any of these four rules. They are Side Angle Side(SAS) Rule, Angle Side Angle(ASA) Rule, Side Side Side(SSS) Rule, Right angle Hypotenuse Side(RHS) Rule.

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SAS Congruence Rule: If two sides and included angle of one triangle are equal to the corresponding two sides and included angle of other triangle then the two triangles are congruent.

ASA Congruence Rule: If two angles and included side of one triangle are equal to the two angles and included side of other triangle then the two triangles are congruent.

SSS Congruence Rule: Three sides of one triangle are equal to three sides of other triangle then the two triangles are congruent.

RHS congruence rule: Two right triangles are said to be congruent if the hypotenuse and one side of one triangle are equal to the hypotenuse and corresponding side of the other triangle.

Page 17: CBSE maths XI STD

Properties of Triangles

A triangle is said to be an isosceles triangle if two of its sides are equal. The third side is called base.The median from the vertex to the base is perpendicular to the base.The median from the vertex to the base divides the triangle into two congruent right angled triangles

Two objects are said to be congruent if they have same shape and same size.Two triangles are said to be congruent , if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle

We use the symbol for the congruency of two triangles. We write for

congruency of .Further, we follow that the vertices , ,the angles

and the sides AB=XY,BC=YZ,CA=ZXCongruency of triangles can be proved using the four criteria of congruence:

• SAS Congruence rule, • ASA or AAS or SAA congruence rule, • SSS congruence rule and • RHS congruence rule.

,

AB=XY, BC=YZ, CA=ZX

Page 18: CBSE maths XI STD

Theorem: Angles opposite to equal sides of an isosceles triangle are equal.

In the , If the sides AB=AC then Theorem: The sides opposite to equal angles of a triangle are equal.In the , If the angles then AB=AC Theorem: The three angles of an equilateral triangle are equal.

Inequalities in a Triangle

We know that a triangle contains three sides and three angles. we also know that the if two sides of a triangle are equal then the angles opposite to them are also equal and vice versa. What happens when two sides of a triangle are unequal? In a triangle the angles and the lengths of the sides are proportional.

Theorem: If two sides of a triangle are unequal, the angle opposite to the longer side is largerTheorem: In any triangle, the side opposite to the larger angle is longer.Theorem: The sum of any two sides of a triangle is greater than the third side.

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Quadrilateral and Its Types

We know that a quadrilateral has four sides, four angles, four vertices and two diagonals. Quadrilaterals can be classified into different types based on their sides and angles.In case of rectangle and square, all the angles are right angles, the opposite sides are parallel, and the diagonals bisect each other. In a rectangle, the opposite sides are equal, whereas in a square, all the sides are equal. Hence, a square is a rectangle with adjacent sides equal. A parallelogram is a quadrilateral in which the opposite sides are parallel and equal in length. In a parallelogram, the opposite angles are equal and the diagonals bisect each other. In a parallelogram, the angles are not right angles. When we compare a parallelogram with a rectangle, we see that it is different from the rectangle in terms of the measure of its angles.A rhombus is a quadrilateral in which all the sides are equal in length, the opposite sides are parallel, the opposite angles are equal and the diagonals bisect each other at right angles. A square is a rhombus in which, each angle measures 900. Squares, rectangles and rhombuses are all examples of parallelograms.

A trapezium is a quadrilateral in which one pair of opposite sides is parallel. A trapezium with the non-parallel sides equal and the base angles equal is known as an isosceles trapezium.

A kite is a quadrilateral in which two pairs of adjacent sides are equal in length and one pair of opposite angles, “the ones that are between the sides of unequal length,” are equal in measure and the diagonals intersect at right angles.

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Properties of a Parallelogram

A quadrilateral is a closed figure which has four sides, four angles and four vertices. There are different types of quadrilaterals such as Parallelogram, Rectangle, Square, Rhombus, Trapezium and Kite. A parallelogram is a quadrilateral in which

• opposite sides are parallel and equal, • opposite angles are equal, • the diagonals bisect each other, • each diagonal divides it into two congruent triangles, • adjacent angles are supplementary.

A square is a parallelogram in which

• all the sides are equal, • each angle measures 900, • diagonals are equal and bisect at right angles.

A rectangle is a parallelogram in which

• diagonals are equal and bisect each other, • each angle measures 900.

A rhombus is a parallelogram in which

• all four sides are equal, • diagonals bisect each other at right angles.

Theorem: A diagonal of a parallelogram divides it into two congruent triangles.

Theorem: If each pair of opposite sides is equal in a quadrilateral, then it is a parallelogram.

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Theorem: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Mid-Point Theorem

The midpoint of the line segment is the geometric centre of the line segment. Midpoint of line segment divides the line into two equal halves.The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.The line drawn through the mid-point of one side of a triangle and parallel to another side bisects the third side.

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Areas of Parallelograms

We know that Area of a parallelogram is equal to the product of its base and height.The diagonals of a parallelogram bisect each other.A diagonal of the parallelogram divides it into two congruent triangles.Opposite sides and angles of a parallelogram are equal.Parallelograms on the same base and between the same parallels are equal in area.Parallelograms on the same base that have equal areas are between the same parallels.All figures that are on the same base and between same parallels need not have equal areas.Two congruent figures have equal areas but the converse need not be true.Parallelograms on the same base or equal bases and lie between the same parallels are equal in area.Parallelograms on the same base or equal bases that have equal areas lie between the same parallels.If a parallelogram and a triangle are on the same base and they are between the same parallels, then area of the triangle is half the area of the parallelogram.

Areas of Triangles

Two figures are said to be on the same base and between the same parallel lines if they have a common base and the vertices opposite to this common base of each figure lie on a line parallel to the base. Out of these two parallel lines, one should be the line containing the common base and the other is the line passing through the vertices of both the figures opposite to the base. We know that any two congruent figures have same area but the converse is not true. Median of a triangle divides it into two triangles of equal area. Two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.

Lesson Demo

Page 23: CBSE maths XI STD

Area of triangle is half the product of the base and the corresponding altitude. Also, Triangles on the same base and between the same parallels are equal in area.

Triangles on the same base and having equal areas lie between the same parallels.

Basic Concepts of a Circle

You might have seen many objects that are circular in shape. A circle is defined as the collection of all the points on a plane that are at equal distance from a given fixed point on the plane. This fixed point is called centre of the circle and the fixed distance is called the radius.A line segment joining the centre of a circle with any point on its circumference is called the radius of the circle.A line that joins two points on the circumference of a circle is called a chord. A chord that passes through the centre of a circle is called the diameter of the circle. A diameter divides a circle into two equal parts, each is called a semicircle. Diameter is the longest chord of a circle. The diameter of a circle is twice the radius.The part of the circumference of a circle between two given points is called an arc. A chord divides a circular area into two parts called segments they are major segment and minor segments. The region between two radii of a circle and any of the arcs between them is called a sector. The diameter of a circle divides it into two equal segments.

Page 24: CBSE maths XI STD

Chords of a Circle

You know that the perpendicular from a point to a line segment is the shortest distance between them. A line that joins two points on the circumference of a circle is called a chord. A chord passing through the centre of a circle is called the diameter. The longest chord of a circle is the diameter.

Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.

Given: OB AC.⊥

To prove: AB = BC.

Construction: Join OA and OC.

Proof: In triangles OBA and OBC,

OBA = OBC = 90∠ ∠ o (Since OB AC.)⊥

OA = OC (Radii of same circle)

OB = OB (Common side)

ΔOBA ΔOBC (By RHS congruence rule)

AB = BC

Thus, OB bisects chord AC.

Hence, the theorem is proved.

Theorem: The line drawn from the centre of a circle to bisect a chord is perpendicular to the

Lesson Demo

Page 25: CBSE maths XI STD

chord.

Given: AB = BC

To prove: OB AC.⊥

Construction: Join OA and OC.

Proof: In triangles OBA and OBC,

AB = BC (Given)

OA = OC (Radii of same circle)

OB = OB (Common side)

Δ OBA Δ OBC (SSS congruence rule)

OBA = OBC∠ ∠

OBA + OBC = ABC = 180∠ ∠ ∠ o

Since OBA = OBC,∠ ∠

2 x OBC = 180∠ o

OBC = ∠ = 90o

OBC = 90∠ o = OBA∠

OB AC∴ ⊥

Arcs of a Circle

Every day, you come across many things circular in shape. The collection of all that points in a plane that are at a fixed distance from a fixed point in the plane is called a circle. The fixed point is called the centre of the circle, and the fixed distance is called the radius of the circle.

Lesson Demo

Page 26: CBSE maths XI STD

A part of a circle is called an arc. Arcs of a circle that superimpose each other completely are called congruent arcs. A segment with its endpoints on a circle is called a chord. A diameter is the longest chord. If two arcs of a circle are congruent, then their corresponding chords are equal. Conversely, if two chords of a circle are equal, then their corresponding arcs are congruent.

Corresponding Arcs Of Two Equal Chords Of A Circle Are Congruent

Theorem: Congruent arcs of a circle subtend equal angles at the centre.

Given: Two congruent arcs AB and CD.

To prove: AOB = COD∠ ∠

Construction: Draw chords AB and CD.

Proof: The angle subtended by an arc at the centre is equal to the angle subtended by its corresponding chord at the centre.

In the given figure,

AB = CD (Chords corresponding to congruent arcs of a circle are equal)

AOB = COD (Equal chords subtend equal angles at the centre)∠ ∠

Hence, the theorem is proved.

Cyclic Quadrilaterals

You can draw a circle passing through three non-collinear distinct points. The points that lie on a circle are called concyclic points.

Lesson Demo

Page 27: CBSE maths XI STD

So we can say three non-collinear points are always concyclic.

Theorem: If a line segment joining two points subtends equal angles at two other points on the same side of the line segment then all the four points are concyclic.

Given: Line segment AB.

Mark two points C and D such that ACB = ADB.∠

To prove: A, B, C and D are concyclic points.

Draw a circle through points A, B and C.

Assume that the circle drawn through points A, B and C does not pass through D, and intersects AD at D’.

Proof: If A, B, C and D’ are concyclic:

ACB = AD’B (Angles subtended by a ∠ ∠ chord in the same segment of a circle)

ACB = ADB (Given)∠ ∠

AD’B = ADB∴∠ ∠

Or D’ coincides with D.

Thus, A, B, C and D are concyclic points.

Hence, the theorem is proved.

A quadrilateral whose vertices lie on a circle is called a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of the opposite angles is always equal to 180o.

If the sum of the opposite angles of a quadrilateral is 180o, then the quadrilateral is cyclic.

Basic Constructions

In earlier classes you might have learnt to construct perpendicular bisector of a line segment, angles of 300,450 600, 900, 1200 and also to draw bisector of the given angle. An angle bisector is a ray, which divides an angle in to two equal parts. A line bisector is a line that cuts a line segment into two equal halves. A perpendicular bisector is a line, which divides a given line

Page 28: CBSE maths XI STD

segment into two equal halves and is also perpendicular to the line segment.

To construct the bisector of a given angle.Let’s consider angle DEF, we want to construct the bisector of angle DEF.Steps of construction:

1. With E as centre and small radius draw arcs on the rays DE and EF.

2. Let the arcs intersect the rays DE and EF at G and H respectively.

3. With centres G and H draw two more arcs with the same radius such that they intersect at a point . Let the intersecting point be I

4. Now draw a ray with E as the starting point passing through I

5. EI is the bisector of the angle DEF.

To construct a perpendicular bisector of a line segment Lets consider the line segment as PQ. We have to construct the perpendicular bisector of PQ.Steps of Construction:

1. Draw a line segment PQ.

2. With P as centre draw two arcs on either sides of PQ with radius more the half the length of the given line segment.

3. Similarly draw two more arcs with same radius from point Q such that they intersect the previous arcs at R and S respectively.

