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ISBN : 9789389187991
Geometry
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POINTS, LINES, LINE SEGMENT, RAY AND PLANE
Point: A Point is like a dot marked by a very sharp Pencil on a Plane Paper.Line: A line is defined as a group of points. Which are straight one after another. Each line is extended infinitely in two directions.Examples:
A B
(i) (ii)
l
A line is named by either any two points on it or by a single small letter. Arrows on both sides of a line indicate that the line is extended both sides infinitely. A line has infinite length.Line Segment: If a part of the line is cut out, then this cut out piece of the line is called a line segment. This means that line segment has a fixed length.Ray: A ray is a part of a line extended infinitely in any one direction only.Example:
A
B
A ray is named by two points, one of which is the end point on the ray called initial point and other point is any point on the ray.Plane: It is a flat surface extended infinitely. It has only length and breadth but no thickness.
Lines and Angles
Intersecting Lines: If two or more lines intersect each other, then they are called intersecting lines.
A
BC
D
If two lines intersect at right angles, then two lines are called perpendicular linesConcurrent Lines: If three or more lines pass through a point, then they are called concurrent lines and the point through which these all lines pass is called point of concurrency.
Geometry18
A
BC
E
D
F
O
In the figure, AB, CD and EF are concurrent lines and point O is the point of concurrency.Parallel Lines: Two straight lines are parallel if they lie in the same plane and do not intersect even if they are produced infinitely.Perpendicular distances between two parallel lines are the same at all places.
C D
A B
Symbol for parallel lines is ||.Transversal Line: A line which intersects two or more given lines at distinct points is called a transversal of the given lines.
AB
DC
4 1
23
8 5
67
P
Q
In figure straight lines AB and CD are intersected by a transversal PQ.(i) Corresponding angles: In the figure ∠1 and ∠5, ∠4 and
∠8, ∠2 and ∠6, ∠3 and ∠7 are four pairs of corresponding angles.
(ii) Alternate interior angles: ∠3 and ∠5, ∠2 and ∠8, are two pairs of alternate interior angles.
(iii) Alternate exterior angles: ∠1 and ∠7, ∠4 and ∠6 are two pairs of alternate exterior angles.
Note that(i) If two angles of any pair of corresponding angles are equal,
then the two lines are parallel.(ii) If two angles of any pair of alternate interior angles are
equal, then the two lines are parallel.(iii) If two angles of any pair of alternate exterior angles are
equal, then the two lines are parallel.(iv) If any two consecutive interior angles are supplementary
(i.e. their sum is 180°), then the two lines are parallel.
3
Acute angle: An angle is said to be acute angle if it is less than 90°.Right angle: An angle is said to be right angle if it is of 90°.Obtuse angle: An angle is said to be obtuse angle if it is of more than 90°.Straight angle: An angle is said to be straight angle if it is of 180°.Reflex angle: An angle is said to be reflex angle if it is of value greater than 180°.
��
A
B
O
Here θ is the reflex angle.Complementary angles: Two angles, the sum of whose measures is 90°, are called the complementary angles.Supplementary angles: Two angles, the sum of whose measures is 180°, are called the supplementary angles.Adjacent angles: Two angles are called adjacent angles, if(i) they have the same vertex(ii) they have a common arm and(iii) non-common arms are on either side of the common arm.
O
B
A
X
In figure, ∠AOX and ∠BOX are adjacent angles Linear pair of angles: Two adjacent angles are said to form a linear pair of angles.
40°140°
B A
O
C
In figure, ∠AOC and ∠BOC are linear pair angles.Vertically opposite angles: Two angles are called a pair of vertically opposite angles, if their arms form two intersecting lines.
B D
C A
O
In figure, ∠AOC and ∠BOD form a pair of vertically opposite angles. Also ∠AOD and ∠BOC from a pair of vertically opposite angles.Angles on one side of a line at a point on the line: Sum of all the angles on any one side of a line at a point on the line is always 180°.
Here in figure, θ1 + θ2 + θ3 = 180°.Angle around a point: Sum of all the angles around a point is always 360°.
�4
�5
�3
�1
�2
Here θ1, θ2, θ3, θ4 and θ5 are the angles around a point. Hence θ1 + θ2 + θ3 + θ4 + θ5 = 360° Angle bisector: An angle bisector is a ray which bisects the angle whose initial point be the vertex of the angle.
�
�
A
B
C
O
Illustration 1: Three straight lines, X, Y and Z are parallel and the angles are as shown in the figure above. What is ∠AFB equal to?(a) 20° (b) 15° (c) 30° (d) 10°
X
Y
ZF
D E
30°
125°80°
B
C
A
M
Solution: (b) ∠CDE = 180° – 125° = 55° In DDCE, ∠CED = 180° – 55° – 80° = 45° and ∠ABF = 30° (vertically opposite) Also, ∠ABF = ∠BFM = 30° (alternate angle) and, ∠DEF = ∠EFM (alternate angle) ∠EFM = 45° ⇒ ∠EFB + ∠BFM = 45° ⇒ ∠EFB = 45° – 30° ⇒ ∠AFB = 15°Illustration 2: In figure, if AB | | CD, CD | | EF and
y : z =3 : 7, x = ?
A
C
E
B
D
F
z
y
x
(a) 112° (b) 116° (c) 96° (d) 126°
4
Solution: (d) As y + z = 180°, also y : z = 3 : 7 ∴ y = 54° also x + y = 180° ∴x = 180° – 54° = 126°Illustration 3: In the ∆PQR, PS is the bisector of ∠P and PT
D QR, then ∠TPS is equal to
Q RST
P
(a) ∠Q + ∠R (b) 90° + 12
∠Q
(c) 90° – 12
∠R (d) 12
(∠Q – ∠R)
Solution: (d) PS is the bisector of ∠QPR D ∠1 + ∠2 = ∠3 ...(1) ⇒ ∠Q = 90° – ∠1 ∠R = 90° – ∠2 – ∠3 So, ∠Q – ∠R = [90° – ∠1] – [90° – ∠2 – ∠3] ⇒ ∠Q – ∠R = ∠2 + ∠3 – ∠1 = ∠2 + (∠1 + ∠2) – ∠1 [From Eq. (1)]
Q RST
P
1
23
⇒ ∠Q – ∠R = 2∠2 ⇒ 12
(∠Q – ∠R) = ∠TPS
Polygons
Polygons are closed plane figures formed by series of line segments, e.g. triangles, rectangles, etc.
F G
H
I
K
R J
S
A
B
CD
E
P
Q
Convex Polygon Convex Polygon
Convex polygons can be classified into regular and irregular polygons.(a) Regular polygon: A convex polygon whose all the sides are
equal and all angles are equal is called a regular polygon.(b) Irregular Polygon: A convex polygon in which all the sides
are not equal or all the angles are not of the same measure is called an irregular polygon.
Interior and Exterior Angles of a Polygon
An angle inside a polygon between any two adjacent sides at a vertex of the polygon is called an interior angle of the polygon. An angle outside a polygon made by a side of the polygon with the its adjacent side produced is called an exterior angle of the polygon.In the figure ABCDEF is a polygon. ∠FAB, ∠ABC, ∠BCD, ∠CDE, ∠DEF and ∠EFA are interior angles of the polygon ABCDEF.
∠BAG, ∠CBH, ∠DCI, ∠EDJ, ∠FEK and ∠AFL are exterior angles of the polygon ABCDEF.Properties of Polygons
(i) Sum of all the interior angles of a polygon with ‘n’ sides = (n – 2) 180°
(ii) Sum of all the exterior angles of a polygon = 360° ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°
(iii) Perimeter of a regular polygon with a side length of a = n × a
(iv) No. of sides of a regular polygon
= 360
An exterior angle
°
(v) Number of diagonals of a polygon with n sides
= ( 3)
2
n n −
Illustration 4: An interior angle of a regular polygon is 135°. Find the number of sides of the polygon.Solution: Since interior angle of the regular polygon = 135°, hence exterior angle = 180° – 135° = 45°
∴ No. of sides = 360 360
8An exterior angle 45
° °= =
°
∴ No. of sides = 8
5
Illustration 5: An interior angle of a regular polygon is 100° more than its an exterior angle. Find the number of sides the polygon.Solution: Let measure of each exterior angle be x°.Then measure of each interior angle = (x + 100)Now, x + (x + 100) = 180 ⇒ 2x = 80 ⇒ x = 40
Now number of sides = 360 360
9An exterior angle 40
= = .
Triangles
A triangle is a convex polygon having three sides.Triangles can be classified on the basis of their sides or angles.On the basis of sides, triangles are of the following types(a) Equilateral triangle: All the three sides are equal(b) Isosceles triangle: Two sides are equal(c) Scalene triangle: All the three sides are unequal.On the basis of angles, triangles are of the following types(a) Acute angled triangle: Each interior angle is less than 90°.(b) Right angled triangle: One of the interior angle is equal 90°.(c) Obtuse angled triangle: One of the interior angle is more
than 90°.
Basic properties and some important theorems of triangles
1. Sum of measures of the interior angles of a triangle is 180°.C
A B
In ∆ABC, ∠CAB + ∠ABC + ∠ACB = 180° or ∠A + ∠B + ∠C = 180°2. The exterior angle of a triangle is equal to the sum of the
opposite (not adjacent) interior angles
In ∆ABC, ∠CBD = ∠A + ∠C = ∠ABE3. Sum of the lengths of any two sides of a triangle is greater
than the length of the third side.C
A B
(i) AB + AC > BC (ii) AC + BC > AB (iii) AB + BC > AC4. Difference between the lengths of any two sides of a triangle
is smaller than the length of the third side.
A B
C
(i) | AB – BC | < AC (ii) | AC – AB | < BC (iii) | AC – BC | < AB5. In any triangle, side opposite to greatest angle is largest and
side opposite to smallest angle is smallest.A
C B
In ∆ABC, if ∠A > ∠B > ∠C, then BC is the largest side and AB is the smallest side.
6. In any triangle line joining any vertex to the mid point of its opposite side is called a median of the opposite side of the triangle.
C
A BD
In ∆ABC, D is the mid point of AB Hence CD is a median of ∆ABC. Any median of a triangle divides the triangle into two
triangles of equal areas.7. Sides opposite to equal angles in a triangle are equal.
A
B C
In ∆ABC, ∠B = ∠C ∴ AB = AC Converse of this property is also true.8. In an isosceles triangle, if a perpendicular is drawn to
unequal side from its opposite vertex, then (a) The perpendicular is the median (b) The perpendicular bisects the vertex angle.
6C
A BD
∆ABC is an isosceles triangle in which AC = BC. CD is perpendicular to AB, hence CD is a median and ∠ACD = ∠BCD9. In a right angled triangle, the line joining the vertex of the
right angle to the mid point of the hypotenuse is half the length of the hypotenuse.
C
A B
D
In ∆ABC, ∠BAC = 90° and D is the mid point of BC, then
AD = 12
BC = BD = CD
10. Mid-point theorem: In any triangle, line segment joining the mid points of any two sides is parallel to the third side and equal to half of the length of third side.
C
A B
D E
In ∆ABC, D and E are mid points of sides AC and BC, then
DE is parallel to AB i.e. DE || AB and DE = 1
2 AB
11. Angle Bisector Theorem: Bisector of an angle (internal or external) of a triangle divides the opposite side (internally or externally) in the ratio of the sides containing the angle.
For example:
C
A
BD
In figure AD is the bisector of exterior ∠BAC
D AB BD
AC DC=
C
A
BD
In figure AD is the bisector of exterior ∠BAC.
∴ AB BD
AC DC=
Converse of the angle bisector theorem is also true.12. Pythagoras Theorem: In a right angled triangle.
C
A B
D (BC)2 = (AB)2 + (AC)2
Converse of this theorem is also true. List of some common phythagoras triplets. 5, 12, 13 3, 4, 5 6, 8, 10 8, 15, 17 7, 24, 2513. Basic Proportionality Theorem (BPT): If a line is drawn
parallel to one side of a triangle which intersects the other two sides in distinct points, the other two sides are divided in the same ratio.
C
A
B
ED
In ∆ABC, DE || BC,
Then, AD AE
DB EC=
This theorem is also known as Thales theorem. Converse of this theorem is also true.Illustration 6: In a triangle ABC, ∠A = x, ∠B = y, and
∠C = y + 20.If 4x – y = 10, then the triangle is :(a) Right-angled (b) Obtuse-angled (c) Equilateral (d) None of these
7
Solution: (a) We have, x + y + (y + 20) = 180 or x + 2y = 160 ...(1) and 4x – y = 10 ...(2) From (1) and (2), y = 70, x = 20Angles of the triangles are 20°, 70°, 90°. Hence the triangle is a right angled.Illustration 7: In the given figure, CD || AB. Find y.
B CE
A
D
3x°
3x°4x° y°
(a) 79° (b) 72° (c) 74° (d) 77°Solution: (b) In ∆ABC, ∠ABC + ∠BCA + ∠CAB = 180° ⇒ 4x + 3x + 3x = 180° ⇒ 10°x = 180° ⇒ x = 18° Now, ∠ABC = ∠DCE (corresponding angles are equal) ⇒ ∠DCE = y ⇒ 4x° = 4 × 18° = 72°Illustration 8: In the adjoining figure, AE is the bisector of
exterior ∠CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, then CE is equal to
B EC
AD
(a) 6 cm (b) 12 cm (c) 18 cm (d) 20 cm
Solution: (c) BE AB
CE AC= as AE is an exterior angle bisector.
Let CE = x, BE = BC + EC = 12 + x
⇒ 12 10
6x
x
+= ⇒ (12 + x) 6 = 10x
⇒ 72 + 6x = 10x ⇒ 4x = 72 ⇒ x = 18 cmIllustration 9: OB and OC are respectively the bisectors of
∠ABC and ∠ACB. Then, ∠BOC is equal to
B C
O
A
(a) 90° – 12
∠A (b) 90° + ∠A
(c) 90° + 12
∠A (d) 180° – 12
∠A
Solution: (c) In ∆BOC, ∠1 + ∠2 + ∠BOC = 180° ∠A + ∠B + ∠C = 180°. …(1)
⇒ 1 1 1
902 2 2
A B C∠ + ∠ + ∠ = °
⇒ 12
(∠A) + ∠1 + ∠2 = 90° ⇒ ∠1 + ∠2 = 90° – 12
∠A
Put ∠1 + ∠2 in Eq. (1), we get
∠BOC = 180° – 90° – 1
902
A ° − ∠
= 90° + 12
A∠
Important Terms Related to a Triangle
1. Medians and Centroid:
C
A
B
G
P
R Q
AP, BQ and CR are medians of ∆ABC where P, Q and R are mid points of sides BC, CA and AB respectively.
(i) The point of concurrency of three medians is called concurrency of the triangle denoted by G.
(ii) Centroid of the triangle divides each median in the ratio 2 : 1
i.e. AG : GP = BG : GQ = CG : GR = 2 : 1, where G is the centroid of ∆ABC.
2. Altitudes and Orthocentre: A perpendicular drawn from any vertex of a triangle to its opposite side is called altitude of the triangle.
In the figure, AP, BQ and CR are altitudes of DABC. The point of concurrency of altitudes is called Orthocentre
of the triangle, denoted by H.
C
A
B
H
P
R Q
8
In figure, AP, BQ and CR meet at H, hence H is the orthocentre of the triangle ABC.
Note: The angle made by any side at the orthocentre and at the vertex opposite to the side are supplementary angle.
Hence, ∠BAC + ∠BHC = ∠ABC + ∠AHC = ∠ACB + ∠AHB
= 180°.
3. Perpendicular Bisectors and Circumcentre: A line which is perpendicular to a side of a triangle and also bisects the side is called a perpendicular bisector of the side.
(i) Perpendicular bisectors of sides of a triangle are concurrent and the point of concurrency is called circumcentre of the triangle, denoted by 'O'.
(ii) The circumcentre of a triangle is centre of the circle that circumscribes the triangle.
(iii) Angle formed by any side of the triangle at the circumcentre is twice the vertical angle opposite to the side.
L
M N
O
In figure, perpendicular bisectors of sides LM, MN and NL of ∆LMN meets at O. Hence O is the circumcentre of the triangle LMN.
∠MON = 2 ∠MLN.4. Angle Bisectors and Incentre: Lines bisecting the
interior angles of a triangle are called angle bisectors of triangle.
(i) Angle bisectors of a triangle are concurrent and the point of concurrency is called Incentre of the triangle, denoted by I.
(ii) With I as centre and radius equal to length of the perpendicular drawn from I to any side, a circle can be drawn touching the three sides of the triangle. So this is called incircle of the triangle. Incentre is equidistant from all the sides of the triangle.
(iii) Angle formed by any side at the incentre is always 90° more than half the vertex angle opposite to the side.
r r
rI
In figure AI, BI, CI are angle bisectors of DABC. Hence I is the incentre of the DABC and
∠BIC = 90° + 1
2 ∠A,
∠AIC = 90° + 1
2 ∠B
and ∠AIB = 90° + 1
2 ∠C
A
B C
I′
If BI′ and CI′ be the angle bisectors of exterior angles at B and C, then
∠BI′C = 90° – 1
2 ∠A.
Illustration 10: If in the given figure ∠PQR = 90°, O is the centroid of ∆PQR, PQ = 5 cm and QR = 12 cm, then OQ is equal to
(a) 1
3 cm2
(b) 1
4 cm3
(c) 1
4 cm2
(d) 1
5 cm3
Solution: (b) By Pythagoras theorem,
PR = 2 2 2 25 12PQ QR+ = + = 13 cm
∴ O is centroid ⇒ QM is median and M is mid-point of PR.
QM = PM = 132
∴ Centroid divides median in ratio 2 : 1.
∴ OQ = 23
QM = 2 13 133 2 3× =
∴ OQ = 1
4 cm3
Congruency of two Triangles
Two triangles are congruent if they made to super impose the other it will cover exactly.
A
B C
P
Q R
9
If two triangles ABC and PQR are congruent then.(i) ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R(ii) AB = PQ, BC = QR, AC = PRThis is symbolically written as ∆ABC ≅ ∆PQRNote: In two congruent triangles, sides opposite to equal angles are corresponding sides and angles opposite to equal sides are corresponding angles.
Conditions of CongruencyThere are 4 conditions of congruency of two triangles.1. SAS (Side-Angle-Side) Congruency: If two sides and the
included angle between these two sides of one triangle is equal to corresponding two sides and included angle between these two sides of another triangle, then the two triangles are congruent.
A
B C
P
Q R
In ∆ABC and ∆PQR AB = PQ, BC = QR and ∠ABC = ∠PQR ∴ ∆ABC ≅ ∆PQR [by SAS congruency]2. ASA (Angle-Side-Angle) Congruency: If two angles and
included side between these two angles of one triangle are equal to corresponding angles and included side between these two angles of another triangle, then two triangles are congruent.
A
B C
P
Q R
In ∆ABC and ∆PQR ∠A = ∠P ∠B = ∠Q AB = PQ ∴ ∆ABC ≅ ∆PQR [by ASA congruency]3. SSS (Side-Side-Side) Congruency: If three sides of one
triangle are equal to corresponding three sides of another triangle, the two triangles are congruent.
A
B C
P
Q R
In ∆ABC and ∆PQR AB = PQ BC = QR CA = RP ∴ ∆ABC ≅ ∆PQR [by SSS congruency]4. RHS (Rightangle-Hypotenuse-Side) Congruency:
Two right angled triangles are congruent to each other if hypotenuse and one side of one triangle are equal to hypotenuse and corresponding side of another triangle.
A
B C
P
Q R
In ∆ABC and ∆PQR ∠ABC = ∠PQR = 90° AC = PR BC = QR ∴ ∆ABC ≅ ∆PQR [by RHS congruency]
Similarity of Two Triangles
Two triangles are said to be similar, if their shapes are the same but their size may or may not be equal.When two triangles are similar, then(i) all the corresponding angles are equal and(ii) all the corresponding sides are in the same ratio
A
B C
D
E F
If ∆ABC and ∆DEF are similar, then ∠A = ∠D ∠B = ∠E ∠C = ∠F
and AB BC CA
DE EF FD= =
∆ABC ~ ∆DEF, read as triangle ABC is similar to triangle DEF. Here ~ is the sign of similarity.Conditions of SimilarityThere are 4 conditions of similarity.1. AA (Angle-Angle) Similarity: If two angles of one triangle
are respectively equal to two angles of another triangle, then two triangles are similar.
C
A
B
D
E F
10
In ∆ABC and ∆DEF, if ∠A = ∠D ∠B = ∠E then ∆ABC ~ ∆DEF [By AA Similarity]2. SSS (Side–Side–Side) Similarity: Two triangles are said
to be similar, if sides of one triangle are in the same ratio to the sides of the other triangle:
For example:
C
A
B
D
E F
In ∆ABC and ∆DEF, if
AB BC CA
DE EF FD= =
Then ∆ABC ~ ∆DEF [By SSS Similarity]3. SAS (Side–Angle–Side) Similarity: Two triangles are said
to be similar if two sides of a triangle are proportional to the two sides of the other triangle and the angles included between these sides of two triangles are equal.
For example:
C
A
B
D
E F
In ∆ABC and ∆DEF, if
AB BC
DE EF=
and ∠B = ∠E Then, ∆ABC ~ ∆DEF [By SAS Similarity]4. RHS (Rightangle-Hypotenuse-Side) Similarity: Two
triangles are said to be similar if one angle of both triangle is right angle and hypotenuse of both triangles are proportional to any one other side of both triangles respectively.
For example:A
BC
D
F E
In ∆ABC and ∆DEF, if ∠B = ∠E [= 90°]
AC AB
DF DE=
Then ∆ABC ~ ∆DEF [By RHS similarity] Note: In similar triangles,
Ratio of medians = Ratio of corresponding heights = Ratio of circumeradii = Ratio of inradii
Theorem
If two triangles are similar, then ratio of areas of two similar triangle is equal to the ratio of square of corresponding sides.i.e. if DABC in DPQR then –
2 2 2ar ABC AB BC AC
ar PQR PQ QR PR
∆ = = = ∆
Illustration 11: D and E are the points on the sides AB and AC respectively of a ∆ABC and AD = 8 cm, DB = 12 cm, AE = 6 cm and EC = 9 cm, then BC is equal to
(a) DE25
(b) DE52
(c) 32
DE (d) 23
DE
Solution: (b) As in ∆ADE and ∆ABC
AD 2 AE 2
andAB 5 AC 5
= =
D AD AE
AB EC=
and ∠A = ∠A (common) ∴ ∆ADE ~ ∆ABC
∴ 25
DE AD DE
BC AB BC= ⇒ =
⇒ BC = 52
DE
Illustration 12: In a right angled ∆ABC in which ∠A = 90°. If AD D BC, then the correct statement is
B CD
A
(a) AB2 = BD × DC (b) AB2 = BD × AD(c) AB2 = BC × DC (d) AB2 = BC × BDSolution: (d) Clearly, ∆ABD ~ ∆CBA
⇒AB CB
BD BA=
⇒ AB2 = BC × BD
11
Illustration 13: From the adjoining diagram, calculate(i) AB (ii) AP
A
P
B C
6 cm
8 cm
10 cm
40°
40°
Solution: In ∆APC and ∆ABC, ∠ACP = ∠ABC ∠A = ∠A
⇒ ∆ACP ~ ∆ABC ⇒ AP PC AC
AC BC AB= =
∴8 6
6 10AP
AB= =
⇒ AP = 8
610
× = 4.8 and AB = 608
= 7.5
⇒ AP = 4.8 cm and AB = 7.5 cmIllustration 14: In the adjoining figure, DE || BC and AD : DB = 4 : 3
A
B C
D
F
E
L
Find AD
AB and then
DE
BC
Solution: Since the sides of similar triangles are proportional, we have
AD DE
AB BC=
But, 4 4 43 4 3 7
AD AD AD
DB AD DB AB= ⇒ = ⇒ =
+ +
∴ 47
DE AD
BC AB= =
Illustration 15: In the given figure, DE parallel to BC. If AD = 2 cm, DB = 3 cm and AC = 6 cm, then AE is (a) 2.4 cm (b) 1.2 cm (c) 3.4 cm (d) 4.8 cm
CB
A
D E
2
8cm
6
3
Solution: (a) The triangles ADE and ABC are similar.
⇒ AD AE
AB AC=
or 25 6
AE=
∴ AE = 125
= 2.4cm
Illustration 16: The perimeters of two similar triangles ABC and PQR are 36 cm, and 24 cm, respectively. If PQ = 10 cm, then the length of AB is :(a) 16 cm (b) 12 cm (c) 14 cm (d) 15 cmSolution: (d)
∆ ABC and ∆PQR are similar.
Perimeter of
Perimeter of
36
24
∆=
∆
⇒ =
AB ABC
PQ PQR
AB
PQ
or AB = 36
1024
× = 15
Quadrilaterals
Quadrilateral is a plane figure bounded by four straight lines. In figure, PQRS is a quadrilateral and PR, QS are its two diagonals.
S
R
QP
Sum of angles of a quadrilateral = 360° i.e. ∠P + ∠Q + ∠R + ∠S = 360°Types of Quadrilaterals1. Parallelogram: A parallelogram is a quadrilateral with
opposite sides parallel and equal.D C
BA
O
In figure, ABCD is a parallelogram in which AC and BD are diagonals which intersect each other at O.
Properties: (i) Opposite sides are equal i.e. AB = DC, AD = BC
12
(ii) Opposite sides are parallel i.e. AB || DC and AD || BC (iii) Opposite angles are equal i.e. ∠BAD = ∠BCD and ∠ABC = ∠ADC (iv) Diagonals bisect each other, i.e. OA = OC, OB = OD (v) Sum of pair of consecutive angles is 180° i.e., ∠A + ∠B = 180°, ∠B + ∠C = 180°, ∠C + ∠D = 180°, ∠D + ∠A = 180°.2. Rectangle: A rectangle is a parallelogram with all angles
equal to 90°.D C
BA
In figure, ∠A = ∠B = ∠C = ∠D = 90° Properties: (i) In a rectangle Length of diagonals, are equal i.e.