4. Join the Points R and S. RS is the required perpendicular bisector of the given line segment PQ.

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To Construct an angle of 600 at the initial point of a given ray. Let us take ray PQ with P as the initial point. We have to construct a ray PR such that it makes angle of 600 with PQ.Steps of Constructions:

1. Draw a ray PQ

2. With P as centre draw an arc with small radius such that it intersects ray PQ at C.

3. With C as centre and same radius draw another arc to intersect the previous arc at D.

4. Draw a ray PR from point P through D. Hence the angle RPQ is equal to 60 degrees.

Constructions of Triangles

You have already learnt how to construct a triangle when three measurements are given. Even we can construct a triangle when the base, one base angle and the sum of the other two sides are given or given its base, a base angle and the difference of the other two sides or given, its perimeter and two base angles.

Page 30: CBSE maths XI STD

In QR = 'a'cm, and PQ + PR = 'b' cm.Step 1: Draw the base QR = 'a' cm.Step 2: Draw XQR =

Step 3: Mark an arc S on QX such that QS = 'b' cm.Step 4: Join RS.Step 5: Draw the perpendicular bisector of RS such that it intersects QS at P.Step 6: Join PR.

Thus, is the required triangle.

In , given BC = 'a' cm, and difference of two sides AB and AC is equal to 'b' cm.Case I: AB > ACStep 1: Draw the base BC = 'a' cm.Step 2: Make .Step 3: Mark a point D on ray BX such that BD = 'b' cm.Step 4: Join DC.Step 5: Draw perpendicular bisector of DC such that, it intersects ray BX at a point A.Step 6: Join AC.Thus, ABC is the required triangle.

Case II: AB < AC

Step 1: Draw the base BC = 'a' cm.Step 2: Make and extend ray BX in the opposite direction.Step 3: Mark a point D on the extended ray BX such that BD = 'b'cm.Step 4: Join DC.Step 5: Draw a perpendicular bisector of DC such that, it intersects ray BX at point A.Step 6: Join AC.Thus, ABC is the required triangle.

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To construct , given Perimeter ( AB + BC + CA) = 'a' cm, .

Steps of construction:Step 1: Draw XY = 'a' cm.

Step 2: Draw the ray XL at X making an angle of with XY.

Step 3: Draw the ray YM at Y making an angle of with XY.Step 4: Draw angle bisector of Step 5: Draw angle bisector of such that it meets the angle bisector of at a point A.Step 6: Draw the perpendicular bisector of AX such that it meets XY at a point B.Step 7: Draw the perpendicular bisector of AY such that it meets XY at a point C.Step 8: Join AB and AC.Thus, ABC is the required triangle.

Cube and Cuboid

A cuboid is a solid bounded by six faces that are rectangular in shape. A cuboid whose length, breadth and height are equal is called a cube. Any two faces other than the opposite faces are called adjacent faces. Any two adjacent faces meet in a line segment, which is called an edge of the cuboid. The point of concurrency of any three edges of a cuboid is called a vertex of the cuboid. A cuboid has 8 vertices and 12 edges. Any face of a cuboid may be called the base of the cuboid. In that case, the four adjacent faces of the base are called the lateral faces of the cuboid.The lateral surface area of the cuboid = 2 (l + b) h. It is equal to sum of areas of all its lateral faces.Lateral surface area of the cube=4a2Total surface area of the cuboid=2(lb+bh+lh)Total surface area of the cube=6a2Volume of a solid object is the measure of the space occupied by it The volume of substance that can be stored by the object is called its capacity.

Volume of Cuboid = length breadth height=lbh

Volume of a Cube =

L.S.A of a cuboid =2 (l + b) hT.S.A of a cuboid =2(lb+bh+lh)Volume of the cuboid =lbh

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T.S.A of a cube =6a2< Total surface area of a cube, sum areas of all the faces of a cube >

volume of the cube=

Cylinder

A cylinder can be defined as a solid figure that is bound by a curved surface and two flat surfaces. The flat surfaces are made up of two congruent circles that are parallel to each other. These flat surfaces are called the bases of the cylinder. The radius of the circular bases is the radius of the cylinder. The perpendicular line that passes through the centers of the two circular bases is the height of the cylinder or axis of the cylinder. A cylinder is said to be right circular cylinder if axis is perpendicular to the radius of the cylinder.

The curved surface joining the two basses of a right circular cylinder is called its lateral surface area.For a right circular cylinder of radius r and height hLateral surface area of the Cylinder Base surface area of the Cylinder Total surface area of the Cylinder = Volume of the Cylinder =

L.S.A of Cylinder=Base surface area of the Cylinder =T.S.A of the Cylinder=Volume of the Cylinder=

Cone

Cone is solid figure with a circular base that tapers to a point or vertex.A Cone is said to be a right circular cone if its height is perpendicular to the radius of the base.Let "r" be the base radius, "h" be the height, "l" be the slant height of a right circular cone. Then l=

Base area of the cone

Curved surface area of the cone=

Lesson Demo

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Total surface area of the cone=

Volume of Cone =

Base area of a cone=

C.S.A of a cone=

T.S.A of a cone =

Volume of a cone=

Sphere and Hemisphere

A sphere is a three dimensional figure, made up of points that are equidistant from a given point.It does not have an edge or a vertex.The surface of a sphere is uniform and smooth.The centre of a sphere is a point, which is equidistant from all the points on a sphere.The distance between the centre and any point on the surface of the sphere is called the radius of a sphere. Generally the radius is denoted by the letter r.A line segment through the centre of the sphere, and with the end points on the sphere is called a diameter of the sphere.

A plane through centre of a sphere divides the sphere in two equal parts, each of which is called a hemisphere.A hemisphere has two faces. A flat surface which is called the base and a curved surface.

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Collection and Presentation of Data

A systematic record of facts or different values of a quantity is called data. Data is of two types - Primary data and Secondary data. The data collected by a researcher with a specific purpose in mind is called primary data. The data gathered from a source where it already exists is called secondary data.The difference between the highest and lowest values in the given data is called the range of the given data. The number of times a value occurs in the given data is called the frequency of that value.

A table that shows the frequency of different values in the given data is called a frequency distribution table. A frequency distribution table that shows the frequency of each individual value in the given data is called an ungrouped frequency distribution table. A table that shows the frequency of groups of values in the given data is called a grouped frequency distribution table.The groupings used to group the values in given data are called classes or class-intervals. The number of values that each class contains is called the class size or class width. The lower value in a class is called the lower class limit. The higher value in a class is called the upper class limit.

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Graphical Representation of Data

Graphical representation of data helps in faster and easier interpretation of data.A bar graph uses bars or rectangles of the same width but different heights to represent different values of data.

In a bar graph:

1. The bars have equal gaps between them.

2. The width of the bars does not matter.

3. The height of the bars represents the different values of the variable.

In a histogram:

1. The bars do not have gaps between them.

2. The width of the bars is proportional to the class intervals of data.

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3. The height of the bars represents the different values of the variable.

4. The area of each rectangle is proportional to its corresponding frequency.

The area of a histogram is equal to the area enclosed by its corresponding frequency polygon.

Measures of Central Tendency

The mean of a given set of values is equal to the sum of all the values divided by the total number of values. The median is that value of the given number of observations, which divides it into exactly two parts.The value that lies in the very centre of a given set of values arranged in ascending or descending order, is called the median of the given data. If the number of given values is odd, Median = [ (n+1)/2]th value, where n = number of given values.If the number of given values is even, Median = Mean of (n/2)th and (n/2 + 1)th values, where n = number of given values.The value that occurs the most number of times in a given set of values is called the mode of the given data or an observation with maximum frequency is known as mode.Mean, Median and Mode together are called the measures of central tendencies of data.The central tendencies of data depend on distribution of values and must be considered with other information for effective interpretation of data.

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Probability - Experimental Approach

Probability is defined as the likelihood or chance of something occurring. It is widely used in the study of mathematics, statistics, gambling, physical science, Biological science, weather forecasting, finance etc. to draw conclusions.Probability is defined as the numerical method of measuring uncertainty involved in a situation. An experiment is defined as an action or process that results in well defined outcomes.A trial is an action which results in one or several outcome, for example each toss of the coin and each throw of the die are called a trial.An event for an experiment can be defined as the collection of some outcomes of the experiment.Experimental or empirical probability is an estimate that an event will happen based on how often the event occurs after performing an experiment in a large number of trials. The experimental or empirical probability P(E) of an event E happening is given by: P(E) = Number of trials in which the event happened / The total number of trials

Lesson Demo

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Term 1 - Summative Assessment

Mathematics

Question Paper Set - 1

Max. Marks: 80

1. All questions are compulsory. 2. The questions paper consists of 34 questions divided into four sections A, B, C and D.

Section A comprises of 10 questions of 1 mark each, Section B comprises of 8 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions of 4 marks each.

3. Question numbers 1 to 10 in section A are multiple choice questions where you have to select one correct option out of the given four.

Section A

Time: 3 to 3 ½ hours

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Section B

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Section C

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Section D

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Answers

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Lesson Summary

erm 1 - Summative Assessment

Mathematics

Question Paper Set - 2

Max. Marks: 80

1. All questions are compulsory. 2. The questions paper consists of 34 questions divided into four sections A, B, C and D.

Section A comprises of 10 questions of 1 mark each, Section B comprises of 8 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions of 4 marks each.

3. Question numbers 1 to 10 in section A are multiple choice questions where you have to select one correct option out of the given four.

Section A

Time: 3 to 3 ½ hours

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Section B

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Answers

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Lesson Summary

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Term 2 - Summative Assessment

Mathematics

Question Paper Set - 1

Max. Marks: 80

1. All questions are compulsory. 2. The question paper consists of 34 questions divided into four sections – A, B, C and D.

Section A consists of 10 questions of 1 mark each, Section B of 8 questions of 2 marks each, Section C of 10 questions of 3 marks each, and Section D consists of 6 questions of 4 marks each.

3. Question numbers 1 to 10 in section A are multiple-choice questions, wherein you have to select the correct option from among those given.

Section A

1. The coordinates of the point of intersection of the lines x = −8 and y = 5 are:

A. (8, 5)

B. (-8, 5)

C. (8, -5)

D. (-8, -5)

2. x = 2 and y = 5 is a solution of:

A. x + 2y = 12

B. x + y = 12

C. 2x + y = 10

D. x - 2y = 10

3. If the angles of a quadrilateral are in the ratio of 2 : 4 : 5 : 7, then its angles measure:

A.

B.

C.

D.

4. In the figure given here, PQRS is a parallelogram with PR = 7.8 cm and QS = 6 cm. If PQ and QS intersect at O, then OP= and OQ=:

A. 3.7 cm, 3 cm

B. 4.9 cm, 4 cm

C. 3.9 cm, 3 cm

D. 3 cm, 3 cm

5. The region occupied by a simple closed figure in a plane is called:

A. Length

B. Volume

Time: 3 to 3 ½ hours

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C. Perimeter

D. Area

6. One-fourth of a circular disk is called a:

A. Semi-circle

B. Quadrant

C. Sector

D. Arc

7. In the figure given here, if O is the centre of the circle, then x measures:

A.

B.

C.

D.

8. In the figure given here, ABCD is a trapezium in which AB || DC. Then ar (∆AOD)= ____.

A.

B.

C.

D.

9. The measure of the edge of a cube is 5.5 cm. Its surface area is:

A. 172 m2

B. 181.5 m2

C. 160 m2

D. 150.2 m2

10.The probability of getting a number less than 5 in a single throw of a die is:

A.

B.

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C.

D.

SECTION B

11.Write four solutions of the equation 3x + y = 4. 12.In ABC, D and E are the mid-points of AC and BC, respectively. If DE = 11.5 cm, then △

find the length of AB.

13.ABCD is a parallelogram. P is a point on AD such that If Q is a point on BC such

that , then show that AQCP is a parallelogram. 14.In ∆ABC, AD is a median. Prove that ar(ΔABD) = ar(ΔACD).

15.In the figure given here, Find

16.The sides of a triangle are 9 cm, 12 cm and 15 cm in length. Find its area. 17.A hemispherical bowl made of stone is 5 cm thick. If its inner radius is 35 cm, then find its

total surface area. 18.18. A coin is tossed 500 times with the following frequencies:

Head: 245, Tail: 255

Compute the probability of each event.

SECTION C

19.Draw the graph of the equation 2x + 3y = 13, and determine whether x = 4, y = 2 is a solution or not.

20.A shopkeeper makes a profit of 20 per cent on selling a children’s umbrella, and a loss of 10 per cent on selling a normal umbrella, but gains Rs. 10 on selling one of each. If the cost of a children’s umbrella is Rs. x, and that of a normal umbrella is Rs. y, then frame a linear equation that represents the information given.

21.Two opposite angles of a parallelogram measure (60 – x)° and (3x – 4)°. Find the measure of each angle of the parallelogram.

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22.PQRS and ABRS are two parallelograms, and

X is any point on side BR. Show that:

(i) ar(PQRS) = ar(ABRS)

(ii) ar(AXS) =

23.Construct a triangle ABC in which BC = 7 cm, B = 75° and AB + AC = 13 cm. ∠24.A cylinder is 12 cm high, and the circumference of its base is 44 cm. Find its curved surface

area and total surface area. 25.A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm.

Find the area of the sheet required to make ten such caps. 26.This table gives the life of 400 neon lamps:

Life (in hours)Number of lamps

300 – 400 14

400 – 500 56

500 – 600 60

600 – 700 86

700 – 800 74

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800 – 900 62

900 – 1000 48

(i) Represent the given information with the help of a histogram.

(ii) How many lamps have a life of more than 700 hours?

27.In a cricket match, a batsman hits a boundary 6 times off the 30 balls that he plays. Find the probability that he does not hit a boundary off a delivery.

28.Three coins are tossed simultaneously 200 times, with the following frequencies of different outcomes:

Outcome 3 heads2 heads1 headNo head

Frequency23 72 77 28

If the three coins are tossed simultaneously again, compute the probability of two heads coming up.

SECTION D

x–5–13b13

y–2a 257

From the graph, find the values of a and b.

29.Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

30.In the figure given here, CD AB and CBA = 25°. Find x. ∥ ∠31.If each diagonal of a quadrilateral separates

it into two triangles of equal areas, then show that the quadrilateral is a parallelogram.

32.The difference between the semi-perimeter and the sides of triangle ABC are 8 cm, 7 cm and 5 cm, respectively. Find its semi-perimeter.

33.The graph given here shows the histogram and frequency polygon of the daily expenses of seven groups of tourists.

Read the graph and answer the following questions:

(i) What is the least daily expense?

(ii) What is the highest daily expense?

(iii) How many tourists spent an average of Rs. 150 every day?

(iv) Determine the average expenditure of the group that had the maximum daily expenses.

(v) How many tourists belong to Group D?

(vi) Determine the lowest and highest expenditure of the tourists from Group F.

(vii) Which group was the largest? How many tourists were there in this group?

(viii) Which group was the smallest? How many tourists were there in this group?

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Answers

SECTION A

1. Answer: (B)

The lines x = a and y = b always intersect at the point (a, b).

Hence, the point of intersection of the given lines is (-8, 5).

2. Answer: (A)

Substitute x = 2 and y = 5 in the

equation x + 2y = 12.

2 + 2(5) = 2 + 10

= 12 RHS

LHS = RHS∴

Therefore, x = 2 and y = 5 is a solution of x + 2y = 12.

3. Answer: (B)

Let the angles of the quadrilateral measure

Now, 2x + 4x + 5x + 7x = 3600

Therefore, the angles measure .

4. Answer: (C)

The diagonals of a parallelogram bisect each other.

Also,

5. Answer: (D)

The region occupied by a simple closed figure in a plane is called its area.

6. Answer: (B)

One-fourth of a circular disk is called a quadrant.

7. Answer: (A)

From the figure,

In ∆OPQ, OP = OQ = Radius

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( Angles opposite to equal sides in a triangle are equal)∵

We know that the sum of the angles in a triangle is 1800.

x + x + 500 = 1800

8. Answer: (A)

∆ABC and ∆ABD are on the same base AB and between the same parallels AB and DC.

ar(∆ABC) = ar(∆ABD)∴

9. Answer: (B)

Edge = 5.5 cm

Surface area =

=

= 6 × 30.25

= 181.5 m2

10.Answer: (C)

We know that there are only six possible outcomes in a single throw of a die.

It may show any number from 1 to 6. However, we have to get a number less than 5.

The numbers less than 5 on a die are 1, 2, 3 and 4.

SECTION B

11.Given: 3x + y − 4 = 0

y = 4 − 3x⇒

Putting x = 0, we get y = 4 − 3(0) = 4− 0 = 4

Putting x = 1, we get y = 4 −3(1) = 4− 3 = 1

Putting x = 2, we get y = 4−3(2) = 4− 6 = −2

Putting x = 3, we get y = 4 −3(3) = 4−9 = −5

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Hence, the four solutions of the given equation are (0, 4), (1, 1), (2, −2) and (3, −5).

12.Given, DE = 11.5cm

By the mid-point theorem,

DE is parallel to AB and half of it. (∵ The line joining the mid-points of two sides of a triangle is parallel to the third side, and half of it in length)

AB=2×DE⇒

= 2 × 11.5

AB = 23 cm⇒

Hence, the length of AB is 23 cm.

13.Given:

AP= AD ----- (i) CQ= BC ----- (ii)

AD = BC ( Opposite sides of a parallelogram are equal)∵

Multiplying by on both sides, we get

AD= BC

AP = CQ ----- (iii) (from (i) and (ii))⇒

AD BC∥

AP CQ ----- (iv)⇒ ∥

From (iii) and (iv):

AQCP is a quadrilateral, whose one pair of opposite sides AP and CQ are equal and parallel.

Hence, AQCP forms a parallelogram.

14.Given: AD is a median of ΔABC.

To prove: ar(ΔABD) = ar(ΔACD)

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Proof:

ar(ΔABD) - - - - (1)

ar(ΔACD) - - - - - (2)

We have BD = DC - - - - -(3) [ AD is the median on side BC]∵

From equations (1), (2) and (3) we get

ar(ΔABD) = ar(ΔACD)

Hence proved.

15.In ∆PQR,

But

We know that (Angles in the same segment)

= 750

16.If a, b and c are the sides of a triangle, and s is its semi-perimeter, then its area is given by:

Here,

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= 9 × 3 × 2

= 54 sq. cm

Hence, the area of the triangle is 54 sq. cm.

17.17. Let R be outer radius and r be the inner radius of the bowl.

Thickness of ring = 5 cm

Total surface area = CSA of outer hemisphere + CSA of inner hemisphere + Area of ring

= 18935.71 sq. cm

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18.Since the coin is tossed 500 times, the total number of trials is 500.

Let E be the event of getting a head and F be the event of getting a tail.

Then the number of trials in which event E happens is 245.

And the number of trials in which event F happens is 255.

P (Getting a head)

P (Getting a tail)

SECTION C

19.The given equation is 2x + 3y = 13.

If x = 2,

If x = 5,

Thus, we have the following table for 2x + 3y = 13:

x25

y31

Plot the points A (2, 3) and

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B (5, 1) on a graph paper.

Draw a line passing through points A and B.

Line AB represents the graph for 2x + 3y = 13.

We can see that the point (4, 2) does not lie on the graph of the equation 2x + 3y = 13.

Hence, x = 4, y = 2 is not a solution of the equation 2x + 3y = 13.

20.Given that the cost of a children’s umbrella is Rs x and the cost of a normal umbrella is Rs y.

Profit on selling a children’s umbrella

Loss on selling a normal umbrella

Also given that the shopkeeper gains Rs. 10 on every time he sells one children’s umbrella and one normal umbrella.

Net profit, 10 = ∴

Hence, the required linear equation is

21.21. Let ABCD be a parallelogram, where A = (3x – 4)° and C = (60 – x)°.∠ ∠

Then A = C [Opposite angles of a ∠ ∠ gm]

3x – 4 = 60 – x⇒

4x = 64⇒

x = 16⇒

A = (3 × 16 – 4)°∠

= (48 – 4)°

A = 44°∠

C = (60 – 16)° = 44°∠

Now, A + B = 180° [Co-interior angles, AD ∠ ∠ BC]

44° + B = 180°⇒ ∠

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B = 180° – 44° = 136°⇒ ∠

Also, D = B = 136° [Opposite angles of a ∠ ∠ gm]

Thus, the angles of the parallelogram are 44°, 136°, 44°, 136°.

22.(i) Parallelograms PQRS and ABRS are on the same base RS and between the same parallels RS and BP.

ar(∴ gm PQRS) = ar( gm ABRS).

Hence, ar(PQRS) = ar(ABRS).

(ii) ΔAXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR.

[ ar ( gm PQRS) = ar ( gm ABRS), proved in (i)]∥

Hence, ar(AXS) = .

23.Steps of construction:

1. Draw a line segment BC of length 7 cm.

2. At B, construct XBC = 75°.∠

3. With B as the centre and radius equal to 13 cm, draw an arc cutting XB at D.

4. Joint DC.

5. Draw PQ, the perpendicular bisector of DC, and let it intersect DB at A.

6. Join AC.

Then, ΔABC is the required triangle.

Justification:

Since A lies on the perpendicular bisector of DC, it is equidistant from points D and C.

AD = AC … (1)∴

Now, BD = 13 cm [By construction]

BA + AD = 13 cm⇒

BA + AC = 13 cm [Using (1)]⇒

Hence, ΔABC is the required triangle.

24.Let r be the base radius and h be the height of the cylinder.

Then h = 12 cm.

Given: Circumference of base = 44 cm

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2 π r = 44 cm⇒

r = 7 cm … (1)⟹

Curved surface area of a cylinder = 2Πrh

= 44 × 12

= 528 cm2

Total surface area a cylinder = 2Πrh + 2Πr2

= 2Πr(r + h)

= 44(7 + 12) [ r = 7 cm, from (1)]

= 44(19)

= 836 cm2

Hence, the curved surface area of the cylinder is 528 cm2, and its total surface area is 836 cm2.

25.Let r be the base radius, h the height, and l the slant height of the cone.

r = 7 cm and h = 24 cm∴

Slant height (l) =

= 25 cm … (1)

Curved surface area of one cap = Πrl

cm2 [Using (1)]

Curved surface area of 10 caps ∴

= 5500 cm2

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Hence, the area of the sheet required for 10 caps is 5500 cm2.

26.(i) Taking the life (in hours) along the X-axis, and the number of lamps along the Y-axis on suitable scales, we draw rectangles with class intervals as the bases and the corresponding frequencies as the heights to obtain the required histogram.

Note that the scale on the X-axis starts at 300, so a kink ( ), i.e. break is indicated near the origin to signify that the graph is drawn with a scale beginning at 300 and not at the origin.

(ii) 74 + 62 + 48 = 184 lamps have a life of more than 700 hours.

27.The batsman played 30 balls, so the total number of trials is 30.

He hits a boundary 6 times, so he did not hit a boundary (30 – 6) = 24 times.

Let E be the event that the batsman did not hit a boundary.

Then the number of trials in which event E happened = 24

P(Batsman did not hit a boundary) = P(E)

28.28. The three coins are tossed simultaneously 200 times.

Therefore, the total number of trials is 200.

Let E be the event of getting two heads

Then the number of trials in which event E happens = 72

P(Getting two heads) = P(E)

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SECTION D

29.Plot the points A (–5, –2), B (3, 2) and C (13, 7).

Draw a straight line passing through A, B and C.

The y coordinate corresponding to the x coordinate – 1 is 0.

a = 0∴

Through y = 5, draw a horizontal line that meets the graph at a point, say P. Through P, draw a vertical line that meets the X-axis. It will meet the X-axis at x = 9.

b = 9∴

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30.Let ABCD be a quadrilateral such that P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. PR and SQ are joined. Join AC, PQ, QR, RS and SP.