AC = 2 2AB BC+ = BD (ii) In a rectangle diagonals bisect each other. (iii) All rectangles are parallelogram but all parallelograms
are not rectangles.3. Rhombus: A parallelogram is a rhombus if its all sides are
equal.D C
BA
In rhombus ABCD, AB = BC = CD = DA Properties: (i) In a rhombus diagonals bisect each other at right angles
i.e. angle between AC and DB is 90°. (ii) All rhombus are parallelogram but all parallelograms
are not rhombus.4. Square: A parallelogram is a square if all the four sides are
equal and also all the four angles are equal (i.e. 90°).D C
BA
O
In figure, ABCD is a square in which AB = BC = CD = DA and ∠A = ∠B = ∠C = ∠D = 90°
Properties: (i) In a square diagonals are equal i.e. AC = BD (ii) In a square diagonals bisect each other at right angle,
i.e. OA = OC, OB = OD and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.
(iii) All square are rhombus but rhombus may or may not be a square.
5. Trpaezium: A quadrilateral is a trapezium if one pair of opposite sides are parallel.
In trapezium ABCD, AB || DC.D C
d
BA
If lateral sides (i.e. non-parallel sides) of a trapezium are equal, then it is called isosceles trapezium.
Area of trapezium =1
2 (sum of parallel sides) × d
Properties of isosceles trapezium In the figure ABCD is an isosceles trapezium, then
D C
BA
(i) AB || DC (ii) AD = BC (iii) Diagonals are equal i.e. AC = BD
Diagonal Properties of all Parallelograms
Sr.No.
DiagonalProperties
Type of ParallelogramParallelogram Rectangle Rhombus Square
1 Diagonals bisect each other
2 Diagonals areequal
3 Diagonals are at 90° to each other
Circles
A circle is a locus i.e. path of a point in a plane which moves in such a way that its distance from a fixed point (centre) always remains constant.
P
BoAIn figure, 'O' is the fixed point and P is a moving point in the same plane. The path traced by P is called a circle. Fixed point O is the centre of the circle and the constant distance OP is called radius of the circle.A diameter is a line segment passing through the centre and joins the two points on the circle in the figure.AB is the diameter as it passes through the centre and joins the two points on the circle. Diameter = 2 × radius.
13
CircumferencePerimeter of circle is called its circumference.Chord: A line segment joining any two points on the circle is called chord of the circle. A chord which passes through the centre is the diameter of the circle.
P
Q
A
O
B
In the figure, O is the centre of the circle. AB and PQ both are chords.But PQ is the diameter (longest chord) also.Arcs: A piece of a circle between two points is called an arc.Consider two points M and N on the circle. We find that there are two pieces of circle between M and N. One is longer and other is smaller.
M N
The longer piece is called major arc and smaller piece is called minor arc
M N
M N
Major Arc
P
Q
Major arc is denoted by �MPN and minor arc is denoted by�MQN .Segment: The region between a chord and an arc of a circle is called a segment.There are two segments corresponding to two arcs, major segment and minor segment.
P Q
Major
segment
Minor
segment
Centre
Major segment
Minor segment
If two arcs are equal, then both segments are semi-circles. Sector: The region between an arc and the two radii joining the centre to the end point of the arc is called a sector. There are two sectors Minor and Major Sectors.
M N
Major
Sector
Minor
Sector
O
Tangent: A tangent is a straight line which touches the circumference of a circle at only one point. A tangent never intersect the circumference.
OA
B
r
P
Secant: A secant is a straight line of infinite length which intersects the circumference of a circle at two different points. In figure, AB is a secant.
O
A
B
Basic Properties of a Circle1. Equal chord of a circle subtend equal angles at the centre.
O
A
P
Q
B
If AB = PQ, then ∠AOB = ∠POQ The converse is also true.2. The perpendicular from the centre of a circle to a chord of
the circle bisects the chord.
Q
O
PM
In figure, PQ is chord of a circle with centre 'O', OM is perpendicular to PQ therefore PM = MQ. The converse is also true.
3. One and only one circle can pass through given three non-collinear points.
If three or more points lie on a line, then they are called collinear points otherwise called non-collinear points.
14
4. Equal chords of a circle are equidistant from the centre of the circle.
In the figure, if AB = CD, then OP = OQ
Q
O
P
M
B D
A C
The converse is also true.5. Two equal chords have equal corresponding arcs.
QP
M
B
A
L
If PQ = LM then
(a) � �PAQ LBM= (Minor Arc)
(b) � �PBQ LAM= (Major Arc)6. The greater of the two chords is nearer to the centre.
Q
P
D
A
B
CO
If AB > CD, then OP < OQ7. The angle subtended by an arc at the centre is double the
angle subtended by it at any point on the remaining part of the circle.
P1
P2
BA
O
minor arc �AB subtend ∠AOB at the centre O and also subtend ∠APB at point P (situated on remaining part of circle). So ∠AOB = 2 ∠AP1B = 2∠AP2B
8. Angle in a semicircle is a right angle.P
BAO
In figure, AOB is a diameter, hence AOBPA is a semicircle, therefore ∠APB = 90°.
9. Angles in the same segment of a circle are equal. ∠ACB, ∠ADB, ∠AEB are in the same segment ACDEBA
of the circle.
∴ ∠ACB = ∠ADB = ∠AEB.10. If in a plane a line segment joining two points subtends
equal angles at two other points lying on the same side of a line containing the line segment, the four points lie on a circle i.e. they are concyclic.
C D
A B
In figure, if ∠ACB = ∠ADB, then points A, B, D, C lie on a circle.
11. The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
A cyclic quadrilateral is the quadrilateral whose four vertices are concyclic i.e. the four vertices lie on a circle.
In figure, ABCD is a cyclic quadrilateral,
D
B
A
C
∴ ∠A + ∠C = 180° and ∠B + ∠D = 180° The converse is also true.12. If a side of a cyclic quadrilateral is produced the exterior
angle so formed is equal to the interior opposite angle.D
BA
C
P
In figure, ABCD is a cyclic quadrilateral, ∴ ∠CBP = ∠CDA13. Two circles C1 with centre O1, radius r1 and C2 with centre
O2, radius r2 will touch (a) Externally, if and only if O1O2 = r1 + r2
15
r1
r2
O2
O1
(b) Internally, if and only if O1O2 = | r1 – r2 |
14. Two circles are congruent if their radii are equal.Illustration 17: In a circle of radius 17 cm, two parallel chords are drawn on opposite sides of a diameter. The distance between the chords is 23 cm. If length of one chord is 16 cm, then the length of the other one is :(a) 15 cm (b) 23 cm (c) 30 cm (d) 34 cmSolution: (c) Let PQ and RS be two parallel chords of the circle on the opposite sides of the diameter AB = 16 cm.
Now, PN = 8 (Since ON is the perpendicular bisector)In ∆PON ON2 = OP2 – PN2 = (17)2 – (8)2 = 289 – 64 = 225or ON = 15 ⇒ ∴ OM = 23 – 15 = 8 In ∆ORM, RM2 = OR2 – OM2
172 – 82 = 289 – 64 = 225or RM = 15 ⇒ RS = 15 × 2 = 30 cm
Basic Pythagorean Triplets
A Pythagorean triplet is a set of three natural numbers a, b and c, which are length of the sides of a right angled triangle.Hence, if a2 + b2 = c2, then set of natural numbers a, b and c is a Pythagorean triplet.Since 32 + 42 = 52, hence 3, 4, 5 form a Pythagorean triplet.General Rule To Find Pythagorean Triplet: If r and s are two natural numbers such that r > s, r – s is odd and GCD of r and s is 1, then the Pythagorean triplet a, b, c are defined by a = r2 – s2, b = 2rs and c = r2 + s2.
Note: If each term of any Pythagorean triplet is multiplied or divided by such a positive number that the products or quotients obtained respectively are natural numbers then the new products or quotients are also form Pythagorean triplets.Since 3, 4, 5 form a Pythagorean triplet, therefore 9, 12 and 15 also form a Pythagorean triplet.
Determination of Nature of Triangle
Let length of three sides of a triangle are a, b and c.• If c be the length of longest side and c2 = a2 + b2, then the
triangle is right-angled triangle.• If c be the length of longest side and c2 > a2 + b2, then the
triangle is an obtuse-angled triangle.• If c be the length of longest side and c2 < a2 + b2, then the
triangle is an acute-angled triangle.
Important Points
1. In ∆ABC right angled at A, if AD is perpendicular to BC.C
D
A B
∆ABC ~ ∆DBA ~ ∆DAC Now ∆ABC ~ ∆DBA
⇒ AB DB
BC BA= ⇒ AB2 = DB × BC
And ∆ABC ~ ∆DAC
⇒ AC DC
BC AC= ⇒ AC2 = DC × BC
And ∆DBA ~ ∆DAC
⇒ DA DC
DB DA= ⇒ DA2 = DB × DC
2. In a cyclic quadrilateral, product of the diagonals is equal to the sum of the products to the opposite sides,
AC × BD = (AD × BC) + (AB × CD)A B
D
C
3. Bisectors of the angles of a parallelogram or a rectangle form a rectangle.
A B
CD
E
F
G
H
16
In parallelogram ABCD, AG, BG, CE and DE are the bisectors of ∠A, ∠B, ∠C and ∠D respectively. Hence in the figures EFGH is a rectangle.
4. A parallelogram inscribed in a circle is a rectangle. In figure, ABCD is a rectangle.
A B
D C
5. A parallelogram circumscribed a circle is a rhombus. In figure, ABCD is a rhombus.
A B
D C
6. Median of a trapezium is the line segment joining mid-points of non-parallel sides of the trapezium.
In the figure E and F are the mid points of non-parallel sides AB and CD respectively. Hence EF is the median of trapezium ABCD.
A B
E F
D C
EF = 1
2 (AB + CD)
7. Perpendicular bisectors of two chords of a cricle intersect at its centre of the circle.
A
B
DC
O
E
F
In figure, OE and OF are perpendicular bisectors of chords AB and CD, OE and OF meet at point O. Hence O is the centre of the circle.
8. If two circles intersect each other at two points then the line through the centres is the perpendicular bisectors of the common chord.
In figure, two circles with centre P and Q intersect each other at two points A and B.
Hence AB is the common chord of the two circles.
A
B
P Q
Therefore, PQ is the perpendicular bisector of common chord AB.
9. If a circle touches all the four sides of a quadrilateral then the sum of the two opposite sides is equal to the sum of other two.
A
B D
C
AB + DC = AD + BC10. In two concentric circles, if a chord of the larger circle is
also tangent to the smaller circle, then the chord is bisected at the point of contact.
A
BC
Hence in the figure, AC = CB11. Length of two tangents from an exterior point to a circle are
equal.Q
P
R
In figure PQ and PR are two tangents drawn from an exterior point to a circle.
∴ PQ = PR12. Direct common tangent: A tangent to two circles are such
that the two circles lies on the same side of the tangent, then the tangent is called direct tangent to the two circles.
Q
R
P
S
O O�
r1
r2
In the figure, PQ and RS are two direct common tangent to
17
the same two circles. Length of these two common tangents to the same two circles are equal.
i.e. PQ = RS
Also PQ = RS = 2 22 1( ( )OO r r′) + −
Here O, O′ are the centres and r1, r2 are the radii of the two circles respectively. Also r2 > r1.
13. Indirect or Transverse Common Tangent: If a tangent to two circles is such that the two circles lie on opposite sides of the tangent, then the tangent is called indirect tangent.
Length of two indirect tangents to two circles is equal.
QR
P
O
O�
r1
r2
S
In the figure, PQ and RS are two indirect common tangents to the same two circles.
∴ PQ = RS
Also PQ = RS = 2 21 2( ) ( )OO r r′ − +
Here O, O′ are centres r1, r2 are radii of the two circles respectively.
14. Star: A star has a shape like given in the figure.
If a star has n sides, then Sum of its all angles = (n – 4) × 180°.15. In a triangle, the sum of the square of any two sides of a
triangle is equal to twice the sum of the square of the median to the third side and square of half the third side.
A
CBD
In the figure, AD is the median.
∴ AB2 + AC2 = 2
222
BCAD +
16. In a triangle,
Sum of square of Sum of the square of
3 three sides of 4 three medians ofa triangle the triangle
× = ×
A
C
B D
E
F
In figure AD, BE and CF are medians of ∆ABC. ∴ 3 × (AB2 + BC2 + CA2) = 4 × (AD2 + BE2 + CF2)17. In the figure given below, if P is any point inside the
rectangle ABCD, then PA2 + PC2 = PB2 + PD2
A B
D
P
C
18. Diagonals of a trapezium divide each other in the ratio of the parallel sides of the trapezium. In trapezium ABCD, AB || DC
A B
D C
O
∴ AO BO AB
OC OD CD= = .