In ΔABC, P is the mid-point of AB, and Q is the mid-point of BC.

PQ ∴ AC and [By mid-point Theorem] …. (1)

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In Δ ADC, S is the mid-point of DA and R is the mid-point of DC. [Given]

SR∴ AC and [By mid-point Theorem] … (2)

From equations (1) and (2), we get

PQ SR and PQ = SR

Thus, in quadrilateral PQRS, one pair of opposite sides is parallel and equal.

PQRS is a parallelogram.∴

Since the diagonals of a parallelogram bisect each other, PR and SQ bisect each other.

Hence, the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

31.Join AC and BE.

ACB = 90° [Angle inscribed in a semi-∴∠circle] …. (1)

In ΔACB, we have

ACB + CBA + CAB = 180° [Sum of∠ ∠ ∠ angles of a triangle]

90° + 25° + CAB = 180° [Using (1) and⇒ ∠ CBA = 25°, given]∠

CAB = 180° – 115 °⇒ ∠

CAB = 65° … (2)⇒ ∠

CEB = CAB [Angles in the same segment]∠ ∠ …. (3)

CEB = 65° [Using (2) and (3)] … (4)∴∠

CD AB (given) and CB is the transversal∥

DCB = CBA [Alternate angles] …. (5)∴∠ ∠

CBA = 25° [Given] … (6)∠

DCB = 25° [Using (5) and (6)] …. (7)∴∠

DEB = DCB [Angles in the same∠ ∠ segment] … (8)

DEB = 25° [Using (7) and (8) …. (9)∴∠

Now x = CED = CEB – DEB∠ ∠ ∠

x = 65° – 25° [Using (4) and (9)]⇒

x = 40°⇒

Hence, x = 40°.

32.Let ABCD be the given quadrilateral. Diagonal AC separates it into two equal areas such that ar(ΔACD) = ar(CAB), while diagonal BD separates it into two

Page 100: CBSE maths XI STD

equal areas such that ar(DAB) = ar(BCD).

Draw CP AB produced, and DQ AB.⊥ ⊥

ar(quad ABCD) = ar(ΔCAB) + ar(ΔACD)]

ar(quad ABCD) = 2ar (ΔCAB) [ ar(ΔACD) = ar(ΔCAB)] ….. (1)⇒ ∵

Also, ar (quad ABCD) = ar(ΔDAB) + ar(ΔBCD)

ar(quad ABCD) = 2ar (ΔDAB) [ ar(ΔDAB) = ar(ΔBCD) ….. (2)⇒ ∵

From equations (1) and (2), we get

ar(ΔCAB) = ar(ΔDAB)

CP = DQ⇒

Altitude from C of ΔABC = Altitude from D of ΔABD⇒

DC AB⇒ ∥

Similarly, AD BC.∥

Hence, ABCD is a parallelogram.

33.Let the semi-perimeter of the triangle be s.

Let the sides of the triangle be a, b and c.

Given:

s – a = 8

s – b = 7

s – c = 5

Adding all the equations, we get

(s – a) + (s – b) + (s – c) = 8 + 7 + 5

3s – (a + b + c) = 20⇒

3s – 2s = 20 ⇒

s = 20 cm⇒

34.34. (i) The least daily expense is Rs. 100.

(ii) The highest daily expense is Rs. 240.

(iii) The number of tourists who spent on an average of Rs. 150 is 50.

(iv) The average expenditure of the group that had the maximum daily expense is Rs. 230.

(v) A total of 45 tourists belong to Group D.

(vi) The lowest expenditure of tourists belonging to Group F is Rs. 200.

The highest expenditure of tourists belonging to Group F is Rs. 220.

(vii) The largest group was Group C, consisting of 50 tourists.

(viii) The smallest group was Group G, consisting of five tourists.

Page 101: CBSE maths XI STD

Lesson Summary

Mathematics

Question Paper Set - 2

Max. Marks: 80

1. All questions are compulsory. 2. The question paper consists of 34 questions divided into four sections – A, B, C and D.

Section A consists of 10 questions of 1 mark each, Section B of 8 questions of 2 marks each, Section C of 10 questions of 3 marks each, and Section D consists of 6 questions of 4 marks each.

3. Question numbers 1 to 10 in Section A are multiple-choice questions, wherein you have to select the correct option from among those given.

Section A

1. If x = 1 and y = 3 is a solution of 3x + 5y = k, then k =

A. 16

B. 15

C. 18

D. 17

2. Pencils cost Rs 2 each; pens cost Rs 7 each. John spends Rs 20 on these items. Then the linear equation for the given data is:

A. 2x + 7y = 20

B. x + 7y = 20

C. 2x + y = 20

D. 2x – 7y = 20

3. In a quadrilateral, if the opposite sides are equal and the diagonals are also equal, then the quadrilateral is a:

A. Square

B. Rectangle

C. Parallelogram

D. Rhombus

4. If three angles of a quadrilateral measure 100°, 50° and 70°, then the fourth angle measures:

A. 125°

B. 135°

C. 140°

D. 132°

5. The base of a parallelogram is 14 cm, and its distance from the opposite side is 8 cm. Then its area is:

Time: 3 to 3 ½ hours

Page 102: CBSE maths XI STD

A. 144 cm2

B. 64 cm2

C. 76 cm2

D. 112 cm2

6. The part of a plane enclosed by a simple closed figure is called:

A. Volume of the figure

B. Planar region

C. Angular region

D. Area of the figure

7. In the figure given here, the value of x is: A. 120°

B. 100°

C. 105°

D. 102°

8. In the figure given here, ABCD is a parallelogram. A circle through A, B and C meets CD produced at E. Then AE = A. ED

B. AD

C. CD

D. AC

9. The total surface area of a hemisphere of radius 14 cm is:

A. 942 cm2

B. 2464 cm2

C. 1375 cm2

D. 2104 cm2

10.If the mean of 25, 27, 19, 29, 21, 23, p, 30, 28 and 20 is 24.7, then p =

A. 25

B. 31

C. 22

D. 26

SECTION B

11.In the equation 3(x – 3) – 3(y – 1) = 10, if x = 3, then find the value of y.

Page 103: CBSE maths XI STD

12.ABCD is a parallelogram. Diagonals AC and BD intersect at O. If ADB = 40°, ∠

ABD = 25° and DAC = 30°, then find ∠ ∠DOC.∠

13.In the figure given

here, if ABCD, ABFE and CDEF are parallelograms, then prove that ar(ΔADE) = ar(ΔBCF).

14.Diagonals AC and BD of quadrilateral ABCD intersect at O such that ar (ΔBOC) = ar (ΔAOD). Then show that ABCD is a trapezium.

15.Prove that equal chords of a circle subtend equal angles at its centre. 16.Find the area of a triangle whose sides measure 20 cm, 30 cm and 40 cm. 17.A rectangular sheet of paper 44 cm × 18 cm is rolled along its length and a cylinder is

formed. Find the volume of the cylinder. 18.A dice is thrown once. Find the probability of getting

(i) a number greater than 2 (ii) a number less than 4.

SECTION C

19.Draw the graph of the equation 3x – 2y = 4. From the graph, find the coordinates of the point when:

(i) x = 2

(ii) y = 7

20.A railway half ticket costs half the full fare. The reservation charges are the same for both half and full tickets. A family of three adults and five children pay Rs 1200 for their travel from Hyderabad to Bengaluru. If the basic fare is Rs x and reservation charge is Rs y, then find the linear equation that represents the given information.

21.In the figure given here, ABCD is a parallelogram. Compute the values of x and y.

22.In the figure given here, ABCD is a parallelogram. AE DC and CF AD. If CD = 12 cm, AE = 8 ⊥ ⊥cm and CF = 10 cm, find AD.

23.Construct a triangle ABC in which BC = 8 cm, B = 45° and AB – AC = 3.5 cm. ∠

24.Find the area of quadrilateral ABCD in which AB = 7 cm, BC = 6 cm, CD = 12 cm, DA = 15 cm and AC = 9 cm.

25.The thickness of a hollow wooden cylinder is 2 cm. It is 35 cm long and its inner

Page 104: CBSE maths XI STD

radius is 12 cm. Find the volume of the wood required to make the cylinder, assuming it is open at either end.

26.Radhika scored 75, 82 and 90 in three Mathematics tests. How many marks must she obtain in the next test to get an average of exactly 85 for the four tests?

27.The weather forecast from a news channel shows that out of the past 300 consecutive days, its weather forecast was correct 210 times.

(i) What is the probability that on a given day, it was correct?

(ii) What is the probability that on a given day, it was not correct?

28.Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes: Outcome 3 heads2 heads1 headNo head

Frequency23 72 77 28

If the three coins are tossed simultaneously again, compute the probability of two heads coming up.

SECTION D

29.Draw the graph of the equation 2x + y = 8. Read two solutions from the graph and verify the same by actual substitution. Also find the points where the line meets the two axes.

30.Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.

31.Diagonals AC and BD of cyclic quadrilateral ABCD intersect at right angles at E. Line l through E and perpendicular to AB meets CD at F. Prove that F is the mid-point of CD.

32.E, F, G and H are the mid-points of the sides of parallelogram ABCD.

Show that . 33.A cone of height 24 cm has a curved

surface area of 550 cm2. Find its

volume (take ). 34.A die is thrown 500 times with the

frequencies for the outcomes 1, 2, 3, 4, 5, 6 as given in the following table:

Outcome 1 2 3 4 5 6

Frequency8050908510095

Find the probability of getting each outcome.

Answers

1. Answer:(C)

Given: 3x + 5y = k

Substitute x = 1 and y = 3 in this equation, we get

3 (1) + 5 (3) = k

2. Answer:(A)

Page 105: CBSE maths XI STD

Let the number of pencils and pens be x and y, respectively.

Cost of x pencils = 2 × x = 2x

Cost of y pens = 7 × y = 7y

Total cost = 20

Hence, the required linear equation is 2x + 7y = 20.

3. Answer: (B)

In a quadrilateral, if the opposite sides are equal and the diagonals are also equal, then the quadrilateral is a rectangle.

4. Answer:(C)

Let the measure of the fourth angle be x°.

The sum of the angles of a quadrilateral is 360°.

100° + 50° + 70° + x = 360°⇒

220° + x = 360°⇒

x = 140°∴

Hence, the measure of the fourth angle is 140°.

5. Answer:(D)

Let b = 18 cm and h = 8 cm

We know that:

Area of a parallelogram = b × h

= 14 × 8

= 112 cm2

Hence, the area of the parallelogram is 112 cm2.

6. Answer:(B)

The part of the plane enclosed by a simple closed figure is called a planar region.

7. Answer:(B)

Join A and B to any point, say D, on the circle.

ACBD is a cyclic quadrilateral.

= 2 × 500

8. Answer: (B)

Given ABCD is a parallelogram.

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Then AB DC.∥

i.e. AB EC and AC is a transversal.∥

BAC = ACE (Alternate interior angles)⇒∠ ∠

i.e. Angles subtended by AE and BC in corresponding alternate segments are equal.

Arc BC = Arc AE

BC = AE⇒

However, in parallelogram ABCD, BC = AD.

AE = AD∴

9. Answer:(B)

For the hemisphere, r = 14 cm

Total surface area of hemisphere =

= 4 × 22 × 2 × 14

= 2464 cm2

10.Answer: (A)

We know that:

SECTION B

11.Given: 3(x – 3) – 3(y – 1) = 10

Substituting x = 3 in the given equation, we get

3(3 – 3) – 3(y – 1) = 10

12. DOC = ODA + OAD∠ ∠ ∠

( In a triangle, if one side is extended, then∵ the exterior angle soformed is equal to the sum of two opposite interior angles)

13.ABCD is a parallelogram.

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AD = BC⇒

CDEF is a parallelogram.

DE = CF⇒

ABEF is a parallelogram.

AE = BF⇒

In ΔADE and ΔBCF:

AD = BC

DE = CF

AE = BF

By SSS criterion of congruence,

ΔADE ΔBCF≅

ar(ΔADE) = ar(ΔBCF)∴

14.Given: ar (ΔBOC) = ar (ΔAOD)

ar (ΔBOC) + ar (ΔAOB) = ar (ΔAOD) + ar⇒ (ΔAOB)

ar (ΔABC) = ar (ΔABD)⇒

ΔABC and ΔABD are on the same base AB, and between AB and AC.

We know that the areas of two triangles are equal if both are on the same base and between the same parallels.

AB DC⇒ ∥

Hence, ABCD is a trapezium.

15.Let AB and CD be two equal chords of a circle with centre O.

In ΔAOB and ΔCOD,

OA = OC ( OA and OC are radii of same circle)∵

OB = OD ( OB and OD are radii of same circle)∵

Given: AB = CD

ΔAOB ΔCOD (By SSS congruence rule)⇒ ≅

AOB = COD (Corresponding parts of congruent⇒∠ ∠ triangles)

Hence, equal chords of a circle subtend equal angles at its centre.

16.Given that the sides of a triangle are 20 cm, 30 cm and 40 cm in length.

Let a=20cm,b=30cm and c=40cm.

Semi-perimeter

s = 45cm

Page 108: CBSE maths XI STD

We know that:

=75×3.8729

Hence, the area of the triangle is

17.If the sheet of paper is rolled along the length, then the height of the cylinder is 18cm and the circumference of its base is 44cm.

Let r be the radius of the base of the cylinder.

2πr=44⇒

r=7cm⇒

Volume =

Hence, the volume of the cylinder is

18.Total outcomes are 1, 2, 3, 4, 5, 6.

Total number of outcomes = 6

(i) Favourable outcomes are 3, 4, 5, 6

Number of favourable outcomes = 4

P(number greater than 2)

(ii) Favourable outcomes are 1, 2, 3

Number of favourable outcomes = 3

P(number less than 4)

SECTION C

19.Given: 3x – 2y = 4

If x = 0, then

If x = 4, then

Page 109: CBSE maths XI STD

Thus, we have the following table for 3x – 2y = 4:

x0 4

y–24

Plot points A (0, –2) and B (4, 4) on a graph paper.

Draw a line passing through points A and B.

Then line AB represents the graph of 3x – 2y = 4.

From the graph, we see that:

(i) When x = 2, y = 1

(ii) When y = 7, x = 6

20.Given: Basic fare = Rs x

Reservation charge = Rs y

Then fare for one full ticket is Rs (x + y), and for

1 half ticket is

It is given that three adults and five children paid Rs. 1200 for their tickets. Fare of 3 full tickets ⇒+ Fare of 5 half tickets = 1200

Hence, the required linear equation is .

21.We know that the opposite angles of parallelogram are equal.

3x – 20 = x + 40⇒

3x – x = 40 + 20⇒

2x = 60⇒

x = 30⇒

Also, x + 40 + y + 15 = 1800 [Co-interior angles, AB ∥ DC]

30 + 40 + y + 15 = 180 [Using equation 1]⇒

y + 85 = 180⇒

y = 180 – 85 = 95⇒

Hence, x = 30 and y = 95.

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22.Area of a gm = Base × Height or Side × Corresponding height∥

Area of ABCD = CD × AE = 12 cm × 8 cm … (1)∴ ∥

Also,

Area of gm ABCD = AD × CF = AD × 10 cm … (2)∥

From equations (1) and (2), we get

AD × 10 cm = 12 cm × 8 cm

AD = 9.6 cm

23.Steps of construction:

1. Draw a line segment BC = 8 cm

2. At B, construct ΔXBC = 45°

3. With B as the centre and radius equal to 3.5 cm, draw an arc cutting XB at D.

4. Join DC.

5. Draw PQ, the perpendicular bisector of DC, and let itintersect XB at A.

6. Join AC.

Then ΔABC is the required triangle.

Justification:

A lies on the perpendicular bisector of DC, and so it is equidistant from points D and C.

AD = AC … (1)∴

Now, BD = 3.5 cm [By construction]

AB – AD = 3.5 cm⇒

AB – AC = 3.5 cm [Using (1)]⇒

Hence, ΔABC is the required triangle.

24.24. Diagonal AC divides quadrilateral ABCD into two triangles – ABC and ACD.

Area of quadrilateral ABCD = Area of ΔABC + Area of∴ ΔACD.

For ΔABC, we have

Area of ΔABC ∴

Page 111: CBSE maths XI STD

Area of ΔABC = 20.98 cm⇒ 2

For ΔACD, we have

Area of ΔACD ∴

=

= 54 cm2

Area of quadrilateral ABCD = (20.98 + 54) cm2

= 74.98 cm2

Hence, area of quadrilateral ABCD 74.98 cm2.

25.We have:

r = Inner radius of cylinder = 12 cm

Thickness of cylinder = 2 cm

R = Outer radius of cylinder = (12 + 2) cm∴

= 14 cm

h = Height of cylinder = 35 cm

Volume of wood = Π(R2 – r2)h

= 22 × 26 × 2 × 5

= 5720 cm3

Hence, the required volume of the wood is 5720 cm3.

26.Suppose Radhika has to obtain x marks in the next test so as to have an average of exactly 85.

Mean marks = 85∴

340 = 247 + x⇒

x = 340 – 247⇒

x = 93⇒

Page 112: CBSE maths XI STD

Hence, she must obtain 93 marks in the next test to get an average of 85.

27.Total number of days for which weather forecast was made = 300

(i) Let E1 be the event that the forecast was correct on a given day.

Number of days for which forecast was correct = 210

P (Forecast was correct on a given day) = P(E1)

= 0.7

(ii) Let E2 be the event that the forecast was not correct on a given day.

Number of days for which forecast was not correct = 300 – 210 = 90

P (Forecast was not correct on a given day) = P(E2)

= 0.3

28.Since the three coins are tossed simultaneously 200 times, the total number of trials is 200.

Let E be the event of getting two heads.

Then the number of trials in which the event E happens is 72.

P (Getting two heads)

SECTION D

29.Given: 2x + y = 8

Page 113: CBSE maths XI STD

y = 8 – 2x⇒

If x = 0, then

y = 8 – 2(0) = 8

If x = 1, then

y = 8 – 2(1) = 6

Thus, we have the following table for 2x + y = 8:

x01

y86

Plot the points A (0, 8) and B (1, 6) on a graph paper.

Draw a line passing through points A and B.

Then line AB represents the graph of 2x + y = 8.

From the graph, we see that the points (2, 4) and (3, 2) lie on the graph.

Therefore, x = 2, y = 4 and x = 3, y = 2 are solutions of the given equation.

Verification: Substituting x = 2 and y = 4 in the given equation, we get

LHS = 2(2) + 4 = 8 = RHS

x = 2 and y = 4 is a solution.∴

Substituting x = 3 and y = 2 in the given equation, we get

LHS = 2(3) + 2 = 8 = RHS

x = 3 and y = 2 is a solution.∴

The line meets the X-axis at (4, 0) and the Y-axis at (0, 8).

30.Let ABCD be a rectangle in which P, Q, R and S are the mid-points of AB, BC, CD, and DA, respectively. PQ, QR, RS and SP are joined.

Join AC.

In ΔADC, S is the mid-point of AD, and R is the mid-point of CD. [Given]

SR AC and ∴ ∥ [By mid-point theorem] … (1)

In ΔABC, P is the mid-point of AB and Q is the mid-point of BC.

PQ AC and ∴ ∥ [By mid-point theorem] … (2)

From equations (1) and (2), we get

SR PQ and SR = PQ∥

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Thus, PQRS is a quadrilateral whose one of pair of opposite sides is equal and parallel.

PQRS is a parallelogram …. (3)⇒

Now, AD = BC [Opposite sides of a rectangle]

AS = BQ …. (4) [ S and Q are mid-points of AD and BC respectively, given]⇒ ∵

In ΔAPS and ΔBPQ, we have

(i) AP = BP [ P is the mid-point of AB, given]∵

(ii) ΔPAS = ΔPBQ [Each equal to 90°]

(iii) AS = BQ [From equation 4]

ΔAPS ΔBPQ [ By SAS-congruence]∴ ≅

PS = PQ [CPCT] … (5)⇒

PQRS is a parallelogram such that PS = PQ, i.e. PQRS is a parallelogram with two ∴adjacent sides equal.

PQRS is a rhombus.∴

Hence, the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.

31.Let line l through E and perpendicular to AB intersect AB at G such that EGB is a right ∠angle.

In right-angled triangle EGB, we have

1 + 2 + EGB = 180° [Sum of angles of a triangle]∠ ∠ ∠

1 + 2 + 90° = 180° [ EGB = 90°]⇒ ∠ ∠ ∠∵

1 + 2 = 90° … (1)⇒ ∠ ∠

Now, 2 + BEC + 4 = 180° [GEF is a straight line]∠ ∠ ∠

2 + 90° + 4 = 180° [Diagonals intersect at right angle BEC = 90°]⇒ ∠ ∠ ∠∴

2 + 4 = 90° … (2)⇒ ∠ ∠

1 + 2 = 2 + 4 [From equations (1) and (2)]⇒ ∠ ∠ ∠ ∠

1 = 4 … (3)⇒ ∠ ∠

But 1 = 3 [Angles in the same segment] … (4)∠ ∠

3 = 4 [From equations (3) and (4)]⇒ ∠ ∠

EF = FC … (5) [Sides opposite equal angles are equal of a triangle are equal]∴

Similarly, we can prove that

EF = FD … (6)

From equations (5) and (6), we conclude that F is the mid-point of CD.

Hence proved.

32.AD = BC [Opposite sides of a gm]∥

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AH = BF … (1)⇒

[ H and F are the mid-points of AD and BC, respectively]∵

Also, AD BC [Opposite sides of a gm]∥ ∥

AH BF … (2) [ A, H, D are collinear points and B, F, C are collinear points]⇒ ∥ ∵

From equations (1) and (2), we get AH = BF and AH BF∥

ABFH is a parallelogram.∴

Parallelogram ABFH and ΔEFH are on the same base FH and between the same parallels FH and BA.

… (3)

Similarly, we have

… (4)

Adding equations (3) and (4), we get

Hence, .

33.Let r cm be the radius of the base and l cm the slant height. Then,

l2 = r2 + 242 [Using: l2 = r2 + h2]

⇒l2 = r2 + 576

… (i)

Now,

Curved surface area = 550 cm2

Πr⇒ l = 550

r⟹ 2(r2 + 576) = (25 × 7)2

r⟹ 4 + 576r2 – (252 × 72) = 0

r⟹ 4 + 576 r2 – (625 × 49) = 0

r⟹ 4 + 625 r2 – 49r2 – 625 × 49 = 0

r⟹ 2 (r2 + 625) – 49 (r2 + 625) = 0

Page 116: CBSE maths XI STD

(r⟹ 2 + 625) (r2 – 49) = 0

r⟹ 2 – 49 = 0 [ r∵ 2 + 625 ¹0]

r = 7⟹

Volume = ∴

= 1232 cm3

Hence, the volume is 1232 cm3.

34.Let Ei denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6.

Then,

P (Getting 1) = P(E1) =

P (Getting 2) = P(E2) =

P (Getting 3) = P(E3) =

P (Getting 4) = P(E4) =

P (Getting 5) = P(E5) =

Page 117: CBSE maths XI STD

P (Getting 6) = P(E6) =

CBSE - IX Maths Solved Questions

Question 1

If the sides of a triangle are 13 cm, 14 cm and 15 cm respectively then its area is ____cm2.A) 80 B) 82 C) 84 D) 88

Correct Answer : C

Solution : If are the sides of a triangle and is the semi-perimeter, then its area is given by

Here, a = 13, b = 14 and c = 15

s = 21 cm⇒

Page 118: CBSE maths XI STD

= 84 cm2

Hence, the area of the triangle is 84 cm2.