19. If a trapezium is inscribed inside a circle, then the trapezium is an isosceles trapezium i.e. its non-parallel sides are equal.
A B
D C
In the figure, ABCD is a trapezium in which AB || CD ∴ AD = BC20. Area of triangles on the same base and lie between the same
pair of parallel lines are equal.A D E
B C
l
m
In the figure, ∆ABC, ∆DBC and ∆EBC are on the same base BC and lie between the same pair of parallel lines l and m.
∴ area of ∆ABC = area of ∆DBC = area of ∆EBC.
18
21. If a parallelogram and a triangle are on the same base and lie between the same pair of parallel lines, then area of the parallelogram is twice the area of the triangle.
A B
CD
E
In the figure, ABCD a parallelogram and EDC a triangle are on the same base and lie between the same pair of parallel lines AB and CD.
∴ area of parallelogram ABCD = 2 × (area of ∆EDC).22. Concentric circles: Two or more circles in a plane arc said
to be concentric, if they have the same centre.
Concentric circles
23. Intercepts made by three or more parallel lines on two or more lines are in the same ratios.
In the figure three parallel lines AD, BE and CF made intercepts AB, BC and DE, EF on two lines AC and DF respectively.
A D
B E
C F
∴ AB DE
BC EF=
24. (a) In an equilateral triangle centroid, incentre, circumcentre, orthocentre coincide at the same point.
(b) Circumradius = 2 × inradius25. A parallelogram is a rectangle if its diagonals are equal.26. If two chords AB and CD of a circle intersect inside a
circle (or outside a circle when produced) at point E, then AE × EB = CE × ED.
AD
C
B
E
A
D
C
B
E
27. If PB is a secant which intersects the circle at A and B and PT is a tangent at T to the circle, then
A
P
T
B
M N
PA × PB = PT2 = PM × PN28. Angles in the alternate segment:
P A
B
Q
C
In the figure, AB is a chord of a circle. PQ is a tangent at an end point A of the chord to the circle. C is any point on arc AB.
∠BAQ and ∠ACB are angles in the alternate segments Angles in the alternate segments of a circle are equal i.e. ∠BAQ = ∠ACBIllustration 18: In the given figure, chords AB and CD of a circle intersect externally at P. If AB = 6 cm, CD = 3 cm and PD = 5 cm, then PB = ?
(a) 5 cm (b) 6.25 cm (c) 6 cm (d) 4 cmSolution: (d) PA × PB = PC × PD (According to property
of circle) ⇒ (x + 6) × x = 8 × 5 ⇒ x2 + 6x – 40 = 0 ⇒ (x + 10) (x – 4) = 0 ⇒ x = 4 ∴ PB = 4 cmIllustration 19: In the given figure, PAB is a secant and PT is a tangent to the circle from P. If PT = 5 cm, PA = 4 cm and AB = x cm, then x is equal to
(a) 2.5 cm (b) 2.6 cm (c) 2.25 cm (d) 2.75 cm
19
Solution: (c) PA × PB = PT2 ⇒ 4 × (4 + x) = 25
⇒ 4 + x = 254
= 6.25 ⇒ x = 2.25 cm
Illustration 20: Two equal circles pass through each other's centre. If the radius of each circle is 5 cm, what is the length of the common chord?
(a) 5 3 (b) 10 3 (c) 5 3
2 (d) 5
Solution: (a)
O OM
A
B
D OAO' is equilateral triangle and AM is its altitude.
∴ AM = OA 3
2
⇒ AM = 5 3
2 ∴ The length of common chord, AB = 2 × AM
= 5 3
2 5 32
× = cm
Illustration 21: The radius of a circle is 13 cm and XY is a chord which is at a distance of 12 cm from the centre. The length of the chord is(a) 12 cm (b) 10 cm (c) 20 cm (d) 15 cmSolution: (b) From figure,
O
X YM
1213
XM = 2 213 12−
= 169 144− = 5
∴ Length of the chord = 2 × XM = 2 × 5 = 10 cmIllustration 22: Two circles of radii 10 cm and 8 cm. intersect and length of the common chord is 12 cm. Find the distance between their centres.(a) 13.8 cm (b) 13.29 cm (c) 13.2 cm (d) 12.19 cm
Solution: (b) Here, OP = 10 cm; O′P = 8 cm
P
810
O O�
Q
L
PQ = 12 cm
∴ PL = 1/2 PQ ⇒ PL = 1
122× ⇒ PL = 6 cm
In rt. ∆OLP, OP2 = OL2 + LP2
(using Pythagoras theorem) ⇒ (10)2 = OL2 + (6)2 ⇒ OL2 = 64; OL = 8 In∆O′LP, (O′L)2 = O′P2 – LP2 = 64 – 36 = 28 O′L2 = 28 ⇒ O′L = 28 O′L = 5.29 cm ∴ OO′ = OL + O′L = 8 + 5.29 OO′ = 13.29 cm
Locus
The locus of a point is the path traced out by a moving point under given geometrical conditions. Alternatively, the locus is the set of all those points which satisfy the given geometrical conditions.The Locus of a Point in Different Conditions(i) The locus of a point which is equidistant from two fixed
points is the perpendicular bisector of the line segment joining the two fixed points.
A B
Required
Locus
(ii) The locus of a point which is equidistant from two intersecting straight lines is a pair of straight lines which bisect the angles between the two given lines.
Required
Locus
Required Locus
20
(iii) The locus of a point equidistant from two given parallel straight lines is a straight line parallel to the given straight lines and midway between them.
Required Locus
(iv) The locus of a point which is equidistant from a fixed point in a plane is a circle.
(v) The locus of a point, which is at a given distance from a given straight line, is a pair of parallel straight lines either side to the given line at a given distance from it.
Required Locus
d
dRequired Locus
Here d is the given distance.(vi) The locus of the centre of a wheel moving on a straight
horizontal road, is a straight line parallel to the road and at a height equal to the radius of the wheel.
Required Locus
(vii) The locus of mid-points of all parallel chords of a circle, is the diameter of the circle which is perpendicular to the given parallel chords.
Required Locus
(viii) The locus of a point which is equidistant from two concentric circles is the circumference of the circle concentric with the given circles and midway between them.
Required Locus
(ix) If A and B are two fixed points, then the locus of a point P such that ∠APB = 90°, is the circle with AB as diameter.
A B
P
RequiredLocus
(x) The locus of midpoints of all equal chords of a circle is the circumference of the circle concentric with the given circle and radius equal to the distance of equal chords from the centre of the given circle.
RequiredLocus
Distan
ce
of the c
hord
EXAM APPROACHUsing certain results directly, while solving a problem can reduce our calculations a lot.1. In an equilateral triangle, from any point inside it, if
perpendiculars are drawn then -
P3
P2
a a
a
P1
P1 + P2 + P3 = 3
2( )a
2. In a ∆ ABC, if H is orthocentre then -A
BP
C
Q
H
R
∠ ABC = ∠ CHP and ∠ ACB = ∠ BHP also ∠ RPQ = 180° – 2(∠ A) ∠ RQP = 180° – 2(∠ B) ∠PRQ = 180° – 2(∠ C)
21
3. If two triangles ABC and PQR are similar then –
AB
PQ
AC
PR
BC
QR
Perimeter of ABC
Perimeter of PQR= = =
∆
∆
4. In right angled ∆ ACB, if CP is perpendicular to hypotenuse then -
b
a
Pc
p
A
CB
1 1 12 2 2p a b
= +
also AP
PB
b
a=
2
2
also APb
cPB
a
c= =
2 2;
5. Only in an right angled triangle, mid point of hypotenuse is equidistant from all three vertices. (which is also called as circumcentre).
A
B
M
C
BM = AM = CM = R
26. In an right angled triangle with sides “a,b,c” and in radius
“r”A
CB
c
r
a
b
ra c b
=+ −
2
and Rb
=2
D r Ra c
+ =+
2
7. Side of a square with maximum possible length whose one vertex lies at right angle of a triangle, is -
A
c
Ba
bx
C
1 1 1
x a b= +
8. Side of a square, with maximum possible length, whose one side lies on hypotenuse is -
A
c
Ba
b
C
y
yabc
a b ab=
+ +2 2
9. Case of Internal and external angle bisectors -A
B C
B/2
B/2C/2
C/2
I
∠ BIC = 902
° +∠A
A
B C
O
∠ BOC = 902
° −∠A
22
10. Sum of all exterior angles of any convex polygon is 360°
and in case of regular polygon of n sides, each external angle
is given by 360ϒn
11. Sum of internal angle and exteternal angle is always 180°.
12. In a ∆ ABC with sides”a,b,c” and medians” ma, mb, mc”
from vertices A,B and C respectively then -
m b c aa = + −1
22 2 2 2( )
and 4(ma2 + mb
2 + mc2) = 3(a2 + b2 + c2)
13. In given condition, if AB || PQ || CD then -
A
B
Q
D
C
P
x
a b
1 1 1
x a b= +
14. For an equilateral ∆ of side a,
r (in radius) = a
2 3 and R (Circumradius) =
a
3
or 2r = R.
15. If M is any interior point of rectangle ABCD, then
AM2 + CM2 = BM2 + DM2
A
D
B
C
M
16. In radius (r) of circle inscribed in any polygon can be
calculated by formula
rs
=∆
; where “∆” is area of polygon and “s” is semi
perimeter of polygon
17. When 2 circles of equal radius passes through each others centre, then they form rhombus with angles 60° and 120°.
B
D
A C
r r ABCD is rhombus ∠ ABC = 60° ∠ BCD = 120°18. In an equilateral triangle of area “6a”, medians divide it as-
F E
CB D
a
a
a
a
a
a
O
A
i.e. area of each individual small triangle will be “a”19. Circum radius of a triangle with sides a,b,c is calculated as -
rabc
=
4∆
; where is ∆ is area of triangle
20.
C1 C2
C3
r1
r2
r3
for such arrangement r2 = r r1 3.
21. If four sides of a quadrilateral ABCD are tangential to a circle then -
B
C
A
D
AB + CD = BC + AD i.e. sum of opposite pair of sides is equal.
23
22. If M, N are mid points of non - parallel sides of trapezium PQRS, with PQ || RS, then -
2MN = PQ + RSP Q
N
R
M
S 23. If M,N are mid points of diagonals of trapezium PQRS with
PQ || RS then
MN = RS PQ−
2
M N
Q
RS
P
24. Area of the triangle formed by joining mid points of the sides of bigger triangle, will always be one-fourth of the area of bigger triangle.
QR
P
A
B C
or ∆ PQR = 1
4 (or ∆ ABC)
25. In rhombus, if one diagonal is equal to its side, then that rhombus is of 60°, 120°, 60°, 120° angles and ratio of its diagonal is 1 3:
B
C
A
D
120°
120°
60°
60°
BD : AC = 1 3:
24
1. In triangle ABC, angle B is a right angle. If (AC) is 6 cm, and D is the mid-point of side AC. The length of BD is
C
D
A
B
(a) 4 cm (b) 6cm (c) 3 cm (d) 3.5 cm2. AB ⊥ BC and BD ⊥ AC. And CE bisects the angle C.
∠A = 30º. The, what is ∠CED.A
B C
DE
30°
(a) 30° (b) 60° (c) 45° (d) 65°3. If two parallel lines are cut by two distinct transversals, then
the quadrilateral formed by these four lines will always be a :
(a) parallelogram (b) rhombus (c) square (d) trapezium4. In the adjoining the figure, points A, B, C and D lie on the
circle. AD = 24 and BC = 12. What is the ratio of the area of the triangle CBE to that of the triangle ADE
B
D
C A
E
(a) 1 : 4 (b) 1: 2 (c) 1 : 3 (d) Insufficient data5. In ∆ABC, AD is the bisector of ∠A if AC = 4.2 cm., DC =
6 cm., BC = 10 cm., find AB. (a) 2.8 cm (b) 2.7 cm (c) 3.4 cm (d) 2.6 cm6. How many sides a regular polygon has with its sum of
interior angles eight times its sum of exterior angles? (a) 16 (b) 24 (c) 18 (d) 307. A point P is 26 cm away form the centre O of a circle and
the length PT of the tangent draw from P to the circle is 10cm. Find radius of the circle
(a) 24 cm (b) 32 cm (c) 22 cm (d) 42 cm
8. In the given figure, AB || CD, ∠BAE = 45º, ∠DCE = 50º and ∠CED = x, then find the value of x.
45º
xº
B
A
D
C
50º
E
(a) 85º (b) 95º (c) 60º (d) 20º9. Given the adjoining figure. Find a, b, c
C
D
BA
50º
b
ac36º 70º
(a) 74º, 106º, 20º (b) 90º, 20º, 24º (c) 60º, 30º, 24º (d) 106º, 24º, 74º 10. In the figure given below, AB is a diametre of the semicircle
APQB, centre O, ∠POQ = 48º cuts BP at X, calculate ∠AXP.