Question 2

If the sides of a triangle are 150 cm, 120 cm and 200 cm, then its area is ____ cm2.A) B)

C) D)

Correct Answer : A

Solution : If are the sides of a triangle and is the semi-perimeter, then its area is given by

Here, and cm

Hence, the area of the triangle is 8966.56 cm2.

Page 119: CBSE maths XI STD

Question 3

If two sides of a triangle are 8 cm, 11 cm and the perimeter is 32 cm, then its area is ____ cm2.

A) B)

C)

D)

Correct Answer : A

Solution : If are the sides of the triangle and is the perimeter, then and cm

We know that,

Hence, the area of the triangle is

Question 4

If two sides of triangle are 8 cm and 11 cm and its perimeter is 64 cm, then its area is ____cm2.

A)

B)

C)

D)

Correct Answer : B

Solution : If are the sides of the triangle and is the perimeter, then

Page 120: CBSE maths XI STD

and

We know that,

Hence, the area of the triangle is

Question 5

If two sides of a triangle are and the perimeter is 42 cm, then its area is ____cm2.

A)

B)

C)

D)

Correct Answer : B

Solution : Let and

We know that,

Page 121: CBSE maths XI STD

Hence, the area of the triangle is

Question 6

If the sides of a triangle are 40 cm, 32 cm and 24 cm, then its area is ____cm2.A) B)

C) D)

Correct Answer : D

Solution : If are the sides of the triangle and is its semi-perimeter, then its area is given by the formula:

Here, and cm

Page 122: CBSE maths XI STD

Hence, the area of the triangle is 384 cm2.

Question 7

If the sides of a triangle are 25 cm, 17 cm and 12 cm respectively, then its area is ____cm2.A) 80 B) 82 C) 90 D) 84

Correct Answer : C

Solution : If are the sides of the triangle and is its semi-perimeter, then its area is given by

Here, and cm

Hence, the area of the triangle is 90 cm2.

Question 8

If the sides of a triangle are 13 cm, 12 cm and 5 cm, then its area is ____cm2.A) 80 B) 82 C) 30 D) 84

Correct Answer : C

Solution : If are the sides of the triangle and is the semi-perimeter, then its area is

Page 123: CBSE maths XI STD

given by

Here, and

Hence, the area of the triangle is 30 cm2.

Question 9

The sides of a triangle are 12 cm, 13 cm and 15 cm respectively, then its area is ____ cm2.

A) B)

C) D)

Correct Answer : A

Solution : If are the sides of a triangle and is the semi-perimeter, then its area is given by

Here, and

Page 124: CBSE maths XI STD

Hence, the area of the triangle is

Question 10

If the perimeter of an isosceles triangle is 30 cm and each of the equal sides is 12 cm, then its area is ____cm2.

A) B)

C)

D)

Correct Answer : D

Solution : Let the length of the unequal side be x cm.

Given, perimeter

We have,

Hence, the area of the triangle is

Page 125: CBSE maths XI STD

Lesson Summary

CBSE - IX Maths Solved Questions

Question 1

If then _____.

A) B) 4 C) 2

D)

Correct Answer : B

Solution : Given,

Question 2

is a polynomial in one variable.A) TrueB) False

Correct Answer : A

Solution : is a one variable polynomial in

Question 3

In the third quadrant, the nature of the x and y coordinates is ____.A) x > 0 and y > 0B) x < 0 and y < 0C) x < 0 and y > 0D) x > 0 and y < 0

Correct Answer : B

Page 126: CBSE maths XI STD

Solution :

In the third quadrant, the nature of the x and y coordinates is x < 0 and y < 0.

Question 4

If x = 2 and y = 2 is a solution of 2x + ay = 8, then a = ____.A) 3B) 4C) 5D) 2

Correct Answer : D

Solution : Given, 2x + ay = 8 Given, x = 2 and y = 2 is a solution.

2(2) + a(2) - 8 = 0

4 + 2a - 8 = 0⇒

2a = 4⇒

a = 2⇒

Hence, the value of a = 2.

Question 5

If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.A) TrueB) False

Correct Answer : A

Solution : If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.

Page 127: CBSE maths XI STD

Question 6

Any point on the perpendicular bisector of a line segment is equidistant from its end points.A) TrueB) False

Correct Answer : A

Solution : Any point on the perpendicular bisector of a line segment is equidistant from its end points.

Question 7

In a right triangle ABC, right angled at B, M is the midpoint of the hypotenuse AC. If the length of the hypotenuse is 12 cm, then the length of BM is ____.

A) 6 cmB) 12 cmC) 4 cmD) 3 cm

Correct Answer : A

We know that for a right triangle, the midpoint of its hypotenuse is the circumcentre

So, M is the circumcentre of triangle ABC

Circumcentre is equidistant from the vertices of the triangle

So, AM = BM = CM

As M is the midpoint of AC

We have,

AM = BM = 6 cm ∴

Page 128: CBSE maths XI STD

Question 8

ABCD is a trapezium in which AB is parallel to CD. BD is a diagonal and E is the midpoint of AD. A line parallel to AB is drawn through E intersecting BC at F. If BC is 6 cm, then the length of BF is ____.A) 6 cmB) 2 cmC) 3 cmD) 9 cm

Correct Answer : C

Solution :

Given, BC = 6 cm Let the diagonal BD intersect line EF at point P.

In ΔDAB,

E is the mid-point of AD and EP||AB (EF||AB and EP is the a part of EF)

P is the mid-point of other side BD of ΔDAB (The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.)

Now in ΔBCD,

P is the mid-point of BD and PF||DC (EF||AB || DC and AB||DC and PF is a part of EF)

F is the mid-point of BC (By Converse of Mid-Point Theorem)⇒

EF bisect BC ⇒

F is the midpoint of BC. ⇒

BF = 3 cm∴

Question 9

If the area of an equilateral triangle is then the perimeter is ____.A) 36 cmB) 37 cmC) 37.2 cmD) 36.2 cm

Correct Answer : A

Page 129: CBSE maths XI STD

Solution : Given, Area

a = 12⇒

Perimeter

Hence, the required perimeter is 36 cm.

Question 10

If the sides of a triangle are 13 cm, 14 cm and 15 cm respectively then its area is ____cm2.A) 80 B) 82 C) 84 D) 88

Correct Answer : C

Solution : If are the sides of a triangle and is the semi-perimeter, then its area is given by

Here, a = 13, b = 14 and c = 15

s = 21 cm⇒

Page 130: CBSE maths XI STD

= 84 cm2

Hence, the area of the triangle is 84 cm2.

Lesson Summary

CBSE - IX Maths Solved Questions

Question 1

If then ____.A) 3

B) C) 0 D)

Correct Answer : B

Solution : Given,

Page 131: CBSE maths XI STD

Question 2

If then the value of at is ____.A) -1B) 2C) 1D) 0

Correct Answer : D

Solution : Given polynomial is

Hence, the value of at is 0.

Question 3

In the second quadrant, the nature of the x and y coordinates is ____.A) x > 0 and y > 0B) x > 0 and y < 0C) x < 0 and y < 0D) x < 0 and y > 0

Correct Answer : D

Solution :

Page 132: CBSE maths XI STD

In the second quadrant, the nature of the x and y coordinates is x < 0 and y > 0.

Question 4

If x = 2 and y = 2 is a solution of 6x + 2ay = 5a, then the value of a is ____.A) 12B) 4C) 5D) 6

Correct Answer : A

Solution : Given, x = 2 and y = 2 is a solution of 6x + 2ay = 5a

Question 5

If a point Q lies between two points P and R such that PQ = QR, then PQ = ____.

A)

B)

C)

D)

Correct Answer : C

Solution : Given, Q lies between P and R and PQ = QR ----- (1) PQ + PQ = PQ + QR (From (1))⇒

2PQ = PR⇒

Page 133: CBSE maths XI STD

Question 6

In the figure below, if ABCD is a quadrilateral, then and

A) TrueB) False

Correct Answer : A

Solution :

From the figure, we have A = 110∠ o, B = 70∠ o, C = 110∠ o and D = 70∠ o

For the lines, and is a transversal and

For the lines and is a transversal and

and

Question 7

In and and By which of the following congruence conditions are the two triangles congruent?

Page 134: CBSE maths XI STD

A) ASAB) SSSC) SASD) RHS

Correct Answer : A

Solution :

In and

Hence, ΔABC ΔDEF (By ASA congruence condition)

Question 8

ABCD is a quadrilateral. P, Q, R and S are the points on the sides AB, BC, CD and DA respectively

such that , , and Then PQRS is a ____.A) trapeziumB) parallelogramC) rhombusD) rectangle

Correct Answer : B

Solution :

Page 135: CBSE maths XI STD

Given , , and

PQ || AC and SR || AC⇒

PQ || SR⇒

Similarly, we can prove QR || SP

Hence, PQRS is a parallelogram.

Question 9

If the height of an equilateral triangle is 8 cm, then its area is ____.A) 42.8 cm2

B) 36.9 cm2

C) 92.2 cm2

D) 34.8 cm2

Correct Answer : B

Solution : Given, the height of the equilateral triangle = 8 cm

Height

Area of an equilateral triangle

Page 136: CBSE maths XI STD

Hence, the area of the equilateral triangle is 36.9 cm .

Question 10

If a cuboid of dimensions 20 cm, 10 cm and 5 cm is melted and cast into a cube, then the length of the edge of the cube so formed will be ____.A) 5 cmB) 10 cmC) 15 cmD) 25 cm

Correct Answer : B

Solution : Given, the dimensions of cuboid are 20 cm, 10 cm and 5 cm. Volume of cuboid

Also given that the cuboid is melted and made into a cube.

Volume of cube

Hence, the length of the edge of the cube will be

CBSE - IX Maths Solved Questions

Question 1

If then ____.

A) B) 2

C) D) 4

Correct Answer : C

Solution : Given,

Page 137: CBSE maths XI STD

Question 2

The algebraic expression is a polynomial.A) TrueB) False

Correct Answer : B

Solution : Given algebraic expression is

The exponent of the second term, i.e., is -1, which is not a whole number.

Hence, the algebraic expression is not a polynomial.

Question 3

If the abscissa and ordinate of a point are negative and positive respectively, then the point lies in the third quadrant.A) TrueB) False

Correct Answer : B

Solution : If the abscissa and ordinate of a point are negative and positive respectively, then the point lies in the second quadrant.

Question 4

Which of the following graphs represents the line x - 2y = 3?

Page 138: CBSE maths XI STD

A)

B)

C)

D)

Correct Answer : A

Solution : Given, x - 2y = 3

2y = x - 3⇒

when x = 0,

Page 139: CBSE maths XI STD

y = -1.5⇒

when y = 0,

x = 3⇒

Plot the points (0, -1.5) and (3, 0) on a graph paper and join the points to get the graph of x - 2y = 3.

Question 5

The measure of an angle is 24o less than its complement, then the angle is ____.A) 66o

B) 156o

C) 33o

D) 78o

Correct Answer : C

Solution : Let the measure of the required angle be x. Measure of its complement = 90o - x Given, (90o - x) - x = 24o

= 33o

Hence, the measure of the angle is 33o.

Question 6

In the figure below, AOC and BOC form a linear pair. If x = 75∠ ∠ o, then the value of y is ____.

Page 140: CBSE maths XI STD

A) 105o

B) 115o

C) 90o

D) 60o

Correct Answer : A

Solution :

Given, x = 75o

Since AB is a straight line, the sum of all the angles on the same side of AB at a point O on it is 180o.

x + y = 180⇒ o

75 + y = 180⇒

y = 180 - 75⇒

y = 105⇒ o

Hence, the value of y is 105o.

Question 7

In and and By which of the following congruence conditions are the two triangles congruent?