X
QP
A BO
48º
(a) 50º (b) 55º (c) 66º (d) 40º
11. In the figure , if NT
AB =
9
5 and if MB = 10, find MN.
M
B
A
N T065085
095
(a) 5 (b) 4 (c) 28 (d) 1812. In the given figure, AB | | CD , ∠ABO = 40° and
∠CDO = 30°. If ∠DOB = x, then find the value of x.
LEVEL – I
EXERCISE
25
A B
C D
OE E'
40o
30o
(a) 10° (b) 70° (c) 110° (d) 20°13. M and N are points on the sides PQ and PR respectively
of a ∆ PQR. For each of the following cases state whether MN is parallel to QR
A. PM = 4, QM = 4.5, PN = 4, NR = 4.5 B. PQ = 1.28, PR = 2.56, PM = 0.16, PN = 0.32 (a) only in case A (b) only in case B (c) both in the case A & B (d) None of these14. The perimeters of two similar ∆s ABC and PQR are
respectively 36 cm and 24 cm. If PQ = 10 cm, then AB is equal to
(a) 5 cm (b) 10 cm (c) 15 cm (d) 9 cm15. In the triangle ABC, AD bisects ∠BAC, BC = 6.4, AB = 5
and AC = 3, then the length of BD is equal to (a) 3.5 (b) 5.5 (c) 3.2 (d) 416. In the given figure, m ∠EDC = 54°, m∠DCA = 40°. Find
x, y and z respectively.
D
E
XY
54°
y°z°
x°
40°
A BC
(a) 20°, 27°, 86° (b) 40°, 54°, 86° (c) 20°, 27°, 43° (d) 40°, 54°, 43°17. In the adjoining figure, ABCD is a cyclic quadrilateral. If AB
is a diameter, BC = CD and ∠ABD = 40°, find the measure of ∠DBC.
CD
BA
Xd
40°a
X
(a) 65 (b) 25 (c) 45 (d) 6018. In the cyclic quadrilateral ABCD, ∠BCD = 120°, m (arc
DZC) = 70°, find ∠DAB and m(arc CXB).
D
A
B
C
Z X
(a) 60°, 70° (b) 60°, 40° (c) 60°, 50° (d) 60°, 60°19. If two tangents inclined at an angle 60° are drawn to a circle
of radius 3 cm, then length of each tangent is equal to
Q
P
O
30°
60°T
3
(a) 3
22 cm (b) 6 cm (c) 3 cm (d) 3 3 cm
20. In the given fig. PQ is a chord of a circle and PT is the tangent at P such that ∠QPT = 60°. Then ∠PRQ is equal to
R
O
P
Q
T
(a) 135° (b) 150° (c) 120° (d) 110°21. If four sides of a quadrilateral ABCD are tangential to a
circle, then. (a) AC + AD = BD + CD (b) AD + BC = AB + CD (c) AB + CD = AC + BC (d) AC + AD = BC + DB22. In the given figure, AB || CD, ∠ALC = 60°, EC is the bisector
of ∠LCD and EF || AB. Then, find the measure of ∠CEF.
CD
A L B
E F60
o
(a) 80° (b) 130° (c) 120° (d) 150°23. D, E, F are midpoints of BC, CA and AB respectively. G, H,
I are midpoints of FE, FD, DE respectively. Areas of ∆DHI and ∆AFE are in the ratio
26A
H
F E
I
B D C
G
(a) 1 : 3 (b) 1 : 4 (c) 1 : 9 (d) 1 : 1624. John wishes to determine the distance between two objects A
and B, but there is an obstacle between the two objects which prevents him from making a direct measurement. He designed an ingenious way to overcome this difficulty. First, he fixes a pole at convenient point O so that from O, both ends are visible. Then he fixes another pole at a point D on the line AO (produced) such that AO = DO. In a similar way, he fixes a third pole at a point C on the line BO (produced) such that BO = CO. Then he measures CD and finds that CD = 170 cm. Find the distance between the objects A and B.
(a) 90 cm (b) 170 cm (c) 140 cm (d) 150 cm25. In the adjoining figure, ABCD is a cyclic quadrilateral. Then
r + s is equal to R
Dc1
r
d
bA
B
c
S
s
C
(a) 180° (b) 2c (c) 180° + 2c (d) 180° – 2c26. P is the centre of the circle m ∠ACB = 65°. Find m (arc AXB )
B
C
A
P X 65°
(a) 105° (b) 115° (c) 65° (d) 245°27. The centroid, circumcenter, orthocenter in a triangle– (a) are always coincident. (b) are always collinear. (c) are always the inside the triangular area. (d) always coincide in a equilateral triangle and otherwise
collinear.
28. In the given figure AB || CD and AC || BD. If ∠EAC = 40°, ∠FDG = 55°, ∠HAB = x°, then find the value of x.
A B
CE
x°
K
H
D
G
F
(a) 85° (b) 75° (c) 65° (d) 55°29. Which one of the following cannot be the ratio of angles in
a right angled triangle? (a) 1 : 2 : 3 (b) 1 : 1 : 2 (c) 1 : 3 : 6 (d) None of these30. In the adjoining figure ABCD is a rectangle and DF = CF
also, AE = 3BE. What is the value of ∠EOF, if ∠DFO = 28° and ∠AEO = 42°?
O
A B
CD
E
F
28°
42°
(a) 14° (b) 42° (c) 70° (d) 90°31. Each interior angle of a regular polygon exceeds its exterior
angle by 132°. How many sides does the polygon have? (a) 9 (b) 15 (c) 12 (d) None of these32. In a triangle ABC, O is the centre of incircle PQR, ∠BAC
= 65°, ∠BCA = 75°, find ∠ROQ:
65° 75°
A
B
CP
QR
O
(a) 80° (b) 120° (c) 140° (d) can’t be determined
27
33. ABC and CDE are right angled triangle. ∠ABC = ∠CDE = 90°. D lies on AC and E lies on BC. AB = 24 cm, BC = 60 cm. If DE = 10 cm, then CD is:
A B
C
D
E
(a) 28 cm (b) 35 cm (c) 25 cm (d) can’t be determined34. The largest angle of a triangle of sides 7 cm, 5 cm and 3 cm
is (a) 45° (b) 60° (c) 90° (d) 120°35. ABCD is a paralellogram in which ∠B = 70°. Find the
number of points X in the plane of the parallelogram such that it is equidistant from its vertices.
(a) zero (b) one (c) two (d) three36. PQRS is trapezium, in which PQ is parallel to RS, and PQ
= 3 (RS). The diagonal of the trapezium intersect each other at X, then the ratio of ∆PXQ and ∆RXS is
(a) 6 : 1 (b) 3 : 1 (c) 9 : 1 (d) 7 : 137. C is the midpoint of DE. Area of parallelogram ABCD = 16
sq. cm. Find the area of ∆BCE. (a) 8 sq.cm (b) 16 sq. cm (c) 32 sq. cm (d) 24 sq. cm38. In the figure (not drawn to scale) given below, if AD = CD
= BC, and ∠BCE = 96°, how much is ∠DBC?
AB
C
D
E
�96
(a) 32° (b) 84° (c) 64° (d) Cannot be determined39. In a trapezium ABCD, AB || CD and AD = BC. If P is point
of intersection of diagonals AC and BD, then all of the following is wrong except.
(a) PA.PB = PC.PD (b) PA.PC = PB.PD (c) PA.AB = PD.DC (d) PA.PD = AB.DC
40. ABCD is a quadrilateral in which diagonal BD = 64 cm, AL ⊥ BD and CM ⊥ BD, such that AL = 13.2 cm and CM = 16.8 cm. The area of the quadrilateral ABCD in square centimetres is
(a) 537.6 (b) 960.0 (c) 422.4 (d) 690.041. The area of the shaded region in the following graph is
(2, – 4)
(1, 0)
(2, 0)
(1, –2)
y
x
(a) 2 sq. units (b) 4 sq. units (c) 6 sq. units (d) 8 sq. units42. In the figure below, if AB || CD and CE ⊥ ED, then the value
of x is
A B
C D
E
37°90°
x°
(a) 37 (b) 45 (c) 53 (d) 6343. PA and PB are two tangents drawn from an external point
P to a circle with centre O where the points A and B are the points of contact. The quadrilateral OAPB must be
(a) a square (b) concylic (c) a rectangle (d) a rhombus44. In the following figure, if OA = 10 and AC = 16, then OB
must be
O
A B C
(a) 3 (b) 4 (c) 5 (d) 645. Triangle PQR circumscribes a circle with centre O and
radius r cm such that ∠PQR = 90°. If PQ = 3 cm, QR= 4 cm, then the value of r is:
(a) 2 (b) 1.5 (c) 2.5 (d) 1
28
46. In the following figure. AB be diameter of a circle whose centre is O. If ∠AOE = 150°. ∠DAO = 51° then the measure of ∠CBE is:
OA
B C
DE
150°51°
(a) 115° (b) 110° (c) 105° (d) 120° 47. The areas of two similar triangles ABC and DEF are 20 cm2
and 45 cm2 respectively. If AB = 5 cm. then DE is equal to:
(a) 6.5 cm (b) 7.5 cm (c) 8.5 cm (d) 5.5 cm 48. In a triangle ABC, BC is produced to D so that CD = AC.
If ∠BAD = 111° and ∠ACB = 80°, then the measure of ∠ABC is:
(a) 31° (b) 33° (c) 35° (d) 29° 49. In ∆ABC, ∠A + ∠B = 145° and ∠C + 2∠B = 180°. State
which one of the following relations is true? (a) CA = AB (b) CA < AB (c) BC > AB (d) CA > AB
50. In a ∆ ABC, AB
AC
BD
DC= , ∠B = 70° and ∠C = 50°, then
∠BAD = [SSC-Sub. Ins.-2014] (a) 60° (b) 20° (c) 30° (d) 50°51. In a ∆ ABC, AD, BE and CF are three medians. The
perimeter of ∆ABC is always [SSC-Sub. Ins.-2014] (a) equal to AD BE CF+ +( )
(b) greater than AD BE CF+ +( )
(c) less than AD BE CF+ +( )
(d) None of these52. Two circles with radii 25 cm and 9 cm touch each other
externally. The length of the direct common tangent is [SSC-Sub. Ins.-2014]
(a) 34 cm (b) 30 cm (c) 36 cm (d) 32 cm53. If AB = 5 cm, AC = 12 and AB ⊥ AC, then the radius of the
circumcircle of ∆ABC is [SSC-Sub. Ins.-2014] (a) 6.5 cm (b) 6 cm (c) 5 cm (d) 7 cm54. ABC is a right angled triangle, right angled at C and p is
the length of the perpendicular from C on AB. If a, b and c are the lengths of the sides BC, CA and AB respectively, then
(a) 1 1 12 2 2p b a
= − (b) 1 1 12 2 2p a b
= +
(c) 1 1 12 2 2p a b
+ = (d) 1 1 12 2 2p a b
= −
55. From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to diameter of the circle, then ∠APB is
(a) 60° (b) 45° (c) 90° (d) 30°56. A chord 12 cm long is drawn in a circle of diameter 20 cm.
The distance of the chord from the centre is (a) 16 cm (b) 8 cm (c) 6 cm (d) 10 cm57. If in ∆ ABC, ∠ABC = 5∠ACB and ∠ BAC = 3 ∠ACB,
then ∠ABC = (a) 120° (b) 130° (c) 80° (d) 100°58. The perpendiculars, drawn from the vertices to the opposite
sides of a triangle, meet at the point whose name is (a) orthocentre (b) incentre (c) circumcentre (d) centroid59. In ∆ABC, D and E are two points on the sides AB and
AC respectively so that DE||BC and AD
BD=
2
3. Then
the area of trapezium DECB
the area of ADE∆ is equal to [SSC 10+2-2014]
(a) 5
9 (b) 21
25 (c) 1
4
5 (d) 5
1
4
60. One of the angles of a parallelogram is 45°. What will be the sum of the larger angle and twice the smaller angle of the parallelogram ?
(a) 228° (b) 224° (c) 225° (d) 222° (e) None of these61. In the adjoining figure, chord AD and BC of a circle are
produced to meet at P, PA = 10 cm, PB = 8 cm, PC = 5 cm, AC = 6 cm. Find PD.