A) SSS

Page 141: CBSE maths XI STD

B) ASAC) SASD) RHS

Correct Answer : B

Solution :

In ΔABC, ACB + ABC + BAC = 180∠ ∠ ∠ o

70⇒ o + 50o + BAC = 180∠ o (:- In a triangle sum of the three angles is 180o) ⇒ 120o + BAC = 180∠ o ⇒ BAC = 60∠ o

In ΔDEF, EDF + DEF + DFE = 180∠ ∠ ∠ o

60⇒ o + 70o + DFE = 180∠ o (:- In a triangle sum of the three angles is 180o) ⇒ DFE = 50∠ o

In ΔABC and ΔDEF, we haveAB = DF, ABC = DFE = 50∠ ∠ o and BAC = EDF = 60∠ ∠ o

Hence, ΔABC ΔDFE (By ASA congruence)

Question 8

Following table shows the marks scored by a group of 90 students in a mathematics test of 100 marks.

The probability that a student obtained less than 20 marks is ____.

A)

B)

C)

D)

Correct Answer : A

Solution : Let S be the event of all students in the class

Page 142: CBSE maths XI STD

n(S) = 90 ⇒

Let E be the event of number of students scoring less than 20 marks

n(E) = 7⇒

Probability that a student scores less than 20 marks =

Question 9

The below are the temperature details that were recorded in Alaska in 11 days: -21oC, 0oC, 2oC, -15oC, 0oC, 5oC, -18oC, 7oC, 0oC, -32oC and 21oC.

The mode temperature of the data is ____.

A) 2oCB) 21oCC) -18oCD) 0oC

Correct Answer : D

Solution : Given data set is -21oC, 0oC, 2oC, -15oC, 0oC, 5oC, -18oC, 7oC, 0oC, -32oC and 21oC. In the above data, 0oC occurs more number of times. i.e., 3 times.

Hence, the mode temperature of the data is 0oC.

Question 10

The diameter of base and the height of a conical house are 14 m and 18 m respectively and the capacity of a rice bag is The number of such rice bags required to fill the conical house is ____.A) 14B) 24C) 28D) 25

Correct Answer : C

Solution : Let r be the radius of the base of conical house Given, height = 18 m Diameter of the base = 14 m

Volume of the rice bag

Volume of the house

Page 143: CBSE maths XI STD

Number of bags required to fill the house

Hence, the number of rice bags required is .

CBSE - IX Maths Solved Questions

Question 1

The value of is ____.A) 1 B) 3 C) 2 D) 5

Correct Answer : C

Solution :

Hence, the value of is 2.

Question 2

If the polynomial is divided by then the remainder is ____.

A) B) 92 C) 10

D)

Correct Answer : A

Page 144: CBSE maths XI STD

Solution : Given polynomial is

By Remainder Theorem, when is divided by the remainder is equal to

Hence, the required remainder is

Question 3

Which of the following is the correct representation of locating the points (2, 4), (- 5, 3), (- 1, - 3), (2, - 1), (- 3, 0) and (0, - 2) on the graph paper?A)

B)

C)

D)

Page 145: CBSE maths XI STD

Correct Answer : A

Solution : Plotting the points (2, 4), (- 5, 3), (- 1, - 3), (2, - 1), (- 3, 0) and (0, - 2) on graph paper, we get

Question 4

A pineapple is priced at cost Rs 7 and a watermelon cost Rs 5. Stella spends Rs 40 on these fruits. The linear equation which satisfies the given data is ____.A) 7x + y = 40B) x + 5y = 40C) 7x - 5y = 40D) 7x + 5y = 40

Correct Answer : D

Solution : Let the number of pineapples and watermelons purchased be x and y respectively. Cost of x pineapples = Rs 7x

Cost of y watermelons = Rs 5y

Total cost = Rs 40

Hence, the linear equation which satisfies the given data is 7x + 5y = 40.

Question 5

An angle is 14o more than its complement, then the angle is ____.A) 62o

B) 72o

C) 52o

D) 42o

Correct Answer : C

Page 146: CBSE maths XI STD

Solution : Let the required angle be θ, then its complement is 90o - θ As per the problem,

θ = (90o - θ) + 14o

θ = 90 - θ + 14⇒

2θ = 104⇒

θ = 52⇒ o

Hence, the required angle is 52o.

Question 6

In the figure below, OA and OB are opposite rays. Then the value of x is ____.

A) 40o

B) 35o

C) 30o

D) 25o

Correct Answer : C

Solution :

In the given figure, AOC and BOC form a linear pair. ∠ ∠ AOC + BOC = 180⇒∠ ∠ o

4x + 2x = 180⇒

6x = 180⇒

Page 147: CBSE maths XI STD

x = 30⇒ o

Hence, the value of x is 30o.

Question 7

In the figure below, AB is a line segment and P is its midpoint, D and E are points on the same side of AB such that BAD = ABE and EPA = DPB. If AD = 4.7 cm, then BE = ____.∠ ∠ ∠ ∠

A) 4 cmB) 7 cmC) 4.7 cmD) 2.7 cm

Correct Answer : C

Solution :

In APD and BPE,

PAD = PBE (Given)∠ ∠

APD = EPB (:- EPA + EPD = DPB + EPD)∠ ∠ ∠ ∠ ∠ ∠

AP = BP (:- P is mid-point of AB)

APD ⇒ BPE (By ASA congruence rule)

AD = BE = 4.7 cm (By CPCT)

BE = 4.7 cm∴

Question 8

A coin is tossed 1000 times with the following frequencies :

Page 148: CBSE maths XI STD

Head : 455 and Tail : 545

Then the probability of getting a head is ____.

A) 0.455B) 0.055C) 0.545D) 0.0455

Correct Answer : A

Solution : Let A be the event of getting a head. n(A) = 455 ⇒

Let S be the total number of events

n(S) = 1000⇒

P(A) = 0.455∴

Question 9

The marks scored by 20 students is 2, 50, 4, 50, 6, 20, 50, 13, 15, 60, 50, 69, 43, 12, 50, 39, 65, 50, 13, 40 The frequency of 50 in the above data is ____.

A) 3B) 4C) 6D) 1

Correct Answer : C

Solution : Given, the marks scored by 20 students is 2, 50, 4, 50, 6, 20, 50, 13, 15, 60, 50, 69, 43, 12, 50, 39, 65, 50, 13, 40 Frequency of 50 = 6

Question 10

The ratio of the slant heights of two cones is 5 : 4. If the radius and curved surface area of the first cone are 7 cm and 4070 respectively, then the curved surface area of the second cone whose radius is 14 cm is ____.A) 6512 B) 6412 C) 6262 D) 3451

Correct Answer : A

Solution : Let the slant heights of first and second cones be l1 = 5 cm and l2 = 4 cm

Page 149: CBSE maths XI STD

Given, radius of the first cone, r1 = 7 cm,

Radius of the second cone, r2 = 14 cm,

Curved surface area of 1st cone = 4070 cm2

Slant height of the 2nd cone = 37 x 4

= 148 cm

Curve surface area of the 2nd cone

= 6512 cm2

Hence, the curved surface area of the 2nd cone is 6512

CBSE - IX Maths Solved Questions

Question 1

If then the values of is ____.A) 1B) 2C) 3D) 4

Correct Answer : C

Question 2

If the polynomial is divided by then the remainder is ____.A) 28

Page 150: CBSE maths XI STD

B) 1 C) 7 D) 20

Correct Answer : A

Solution : Given polynomial is

By the Remainder Theorem, when is divided by then the remainder is

= 28

Hence, the required remainder is 28.

Question 3

The correct representation of locating the points (2, 3), (-4, 5), (-5, -6) and (3, -5) on the graph paper is ____.A)

B)

Page 151: CBSE maths XI STD

C)

D)

Page 152: CBSE maths XI STD

Correct Answer : A

Solution : Plotting the points (2, 3), (-4, 5), (-5, -6) and (3, -5) on graph paper, we get

Question 4

On comparing 2x = y with ax + by + c = 0. The values of a, b and c are ____.A) 1, 3 and 0B) 2, -1 and 0C) -1, 0 and 2D) 2, 5 and -2

Correct Answer : B

Solution : Given 2x = y 2x - y = 0⇒

Page 153: CBSE maths XI STD

Comparing 2x - y = 0 with ax + by + c = 0

We get,

a = 2, b = -1 and c = 0.

Question 5

The measure of an angle which is 32o less than its supplement is ____.A) 72o

B) 74o

C) 75o

D) 78o

Correct Answer : B

Solution : Let the measure of the required angle be x Measure of its supplement = (180o - x)

(180 - x) - x = 32⇒

2x = 148⇒

x = 74⇒ o

Hence, the measure of the required angle is 74o.

Question 6

In the figure below, OA and OB are opposite rays, then ____.

A) 50o

B) 90o

C) 30o

D) 60o

Correct Answer : D

Solution :

Page 154: CBSE maths XI STD

AOB is a straight line, the sum of all the angles on the same side of AOB at a point O on it is 180o. AOC + COD + BOD = 180∠ ∠ ∠ o

(x + 10) + x + (x + 20) = 180⇒

3x + 30 = 180⇒

3x = 150⇒

x = 50⇒ o

AOC = x + 10∠

= 50 + 10

AOC = 60∴∠ o

Question 7

In the figure below, ABC and DEF are two triangles in which AB = DF. If ACB = 70, ABC = ∠ ∠50, DEF = 70 and EDF = 60. Then the congruence condition of the two triangles are ____.∠ ∠

A) SSSB) RHSC) AASD) SAS

Correct Answer : C

Solution : Given, ACB = 70∠ o, ABC = 50, DEF = 70 and EDF = 60.∠ ∠ ∠

In ΔABC,

( The sum of angles in a triangle is 180∵ o)

Page 155: CBSE maths XI STD

In ΔABC and ΔDEF,

AB = DF, and

Hence, the two triangles are congruent by AAS criterion.

Question 8

There are 12 packets of salt, each marked 2 kg. Actually each packet contains the following weights (in kg) of salt: 1.950, 2.020, 2.060, 1.980, 2.030, 1.970, 2.040, 1.990, 1.985, 2.025, 2.000 and 1.980.

Out of these packets, one packet is chosen at random. The probability that a packet contains more than 2 kg of salt is ____.

A)

B)

C)

D)

Correct Answer : D

Solution : Total number of packets of salt = 12 Number of packets containing more than 2 kg of salt = 5

The probability that a packet contains more than 2 kg

Question 9

The weights of newly born babies (in kg) in a hospital on a particular day are as follows: 2.3, 2.2, 2.7, 3.1, 2.9, 2.8, 3.0 and 2.4

The range of the weights is ____.

A) 0.7 kgB) 0.8 kgC) 1 kgD) 0.9 kg

Correct Answer : D

Solution : Given data is 2.3, 2.2, 2.7, 3.1, 2.9, 2.8, 3.0 and 2.4 Arranging the above data in ascending order, we get

Page 156: CBSE maths XI STD

2.2, 2.3, 2.4, 2.7, 2.8, 2.9, 3.0, 3.1

Highest value = 3.1

Lowest value = 2.2

Range = 3.1 - 2.2

= 0.9 kg

Hence, the range of the given data is 0.9 kg

Question 10

The volume of an iron sphere is The sphere is melted and drawn into a long wire of uniform circular cross section. If the length of the wire is then the radius of the wire is ____.A) 1 cm B) 3 cm C) 2 cm D) 5 cm

Correct Answer : C

Solution :

Given, volume of the iron sphere

Length(height) of the wire, h = 9 cm

The wire is in the form of a cylinder

Volume of the wire

Also, given that

Volume of the iron sphere = Volume of the wire

Page 157: CBSE maths XI STD

Hence, the radius of the wire is 2 cm

CBSE - IX Maths Solved Questions

Question 1

If then ____.A) 25 B) 125 C) 625 D) 1625

Correct Answer : A

Solution : Given,

= 25

Question 2

is a factor of A) TrueB) False

Correct Answer : B

Solution : Let

By the Factor Theorem, is a factor of if

Page 158: CBSE maths XI STD

Hence, is not a factor of

Question 3

If both the abscissa and ordinate of a point are negatives, then the point lies in the ____ quadrant.A) firstB) secondC) thirdD) fourth

Correct Answer : C

Solution : In the third quadrant, the coordinates of a point are of the form (-, -). Hence, if both the abscissa and ordinate of a point are negatives, then the point lies in the third quadrant.