B
A
P
D
C
K
(a) 3 (b) 5 (c) 6 (d) 462. In the figure below, which of the following is the relationship
between ‘x’ and ‘y’ if the equal circles shown are tangents to each other and to the sides of the rectangle
y
x
2
(a) x = 1
4y (b) x = (c) x = Dy2 (d) x = 2Dy
29
63. In the given figure, ∠ABC and ∠DEF are two angles such that BA ⊥ ED and EF ⊥ BC, then find value of ∠ABC + ∠DEF.
DA
P
B
E
C
F
(a) 120º (b) 180º (c) 150º (d) 210º
64. Find the perimeter of the given figure.
6 cm
10
cm
o
(a) (32 + 3π) cm (b) (36 + 6π) cm (c) (46 + 3π) cm (d) (26 + 3π) cm
65. The length of tangent (upto the point of contact) drawn from an external point P to a circle of radius 5 cm is 12 cm. The distance of P from the centre of the circle is
(a) 11 cm (b) 12 cm (c) 13 cm (d) 14 cm
66. Two circles of equal radii touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. The relation of TQ and TR is
(a) TQ < TR (b) TQ > TR (c) TQ = 2TR (d) TQ = TR
67. When two circles touch externally, the number of common tangents are
(a) 4 (b) 3 (c) 2 (d) 1
68. D and E are the mid-points of AB and AC of ∆ABC. If ∠A = 80°, ∠C = 35°, then ∠EDB is equal to
(a) 100° (b) 115° (c) 120° (d) 125°
69. Let C1 and C2 be the inscribed and circumscribed circles of a triangle with sides 3 cm, 4 cm and 5 cm then area of C1 to area of C2 is [SSC CGL 1st Sit. 2015]
(a) 9
16 (b) 9
25
(c) 4
25 (d) 16
2570. If the three angles of a triangle are :
6x 2x
(x 15 ), 6 and 305 3
+ ° + ° + ° then the triangle is :
[SSC CGL 1st Sit. 2015] (a) scalene (b) isosceles (c) right angled (d) equilateral
71. The perimeter of an isosceles triangle is 32 cm and each of the equal sides is 5/6 times of the base. What is the area (in cm2) of the triangle? [SSC CGL 2017]
(a) 39 (b) 48 (c) 57 (d) 64
72. If D is the number of degrees and R is the number of radians in an angle D, then which one of the following is correct ?
[2017-I] (a) DD = 180R (b) DD = 90R (c) DR = 180D (d) DR = 90D
73. What angle does the hour hand of a clock describe in 10 minutes of time? [2018-I]
(a) 1° (b) 5° (c) 6° (d) 10°
74. The locus of a point equidistant from two intersecting lines is [2018-I]
(a) A straight line (b) A circle (c) A pair of straight lines (d) None of the above
75. In the figure given below, PQR is a non-isosceles right-angled triangle, right angled at Q. If LM and QT are parallel and QT = PT, then what is ∠RLM equal to ? [2017-I]
P
T
M
RLQ
(a) ∠PQT (b) ∠LRM (c) ∠ RML (d) ∠QPT
30
76. In triangle ABC, ∠C = 90° and CD is the perpendicular from C to AB.
If (CD)–2 = (BC)–2 + (CA)–2, then which one of the following is correct? [2017-II]
(a) BC . CD = AB . CA (b) AB . BC = CD . CA (c) CA2 + CB2 = 2 (AD2 + CD2) (d) AB . CD = BC . CA77. ABC is a triangle right angled at C with BC = a and
AC = b. If p is the length of the perpendicular from C on AB, then which one of the following is correct? [2018-I]
(a) a2 b2 = p2 (a2 + b2) (b) a2 b2 = p2 (b2 – a2) (c) 2a2 b2 = p2 (a2 + b2) (d) a2 b2 = 2p2 (a2 + b2)78. If each interior angle of a regular polygon is 135°, then the
number of diagonals of the polygon is equal to [2015-I] (a) 54 (b) 48 (c) 20 (d) 1879. A circle of radius r is inscribed in a regular polygon with
n sides (the circle touches all sides of the polygon). If the perimeter of the polygon is p, then the area of the polygon is [2015-II]
(a) (p + n) r (b) (2p – n) r
(c) pr
2 (d) None of the above80. If a point O in the interior of a rectangle ABCD is joined
with each of the vertices A, B, C and D, then OB2 + OD2 will be equal to [2017-II]
(a) 2OC2 + OA2 (b) OC2 – OA2
(c) OC2 + OA2 (d) OC2 + 2OA2
81. A closed polygon has six sides and one of its angles is 30° greater than each of the other five equal angles. What is the value of one of the equal angles? [2017-II]
(a) 55° (b) 115° (c) 150° (d) 175°82. A field is divided into four regions as shown in the given
figure. What is the area of the field in square metres ?[2017-I]
2m 2m
2m
2m
3m
3m
3m
1m
(a) 3
6 54
+ (b) 3
5 32
+
(c) 3
9 154
+ (d) 7 2 2+
83. In the figure given below, D is the diameter of each circle. What is the diameter of the shaded circle ? [2017-I]
(a) D( 2 1)− (b) D( 2 1)+
(c) D( 2 2)+ (d) D(2 2)−
84. The locus of the mid-points of the radii of length 16 cm of a circle is [2018-1]
(a) A concentric circle of radius 8 cm (b) A concentric circle of radius 16 cm (c) The diameter of the circle (d) A straight line passing through the centre of the circle
85. In the figure given below, XA and XB are two tangents to a circle. If ∠AXB = 50° and AC is parallel to XB, then what is ∠ACB equal to? [2018-1]
A
B
C 50° X
(a) 70° (b) 65° (c) 60° (d) 55°
86. In the figure given below, SPT is a tangent to the circle at P and O is the centre of the circle. If ∠QPT = a, then what is ∠POQ equal to? [2018-1]
P
Q
S T
O
a
(a) D (b) 2D
(c) 90° – D (d) 180° – 2D
31
87. What are the respective value of x, y and z in the given rectangle ABCD ?
E
A B
CD
xy
z
9
16
(a) 15, 12, 20 (b) 12, 15, 20 (c) 8, 10, 12 (d) None of these88. Find ∠BOA.
O 50°
C
B A30°
F
(a) 100° (b) 150° (c) 80° (d) Indeterminate89. ABCDEF is a regular hexagon of side 2 feet. The area, in
square feet of the rectangle BCEF is
(a) 4 (b) 4 3
(c) 8 (d) 4 4 3+
90. In ∆ABC, ∠B = 60°, ∠C = 40°. If AD bisects ∠BAC and AE ⊥ BC, then ∠EAD is
(a) 40° (b) 80° (c) 10° (d) 20°91. G is the centroid of ∆ABC. If AG = BC, then ∠BGC is (a) 60° (b) 120° (c) 90° (d) 30°
92. Here XY has been divided into 5 congruent segments and semicircles have been drawn. But suppose XY were divided into millions of congruent segments and semicircles were drawn, what would the sum of the lengths of the arcs be?
X Y
(a) 2yx (b) 5xy (c) xy (d) None of these93. In the adjoining figure the circles touches the side of the
quadrilateral ABCD. If AB= p, express (AD + BC) in terms of p and q
qp
A
B
D
C
Y W
X
Z
(a) p + q (b) 1
2 p + q
(c) 2 (p – q) (d) 3 (p – q)
94. If ABCD is a square and BCE is an equilateral triangle, what is the measure of the angle DEC?
D C
E
BA
(a) 15° (b) 30° (c) 20° (d) 45°
DIRECTIONS (Qs. 95–97) : Answer the questions on the basis of the information given below.
In the adjoining figure, I and II are circles with centers P and Q respectively. The two circle touch each other and have a common tangent that touches them at points R and S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the ratio 4 : 3. It is also known that the length of PO is 28 cm.
OP Q
R
S
III
95. What is the ratio of the length of PQ to that of QO? (a) 1 : 4 (b) 1 : 3 (c) 3 : 8 (d) 3 : 496. What is the radius of the circle II? (a) 2 cm (b) 3 cm (c) 4 cm (d) 5 cm97. The length of SO is (a) 8 3 cm (b) 10 3 cm
(c) 12 3 cm (d) 14 3 cm
LEVEL – II
32
98. What is the inradius of the incircle shown in the figure?
9cm
40 cm
41 cm
A
BC
(a) 9 cm (b) 4 (c) can’t be determined (d) None of these99. In a circle O is the centre and ∠COD is right angle.
AC = BD and CD is the tangent at P. What is the value of AC + CP, if the radius of the circle is 1 metre?
A
B
C
D
P
O
90°
(a) 105 cm (b) 141.4 cm (c) 138.6 cm (d) can’t be determined100. In the triangle ABC, MN is parallel to AB. Area of trapezium
ABNM is twice the area of triangle CMN. What is ratio of CM : AM ?
A B
C
M N
(a) 1
3 1+
(b) 3 1
2
−
(c) 3 1
2
+ (d) None of these
101. ABC is a triangle in which ∠CAB = 80° and ∠ABC = 50°, AE, BF and CD are the altitudes and O is the orthocentre. What is the value of ∠AOB?
A B
C
D
EF
O
(a) 65° (b) 70° (c) 50° (d) 130°102. In the given diagram O is the centre of the circle and CD
is a tangent. ∠CAB and ∠ACD are supplementary to each other ∠OAC = 30°. Find the value of ∠OCB:
AB
C D
O
(a) 30° (b) 20° (c) 60° (d) None of these
103. The sides of a triangle are in the ratio of 1
2
1
3
1
4: : . If the
perimeter is 52 cm, then the length of the smallest side is (a) 9 cm (b) 10 cm (c) 11 cm (d) 12 cm104. The ratio of the area of a square to that of the square drawn
on its diagonal is (a) 1 : 4 (b) 2 : 1 (c) 1 : 2 (d) 1 : 3105. PQ is a tangential to circles with centers A and B at P and Q
respectively. If AB = 10 cm. and PQ = 8 cm, find the radius of the bigger circle. Given that area of triangle APO is four times the area of triangle OQB –
A B
P
O
Q
(a) 2 cm (b) 4 cm (c) 6 cm (d) 8cm106. Two circles touch each other internally. Their radii are
2 cm and 3 cm. The biggest chord of the outer circle which is outside the inner circle is of length
(a) 2 2 cm (b) 3 2 cm
(c) 2 3 cm (d) 4 2 cm
33
107. The sum of the interior angles of a polygon is 1620°. The number of sides of the polygon are :
(a) 9 (b) 11 (c) 15 (d) 12
108. In ∆ABC, DE | | BC and AD
DB=
3
5. If AC = 5.6 cm,
find AE.
B C
ED
A
(a) 2.1 cm (b) 3.1 cm (c) 1.2 cm (d) 2.3 cm109. If one of the diagonals of a rhombus is equal to its side, then
the diagonals of the rhomhus are in the ratio: (a) 3 1: (b) 2 1: (c) 3 : 1 (d) 2 : 1110. If ABCD is a square and BCE is an equilateral triangle, what
is the measure of the angle DEC?
D C
E
BA
(a) 15° (b) 30° (c) 20° (d) 45°111. ABCD is a square, F is the mid-point of AB and E is a point
on BC such that BE is one-third of BC. If area of ∆FBE = 108 m2, then the length of AC is :
(a) 63 m (b) 36 2 m (c) 63 2 m (d) 72 2 m
112. Arc ADC is a semicircle and DB ⊥ AC. If AB = 9 and BC = 4, find DB. (a) 6 (b) 8 (c) 10 (d) 12113. In the given figure given below, E is the mid-point of AB
and F is the midpoint of AD. if the area of FAEC is 13, what is the area of ABCD?
A
F
E
D
B
C
(a) 19.5 (b) 26 (c) 39 (d) None of these114. In the given figure, AB is chord of the circle with centre O,
BT is tangent to the circle. The values of x and y are
B
T A
O
y
P
32ºX
(a) 52º, 52º (b) 58º, 52º (c) 58º, 58º (d) 60º, 64º115. The distance between two parallel chords of length
8 cm each in a circle of diameter 10 cm is (a) 6 cm (b) 7 cm (c) 8 cm (d) 5.5 cm116. The internal bisectors of the angles B and C of a triangle
ABC meet at O. Then find the measure of ∠BOC.
(a) 90° – 2
∠A (b) 180° – 2
∠A
(c) 90° + 2
∠A (d) 180° + 2
∠A
117. In a ∆ABC, angle C is 68°, the perpendicular bisector of AB at R meets BC at P. If ∠PAC = 42° then ∠ABC is equal to
(a) 45° (b) 42° (c) 35° (d) 34°118. A chord of length 14 cm is at a distance of 6 cm from the
centre of a circle. Find the length of another chord at a distance of 2 cm from the centre of the circle.
(a) 18 cm (b) 16 cm (c) 10 cm (d) 12 cm119. In the adjoining figure x is a point on diameter
AB of the circle with centre o, such that AX = 9 cm, XB = 5 cm. Find the radius of the circle (centre Y) which touches the diameter at X and touches the circle, centre O, internally at Z.
A B
Z
X2 5O
Y
r
(a) 33
14cm (b) 3 1
14 cm
(c) 1 1
14cm. (d) 2 3
14 cm.
120. In ∆ABC, AB = AC = 8, PR and PQ are parallel to lines AC and AB respectively. P is the midpoint of BC. Find the perimeter of PRAQ.