Question 4

A man's age is three times the sum of the ages of his two sons, one of whom is twice as old as the other. In 4 years the sum of the son's age will be half of their father's age. The age of younger son, elder son and father are ____.A) 5 years, 10 years and 45 yearsB) 4 years, 8 years and 36 yearsC) 3 years, 6 years and 27 yearsD) 6 years, 12 years and 54 years

Correct Answer : B

Solution : Let the younger son's age be x years. The elder son's age be 2x years. ⇒

The father's age be 3(x + 2x) = 9x years. ⇒

After four years, the father's age will be (9x + 4) years.

The younger son's age = (x + 4) years and

The elder son's age = (2x + 4) years

Given,

2(3x + 8) = 9x + 4⇒

6x + 16 = 9x + 4⇒

9x - 6x = 16 - 4⇒

3x = 12⇒

Page 159: CBSE maths XI STD

x = 4⇒

The age of younger son = x = 4 years

The age of elder son = 2x

= 2 x 4

= 8 years

The age of father = 9x

= 9 x 4

= 36 years

Hence, the age of the younger son, elder son and father are 4 years, 8 years and 36 years respectively.

Question 5

In the figure below, PQ is an incident ray and QR is the reflected ray. QN is normal to the surface AB. If PQR = 124∠ o, then NQR = ____. ∠

A) 62o

B) 28o

C) 32o

D) 64o

Correct Answer : A

Solution :

Given, QN AB ⊥By law of reflection,

Angle of incidence = Angle of reflection

---------(1)

Given PQR = 124∠ o

Page 160: CBSE maths XI STD

NQR + NQR = 124⇒∠ ∠ o (from (1))

2 NQR = 124⇒ ∠ o

NQR = 62∴∠ o

Question 6

In ΔABC, then C = ____.∠A) 45 B) 30 C) 55 D) 40

Correct Answer : C

Solution : Given, We know that,

C = 180 - 125⇒∠

Question 7

In the figure below, ABCD is a trapezium in which AB||CD, M is the mid-point of AC and N is the mid-point of BD. If AB = 12 and CD = 14 cm, then the length of MN is ____.

A) 14 cmB) 13 cmC) 10 cmD) 11 cm

Correct Answer : B

Solution :

Page 161: CBSE maths XI STD

Given, AB = 12 cm and CD = 14 cm In ΔACD, M is the mid-point of AC and ME is parallel CD.

E is the mid-point of AD (By Converse of Mid-Point theorem)

= 7 cm

In ΔABD, N is the mid-point of BD and E is the mid-point of AD.

= 6 cm

MN = ME + EN

= 7 + 6

= 13 cm

MN = 13 cm∴

Question 8

In the figure below, QPRS is a cyclic quadrilateral. R and S are two points on the semi-circle described on PQ where PQ is the diameter. If and then the values of

PQR and QSR are ____ respectively. ∠ ∠

A) 10o and 100o

B) 20o and 80o

Page 162: CBSE maths XI STD

C) 30o and 100o

D) 20o and 90o

Correct Answer : A

Solution :

In the cyclic quadrilateral PQRS, (:- The sum of either pair of opposite angles of a cyclic quadrilateral is

180o)

Given, PQ is the diameter of the circle

(:- The angle in a semi-circle is right angle)

In ΔPQR,

PQR = 180⇒∠ o - 170o

PQR = 10⇒∠ o

and QSR = 100∠ o

Question 9

A quadrilateral ABCD whose sides are 9 cm, 40 cm, 28 cm and 15 cm respectively. If the angle between the first two sides is a right angle, then the area of a quadrilateral is ____.

A) B)

Page 163: CBSE maths XI STD

C) D)

Correct Answer : D

Solution :

Given, ABCD be the given quadrilateral such that ABC = 90∠ o

In ΔABC, (By Pythagoras Theorem)

= 81 + 1600

= 1681

AC = 41 cm⇒

Area of ΔABC

In ΔACD,

AC = 41 cm, CD = 28 cm and DA = 15 cm

Let a = AC = 41 cm, b = CD = 28 cm and c = DA = 15 cm

s = 42 cm

Area of ΔACD

Page 164: CBSE maths XI STD

Area of quadrilateral ABCD = (Area of ΔABC) + ( Area of ΔACD)

= 180 + 126

= 306 cm2

Hence, the area of the quadrilateral ABCD is 306 cm2.

Question 10

A rectangular sheet of paper of size can be made to form the curved surface of a right circular cylinder in two ways. If is the volume of the cylinder made by rolling the paper along the breadth of the sheet and is the volume of the cylinder made by rolling the paper along the length of the sheet, then the ratio of and is ____.A) 4 : 3B) 5 : 3C) 3 : 2D) 4 : 5

Correct Answer : B

Solution :

Let and be the radii of two circular cylinders.

and be the heights of the two cylinders.

Given, circumference of the base of first cylinder is 30 cm

and (:- First cylinder is made by rolling the paper along the breadth)

Circumference of the base of second cylinder is 18 cm

and (:- Second cylinder is made by rolling the paper along the length)

Page 165: CBSE maths XI STD

Now,

Now,

CBSE - IX Maths Solved Questions

Question 1

If then ____.A) 0 B) 2 C) -1 D) -2

Correct Answer : C

Solution : Given,

Page 166: CBSE maths XI STD

Question 2

____.A) B) C) D)

Correct Answer : C

Solution :

Question 3

Which of the following represents the correct plotting of the points (-3, 0), (2, 2), (-2, -3), (2, 4), (0, 4) and (-1, -3) on a graph paper?A)

B)

C)

Page 167: CBSE maths XI STD

D)

Correct Answer : A

Solution : Locate the points (-3, 0), (2, 2), (-2, -3), (2, 4), (0, 4) and (-1, -3) on graph paper.

Question 4

In the graph of the equations and the coordiantes of the point where the two lines intersect are ____.A) (0, 1)B) (2, 2)C) (2, 1)D) (0, 0)

Correct Answer : C

Solution : Graph of the equation

Page 168: CBSE maths XI STD

x = 0

and, x = 4

Thus, the abscissae and ordinates of two points on the line represented by the equation are

By plotting the points (0, -2) and (4, 4) on the graph paper and drawing a line passing through these points, we obtain the graph of the equation as shown in the figure

Graph of the equation

x = 0

y = 3⇒

and, x = 4

y = -1⇒

The abscissae and ordinates of two points on the line represented by the given equation are

By plotting the points (0, 3) and (4, -1) and joining them by a line, we obtain the graph of the equation

Page 169: CBSE maths XI STD

Hence, the lines represented by the equations and intersect at point P whose coordinates are (2, 1).

Question 5

AB and CD are two parallel lines and a transversal intersects AB at X and CD at Y. Then the bisectors of the interior angles form a parallelogram with all its angles right angles.A) TrueB) False

Correct Answer : A

Solution :

AB || CD and is transversal AXY = DYX (:- Alternate angles)∠ ∠

PXA = PXY = QYD = QYX = α ----- (1)⇒∠ ∠ ∠ ∠

BXY = CYX (:- Alternate angles)∠ ∠

⇒ ----- (2)

Page 170: CBSE maths XI STD

(:- Internal angles on the same side of the transversal)

2α + 2β = 180⇒ o

α + β = 90⇒ o ----- (3)

From figure,

PXQ = PYQ = α + β = 90∠ ∠ o ----- (4)

In ΔPXY,

P + PXY + PYX = 180∠ ∠ ∠ o

P + α + β = 180⇒∠ o

P + 90⇒∠ o = 180o [:- From equation (3)]

P = 180⇒∠ o-90o

P = 90⇒∠ o ----- (5)

In ΔQXY,

Q + QXY + QYX = 180∠ ∠ ∠ o

Q + α + β = 180⇒∠ o

Q + 90⇒∠ o = 180o [:- From equation (3)]

Q = 180⇒∠ o-90o

Q = 90⇒∠ o ----- (6)

From (4), (5) and (6), it follows that all four angles of the quadrilateral PXQY are 90o

PXY = QYX = α∠ ∠

PX || QY (:- Alternate angles are equal)⇒

PYX = QXY = β ∠ ∠

PY || QX (:- Alternate angles are equal)⇒

Hence, PXQY is a parallelogram and all its angles are right angles.

Question 6

In ΔPQR, if P = 30∠ o and R = 70∠ o, then QR is the greatest side of the triangle.

Page 171: CBSE maths XI STD

A) TrueB) False

Correct Answer : B

Solution :

Given, P = 30∠ o and R = 70∠ o We know that,

( The sum of three angles in a triangle is 180o)

30 + Q + 70 = 180⇒ ∠

Q = 180 - 100⇒∠

Q = 80⇒∠ o

In the above three angles, we observe that Q is the greatest angle.∠

Hence, PR is the greatest side of the triangle.

Question 7

PQRS is a quadrilateral. A, B, C and D are the mid-points of sides of PQ, QR, RS and SP respectively. If PR = 6 cm and QS = 8 cm, then the lengths of the sides of ABCD are ____.A) 3 cm, 4 cm, 3 cm and 4 cmB) 3 cm, 7 cm, 3 cm and 7 cmC) 5 cm, 8 cm, 5 cm and 6 cmD) 3 cm, 7 cm, 3 cm and 4 cm

Correct Answer : A

Solution :

Given, PR = 6 cm and QS = 8 cm A, B, C and D are the mid-points of sides of PQ, QR, RS and SP respectively.

Page 172: CBSE maths XI STD

In ΔPQR, PR||AB (By Mid-Point Theorem)

= 3 cm

In ΔPSR, PR || DC (By Mid-Point Theorem)

= 3 cm

In ΔQRS, QS || BC (By Mid-Point Theorem)

= 4 cm

In ΔQPS, QS || AD (By Mid-Point Theorem)

= 4 cm

Hence, the lengths of the sides of ABCD are 3 cm, 4 cm, 3 cm and 4 cm.

Question 8

In the figure below, ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle. If then ____.

Page 173: CBSE maths XI STD

A) B) C) D)

Correct Answer : A

Solution :

Given, ADC = 130∠ o In cyclic quadrilateral ABCD,

(:- The sum of either pair of opposite angles of a cyclic quadrilateral is 180o)

ABC = 180⇒∠ o - 130o

From the figure, ACB = 90∠ o (:- Angle in a semi circle is right angle)

In ΔABC, ABC + ACB + BAC = 180∠ ∠ ∠ o

50⇒ o + 90o + BAC = 180∠ o

BAC = 180⇒∠ o - 140o

= 40o

BAC = 40∴∠ o

Page 174: CBSE maths XI STD

Question 9

Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops to suffice the needs of their family. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, then the area each of them will get is ____.A) B) C) D)

Correct Answer : C

Solution :

Given that the field is in the shape of a rhombus and its perimeter is 400 m.

Let the rhombus ABCD represents the field.

Perimeter = 400 m

Each side⇒

= 100 m

AB = AD = BC = CD = 100 m

Let diagonal, BD = 160 m

Then semi-perimeter of ΔABD is given by

= 180 m

Area of ΔABD

Page 175: CBSE maths XI STD

Hence, each of them will get an area of

Question 10

The diameter of a roller is 40 cm and its length is 91 cm. The surface area of a garden is 1,14,40,000 sq. cm. The number of rotations that the roller needed to flatten the surface of the garden when the roller was used twice is ____.A) 1000 B) 3000 C) 2000 D) 1500

Correct Answer : C

Solution : Given, diameter of the roller = 40 cm Radius of the roller, r = 20 cm⇒

Height(length) of the roller = 91 cm

The area of the soil flattened by one rotation of the roller is equal to its curved surface area.

Curved surface area of the roller

Number of rotations of the roller over the garden during one time =

=1000

The number of rotations that the roller needed to flatten the surface of the garden = 2 x 1000

=2000

Hence, the required value is 2000

Lesson Summary


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