(a) 16 (b) 18 (c) 20 (d) 12
34
121. The height of the hexagon whose side is aB
CF
A
DE
O
(a) 3 3
2a (b) 3 3
4 (c) 3 a (d) None of these
122. In ∆ ABC, AB = 8, AC = 6, Altitude AD = 4.8 AE is the diameter of the circumcircle. Find the circumradius.
C
A
B
E
D
(a) 5 (b) 10 (c) 15 (d) Cannot be determined 123. The length of a ladder is exactly equal to the height of the
wall it is resting against. If lower end of the ladder is kept on a stool of height 3 m and the stool is kept 9 m away from the wall the upper end of the ladder coincides with the tip of the wall. Then, the height of the wall is
(a) 12 m. (b) 15 m. (c) 18 m. (d) 11 m.124. Three circles, each of radius 20 and centres at P, Q, R.
further, AB = 5, CD = 10 and EF = 12. What is the perimeter of the triangle PQR?
P Q
R
A
B C
D
EF
(a) 120 (b) 66 (c) 93 (d) 87125. In the diagram given below, ∠ABD = ∠CDB = ∠PQD
= 90° If AB : CD = 3:1 the ratio of CD : PQ isA
B Q D
C
P
(a) 1 : 0.69 (b) 1 : 0.75 (c) 1 : 072 (d) None of these
126. What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm?
(a) 1 or 7 (b) 2 or 14 (c) 3 or 21 (d) 4 or 28127. In the adjoining figure O is the centre of the circle. The
radius OP bisects a rectangle ABCD, at right angle. DM = NC = 2 cm and AR = SB = 1 cm and KS = 4 cm and OP = 5 cm. What is the area of the rectangle?
A R K S B
D M L N C
P
O
(a) 8 cm2 (b) 10 cm2
(c) 12 cm2 (d) None of these128. There are two circles each with radius 5 cm. Tangent AB is
26 cm. The length of tangent CD is:
A B
C
D
(a) 15 cm (b) 21 cm (c) 24 cm (d) can’t be determined129. In the given triangle ABC, the length of sides AB and AC
is same (i.e., b = c) and 60° < A < 90°, then the possible length of BC is
A
bc
B Ca
(a) b < a < 2b (b) 33< <
ca a
(c) 3< <b a b (d) 2< <c a c
130. The angles of a triangle are in the ratio of 4 : 1 : 1. Then the ratio of sine of the largest angle to the smallest angle is
[sin 120° = sin 60°]
(a) 2
3 (b)
1
2 3+ (c) 3
1 (d) 2
1 3+
35
131. What is the sum of all the angles of a 9 pointed star (i.e., ∠1 + ∠2 + ∠3 + .... ∠8 + ∠9):
(a) 909° (b) 900° (c) 720° (d) 540°132. A smaller circle touches internally to a larger circle at A and
passes through the centre of the larger circle. O is the centre of the larger circle and BA, OA are of the diameters of the larger and smaller circles respectively. Chord AC intersects the smaller circle at a point D. If AC = 12 cm, then AD is:
B
C
O
D
A
(a) 4 cm (b) 6 cm (c) 5.6 cm (d) Data insufficient133. Two circles C (O, r) and C (O′, r′) intersect at two points
A and B and O lies on C (O′, r′). A tangent CD is drawn to the circle C (O′, r′) at A. Then
C
D
A
B
O O�
(a) ∠OAC = ∠OAB (b) ∠OAB = ∠AO′O (c) ∠AO′B = ∠AOB (d) ∠OAC = ∠AOB134. ∆ABC has sides AB, AC measuring 2001 and 1002 units
respectively. How many such triangles are possible with all integral sides?
(a) 2001 (b) 1002 (c) 2003 (d) 1004135. One of the angles of a quadrilateral is thrice the smaller
angle of a parallelogram. The respective ratio between the adjacent angles of the parallelogram is 4:5. Remaining three angles of the quadrilateral are in ratio 4 : 11: 9 respectively. What is the sum of the largest and the smallest angles of the quadrilateral?
(a) 255° (b) 260° (c) 265° (d) 270° (e) None of these136. Two circles intersect each other at P and Q. PA and PB are
two diameters. Then ∠AQB is (a) 120° (b) 135° (c) 160° (d) 180°
137. O is the centre of the circle passing through the points A, B and C such that ∠BAO = 30°, ∠BCO = 40° and ∠AOC = x°. What is the value of x ?
(a) 70° (b) 140° (c) 210° (d) 280°138. A and B are centres of the two circles whose radii are 5 cm
and 2 cm respectively. The direct common tangents to the circles meet AB extended at P. Then P divides AB.
(a) externally in the ratio 5 : 2 (b) internally in the ratio 2 : 5 (c) internally in the ratio 5 : 2 (d) externally in the ratio 7 : 2139. A, B, P are three points on a circle having centre O. If ∠OAP
= 25° and ∠OBP = 35°, then the measure of ∠AOB is (a) 120° (b) 60° (c) 75° (d) 150°
140. Side BC of ∆ABC is produced to D. If ∠ACD = 140° and ∠ABC = 3∠BAC, then find ∠A.
(a) 55° (b) 45° (c) 40° (d) 35°141. ABCD is a cyclic quadrilateral, AB is a diameter of the
circle. If ∠ACD = 50°, the value of ∠BAD is (a) 30° (b) 40° (c) 50° (d) 60°142. In the given figure, a smaller circle touches a larger circle
at P and passes through its centre O. PR is a chord of length 34 cm, then what is the length (in cm) of PS?
(SSC CGL 2017)
R
SO
P
Q
(a) 9 (b) 17 (c) 21 (d) 25143. In the given figure , BD passes through centre O, AB = 12
and AC = 8. What is the radius of the circle?(SSC Sub. Ins. 2017)
A
B
O
C D
(a) 3 2 (b) 4 3
(c) 3 5 (d) 3 3
36
144. In the figure given below, PQ is parallel to RS and PR is parallel to QS. If ∠LPR = 35° and ∠ UST = 70°, then what is ∠MPQ equal to ? [2017-I]
T
US
QR
PL
M
(a) 55° (b) 70° (c) 75° (d) 80°
145. If D measured in radians is the angle between the hour hand and the minute hand of a clock when the time is 4 : 36 pm, then which one of the following is correct? [2017-II]
(a) 3 4
5 5
π π< θ < (b)
2 3
5 5
π π< θ <
(c) 2
5 5
π π≤ θ ≤ (d)
7 8
15 15
π π≤ θ ≤
146. Consider the following statements in respect of three straight lines A, B and C on a plane : [2017-II]
1. If A and C are parallel and B and C are parallel; then A and B are parallel.
2. If A is perpendicular to C and B is perpendicular to C; then A and B are parallel.
3. If the acute angle between A and C is equal to the acute angle between B and C; then A and B are parallel.
Which of the above statements are correct? (a) 1, 2 and 3 (b) 1 and 2 only (c) 1 and 3 only (d) 2 and 3 only
147. The sides of a triangle are in geometric progression with common ratio r < 1. If the triangle is a right angled triangle, the square of common ratio is given by [2014-I]
(a) 5 1
2
+ (b) 5 –1
2
(c) 3 1
2
+ (d)
3 –1
2
148. The sides of a triangle are given by 2 2 2 2a b , c a+ + and (b + c) where a, b, c are positive. What is the area of the triangle equal to ? [2016-II]
(a) 2 2 2a b c
2
+ +
(b) 2 2 2 2 2 2a b b c c a
2
+ +
(c) a(b c)
2
+
(d) 2 2 2 2 2 23(a b b c c a )
2
+ +
149. One-fifth of the area of a triangle ABC is cut off by a line DE drawn parallel to BC such that D is on AB and E is on AC. If BC = 10 cm, then what is DE equal to? [2017-II]
(a) 5 cm (b) 2 5 cm
(c) 3 5 cm (d) 4 5 cm
150. A square and an equilateral triangle have the same perimeter.
If the diagonal of the square is 6 2 cm, then what is the area of the triangle? [2018-I]
(a) 212 2 cm (b) 212 3 cm
(c) 216 2 cm (d) 216 3 cm
151. In the equilateral triangle ABC given below, AD = DB and AE = EC. If l is the length of a side of the triangle, then what is the area of the shaded region? [2018-I]
A
B C
D E
(a) 23 3
16
l (b)
23
16
l
(c) 23 3
32
l (d)
23
32
l
152. A square is inscribed in a right-angled triangle with legs p and q, and has a common right angle with the triangle. The diagonal of the square is given by [2016-I]
(a) 2+
pq
p q (b)
2 +pq
p q
(c) 2
+pq
p q (d)
2
+pq
p q
37
153.
CDA
AD is the diameter of a circle with area 707 m2 and AB = BC = CD as shown in the figure above. All curves inside the circle are semicircles with their diameters on AD. What is the cost of levelling the shaded region at the rate of 63 per square metre? [2016-I]
(a) ` 29,700 (b) ` 22,400 (c) ` 14,847 (d) None of the above
154. Two parallel chords of a circle whose diameter is 13 cm are respectively 5 cm and 12 cm in length. If both the chords are on the same side of the diameter, then the distance between these chords is [2017-I]
(a) 5.5 cm (b) 5 cm (c) 3.5 cm (d) 3 cm155. AB and CD are parallel chords of a circle 3 cm apart. If
AB = 4 cm, CD = 10 cm, then what is the radius of the circle? [2017-II]
(a) 7 cm (b) 19 cm (c) 29 cm (d) 14 cm156. In the figure given below, two equal chords cut at point P.
If AB = CD = 10 cm, OC = 13 cm (O is the centre of the circle) and PB = 3 cm, then what is the length of OP?
[2018-1] (a) 5 cm (b) 6 cm (c) 2 29 cm (d) 2 37 cm
38
1. (c) In a right angled ∆, the length of the median is 1
2 the
length of the hypotenuse. Hence 13 cm
2BD AC= =
2. (b) In , 180 90 30 60∆ ∠ = − − = °ABC C
6030
2∴∆ = = °DCE
Again in ,∆DEC 180 90 30 60∆ = − − = °CED
3. (d) The quadrilateral obtained will always be a trapeziam as it has two lines which are always parallel to each other.
D C
BA
4. (a) 24, 12= =AD BC
In &∆ ∆BCE ADE
since ∠ = ∠CBA CDA (Angles by same arc) ∠ = ∠BCE DAE (Angles by same arc) ∠ = ∠BEC DEA (Opp. angles) &∴∠ ∠BCE DAE are similar ∆s with sides in the ratio 1 : 2 DRatio of area = 1:4 ( i.e square of sides)5. (a) A
4.2 cm?
B D C6 cm4 cm
Q AD is angle bisector. D by angle bisector theorem.
4.2
6 4= ⇒ =
AC AB AB
DC BD
2.8 cm∴ =AB
6. (c) Let n be the number of sides of the polygon Now, sum of interior angles = 8 × sum of exterior angles
i.e. (2 – 4) 8 22
π× = × πn
or (2n – 4) = 32 or n = 18
7. (a)
Q
T
r 26-r
M10
Pr
O
26
Q PT2 = PM . PQ ⇒ 100 = (26 – r) (26 + r) ⇒ r = 24
8. (a) ∠EDC = ∠BAD = 45º (alternate angles) ∴ x = ∠DEC = 180º – (50º + 45º) = 85º.
9. (a) a + 36º + 70º = 180º (sum of angles of triangle) ⇒ a = 180º – 36º – 70º = 74º b = 36º + 70º(Ext. angle of triangle ) = 106º c = a – 50º (Ext. angle of triangle ) = 74º – 50º = 24º. ALITER : from figure we see that (a + b) = 180° only option (a) satisfies.
10. (c)
A O B
Q
P
X
48°
= 1
PAQ PAQ2
∠ = ∠∵
PAQ 24⇒∠ = °
APB 90∠ = ° (angle in semicircle) D In DPAX 180° = ∠PAX + ∠APX + ∠PXA ⇒ ∠PXA = 180° – 90° – 24° ⇒ ∠PXA = 66°
11. (d) ∠ MBA = 180º – 95º = 85º ∠ AMB = ∠ TMN ...(Same angles with different names) ∴ ∆MBA – ∆ MNT ......(AA test for similarity)
MB
MN =
AB
NT .......(proportional sides)
10
MN =
5
9
∴ MN = 90
5 = 18.
Hints & Solutions
39
12. (b) Through O draw EOE’ parallel to AB & so to CD.
40°BA
E
C
E’
D30°
O
∴ ∠BOE’ = ∠ABO = 40° (alternate angles) ∠E’OD = ∠CDO = 30° (alternate angles) ∴ ∠BOD = (40° + 30°) = 70°. So, x = 70.13. (c) The triangle PQR is isosceles ⇒ MN || QR by converse of Proportionality Theorem.
P
Q R
M N
(b) Again by Converse of Proportionality theorem, MN || QR.14. (c) Perimeter of ∆ABC = 36 cm. Perimeter of ∆PQR = 24 cm and PQ = 10 cm. We have to find AB. Perimeter of ∆ABC = AB + BC + AC. Perimeter of ∆PQR = PQ + QR + PR. Since ∆ABC ~ ∆PQR. for similar Ds ABC & PQR.
36
24
AB AB BC AC
PQ PQ QR PR
+ +⇒ = =
+ +
⇒36
24=
AB
PQ36 36
10 1524 24
⇒ = × = × =AB PQ cm.
15. (d) AD is the bisector of ∠A. 5
3∴ = =
AB BD
AC DC
A
3
CB
5
D 6.4 cm
3 3 5
5 5
+ +⇒ = ⇒ =
DC DC BD
BD BD
8 5 56.4 4
5 8 8⇒ = ⇒ = × = × =
BCBD BC
BD
16. (b) m ∠ACD = m ∠DEC (Alternate segment) ∴ m ∠DEC = x = 40° ∴ m ∠ECB = m ∠EDC (Alternate segment) ∴ m ∠ECB = y = 54° 54° + x + z = 180° .... (sum of all the angles of a
triangle)
54° + 40° + z = 180° ∴ z = 86° 17. (b) In ∆BCD, BC = CD, ∴ ∠BDC = ∠CBD = x In cyclic quadrilateral ABCD, ∠ABC + ∠ADC = 180° ⇒ 40° + x + 90° + x = 180° ⇒ x = 25°.18. (c) m∠DAB = 180° – 120° = 60° ...(opposite angles of a
cyclic quadrilateral) m(arc BCD) = 2 m∠DAB = 120°.
D
A
B
CZ X
∴ m(arc CXB) = m(arc BCD) – m(arc DZC) = 120° – 70° = 50°.
19. (d) OP
PT = tan30° =
1
3
⇒ PT = 3 OP = 3 3 cm.
20. (c) ∠OPQ = ∠OQP = 30°, i.e., ∠POQ = 120°. Also,
∠PRQ = 1
2 reflex ∠POQ = ( )1
2402
= ° = 120°
21. (b) Since ABCD is a quadrilateral Again AP, AQ are tangents to the circle from the point A.
DS
C
R
BQ
A
P
∴ AP = AQ Similarly BR = BQ CR = CS DP = DS ∴ (AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS) ⇒ AD + BC = AB + CD22. (d) ∠LCD = ∠ALC = 60° (alternate angles)
∠DCE = 1
2∠LCD = 30°. (EC is the angle
bisector)
∴ ∠FEC = (180° – 30°) = 150°.
40
23. (b) We have area of triangle AFE = A/4. (If A = Area of triangle ABC) and area of triangle DHI = (A/4)/4 = A/16. Hence, ratio = 1 : 4.
24. (b) In ∆ AOB and ∆ COD170 cm
A B
C D
O
AO = OD, BO = OC ∠AOB = ∠COD (vertically opposite angles) ∴ ∆AOB ≅ ∆COD ∴ AB = CD = 170 cm. 25. (d) c = c1 (Vert. opp. ∠ s). b = c + s (Ext. ∠). d = c1 + r (Ext. ∠) But b + d = 180° (Opp. ∠s, cyclic quad.) ⇒ c + s + c1 + r = 180° (∵ c = c1) ⇒ r + s + 2c = 180° ⇒ r + s = 180° – 2c.26. (b) m ∠ PAC = m ∠ PBC = 90° ....(Tangent perpendicularity theorem) m ∠ PAC + m ∠ PBC + m ∠ ACB = 360° ∴ m ∠ APB = 360 – (90 + 90 + 65) = 115° ∴ m (AXB) = 115°.27. (d) Basic concept28. (a) ∠DCK = ∠FDG = 55° (corr. ∠s)
A B
CE
x°
K
H
D
G
F
40°
55°
∴ ∠ACE = 180° – (∠EAC + ∠ACE) ∴ ∠HAB = ∠AEC = 85° (corr. ∠s) Hence, x = 85°29. (c) Clearly option (a) shows the angles would be 30, 60
and 90. It can be the ratio of angle in a right angled triangle.
Option (b) shows the angles would be 45, 45 and 90, then it can be the ratio of angle in a right angled triangle.
But option (c) cannot form the ratio of angles of right angled triangle.
30. (c) ∠DFO = ∠FOM and ∠AEO = ∠EOM (since CD || AB)
28°
42°
A B
CD
E
F
MO
42°
28°
∴ ∠FOE = (28° + 42°) = 70°31. (b) we have i + e = 180° and i – e = 132° ⇒ e = 24°
∴ n = 360
24
°°
⇒ n = 15.32. (c) ∠ABC = 180 – (65 + 75) = 40° ∠ORB =∠OQB = 90° ∴ ∠ROQ = 360 – (90 + 90 – 40) ∴ ∠ROQ = 140°33. (c) ∆ABC is similar to ∆EDC
∴ = =
AB BC AC
ED DC EC
∴
24 60
10= ⇒ =
AB BC
DE DC DC
⇒ DC = 25 cm34. (d) Clearly, triangle is obtuse, So (d) is the correct option.35. (a) No such point is possible36. (c)
P Q
RS x
3x
X
( )( )
( )22
2 2
39 :1
∆= = =
∆ar PXQ xPQ
RXS RS x
37. (a) The parallelogram ABCD and ∆ BCE lies between the same parallel lines AB and DE and has base of equal length.
∴ar. (∆BCE) = 1
2ar.(ABCD) = 1
2 × 16 = 8 sq. cm.
38. (c)
AD B
x
x
2x 2x
96°C
E
41
Let ∠CAD = ∠ACD = x At point C, x + (180° – 4x) + 96° = 180° 180 3 96 180⇒ °− + ° = °x
∴ x = 32° Hence, ∠DBC = 2 × 32 = 64º39. (b) A B
CD
P
∆ APD ~ ∆BPC
∴ =PA PD
PB PC
i.e., PA. PC = PB. PD. ∴ option (b)40. (b)
C
D
A
B
L
M
Given : BD = 64 cm AL = 13.2 cm CM = 16.8 cm So, Area (ABCD) = Area (∆ABD) + Area (∆BCD)
= 1
2× AL × BD + 1
2 × CM × BD
= 1
2× BD × (AL + CM)
= 64
2 (13.2 + 16.8)
= 32 × 30 = 960 cm2
41. (b) area = 1
2 42× ×
= 4 sq. units42. (c) ∠AEC = ∠ECD (Alternate interior angles as AB || CD) In ∆CED, ∠ECD + ∠CED + x° = 180° (Sum of angles of ∆ are 180°) 37° + 90° + x° = 180° x° = 180° – 37° – 90° x° = 53°
43. (b)
A
OP
B
OAPB is concyclic because ∠A + ∠B = 180° & ∠O + ∠P = 180°
44. (d)
O
10
A8 8
CB
In OAB, OA2 = OB2 + AB2
[∴ AB = 1
2 AC, because line drawn from centre to a
chord bisect & perpendicular to it] (10)2 = (OB)2 + (8)2
100 – 64 = OB2
OB2 = 36 OB = 6
45. (d) PR2 = PQ2 + PR2 = 32 + 42 = 25
∴ PR = 25 = 5 cm
P
Q R
O
r = Area of triangle
Semi-perimeter of triangle
=
13 4 62 1 cm
3 4 5 62
× ×= =
+ +
42
46. (c)
OA
B C
DE
150°51°
∠AOE = 150° ∠DAO = 51° ∠EOB = 180° – 150° = 30° OE = OB
∴ ∠OEB = ∠OBE = 150
2 = 75°
∴ ∠CBE = 180° – 75° = 105°
47. (b) 2
2
ABC AB
DEF DE
∆=
∆
⇒ 2
20 25
45 DE=
⇒ DE2 = 45 25 225
20 4
×=
∴ DE = 15
2 = 7.5 cm
48. (d) ∠ACB = 80° ∠ACD = 180° – 80° = 100° Q AC = CD ∴ ∠CAD = ∠CDA
A
BC
D
= 80
402= °
∠BAC = 111° – 40° = 71° ∠ABC = 180° – 71° – 80° = 29°49. (d) A
B C
∠A + ∠B = 145° ∠C + 180° – 145° = 35° ∠C + 2∠B = 180° ⇒ 2∠B = 180° – 35° = 145°
⇒ ∠B = 145
2 = 72.5 ° = ∠A
∠B > ∠C ∴ AC > AB
50. (c) A
B C70° 50°
D
According to angle bisector theorem : The angle bisctor, like segment AD, divides the sides of the triangle proportionally.
In ∆ABC ∠A + ∠B + ∠C = 180° ∠A = 180° – 70° – 50° = 60°
∠BAD = 602
= 30°51. (b) A
F E
B D C
Let ABC be the triangle and D, E and F are midpoints of BC, CA and AB respectively.
Hence, in ∆ABD, AD is median AB + AC > 2 AD Similarly, we get BC + AC > 2 CF BC + AB > 2 BE On adding the above inequations, we get (AB + AC + BC + AC + BC + AB) > 2(AD + BE +
CF) 2 (AB + AC + BC) > 2 (AD + BE + CF) ∴ AB + BC + BC > AD + BE + CF Thus, the perimeter of triangle is greater than the sum
of the medians.52. (b)
B
Q P
9R
AC
16
Let the two circles with centre A, B and radii 25 cm and 9 cm touch each other externally at point C. Then AB = AC + CB
= 25 + 9 = 34 cm Let PQ be the direct common tangent i.e. BQ ⊥PQ and
AP ⊥PQ. Draw BR ⊥ AP. Then BRQP is a rectangle. (Tangent ⊥ radius at pt. of contact)
In ∆ABR AB2 = AR2 + BR2
(34)2 = (16)2 + (BR)2
BR2 = 1156 – 256 = 900 BR = 900 = 30 cm
43
53. (a) In ∆ABC, BC2 = AB2 + AC2
BC2 = (5)2 + (12)2
BC2 = 25 + 144 BC2 = 169 BC = 169 = 13 cm
A
B C
512
O
Radius of triangle = BC 13
6.5 cm2 2
= =54. (b) Here, ∠ACB = 90° ∠ADC = 90° ∠BDC = 90°
C
A Bc
ab
D
p
Triangles ACB, ADC and BDC are right angle triangles. Here, Area of ∆ABC = Area of ∆ADC + Area of ∆BDC
⇒ 1 1 1a b p AD p DB
2 2 2× = × × + × ×
⇒ ab = p (AD + DB)
⇒ ab = pc ⇒c = ab
p … (1)
Now, In ∆ABC,
c2 = a2 + b2 2
2 2aba b
p
= +
⇒ 2 2
2 22
a ba b
p= +
2 2 2
1 1 1
p a b= +
Note: students are advised to remember this result.55. (a) Given OP = 2r = Diameter of circle ( )&∵OA PA OB PB⊥ ⊥
r
O
A
2r
P
r
B
�1
�2
∴ In∆OAP, sinθ1 = 1
2 2
r
r=
sinθ1 = sin 30° ⇒ 1 30θ = °
Similary, in ∆OBP, sinθ2 = 1
2 2
r
r=
sinθ2 = sin 30° ⇒ θ2 = 30° ∴ 1 2 30 30 60APB∠ = θ + θ = °+ ° = °56. (b) Given, AB = 12 cm; CD = 20 cm OE = ?
10OD
BA
C10
E
Now, AE = EB = 6cm (The line drawn from centre of circle to the chord bisect the chord)
In ∆OAE, By phythagoras theorem (OA)2 = (OE)2 + (AE)2 ⇒ (10)2 = (OE)2 + (6)2
100 – 36 = (OE)2 ⇒ 64 = OE2 ⇒ OE = 8 cm57. (d) 180A B C∠ +∠ +∠ = °
3 5 180C C C∠ + ∠ +∠ = ° 9 180C∠ = °
20C∠ = ° 100B∠ = °58. (a)59. (d) A
B C
D E
Since DE is parallel to BC ADE ABC∆ ≅ ∆
( )( )
( )( )
2
2
ar ABC AB 25
ar ADE 4AD
∆= =
∆
( )( )
( )( )
ar DECB ar ADE 25
ar ADE ar ADE 4+ =
( )( )
ar DECB 25 211
ar ADE 4 4= − = 1
54
=
60. (c) Second angle of parallelogram = 180° – 45° = 135° ∴ Required value = 135 + 2 × 45 = 135 + 90 = 225°61. (d) Q PA . PD = PC . PB ⇒ 10. PD = 8.5 ⇒ PD = 40/10 = 4 cm. 62. (a) Diameter of circle = x
∴ y = 4x ∴ x = 1
4 y.
44
63. (b) Since the sum of all the angle of a quadrilateral is 360º We have ∠ABC + ∠BQE + ∠DEF + ∠EPB = 360º ∴ ∠ABC + ∠DEF = 180º [Q BPE = EQB = 90º ]64. (d) Perimeter of the figure = 10 + 10 + 6 + 3π= 26 + 3πcm
OA D
B
10 cm
C6 cm
65. (c) 12 cmP
O
A
5cm
AP is a tangent and OA is a radius. Therefore, OA is ⊥at AP. So, In ∆OAP OP2 = 52 + 122
OP2 = 25 + 144 = 169 OP = 13 cm66. (d)
RQ
T
P
TP = TQ [The length of tangents drawn from an external point
to a circle are equal] Similarly, TP = TR Using both equation, we get TQ = TR The relation of TQ and TR is TQ = TR. Note: Students are advised to remember this result.67. (b)
D F C
A E B
There are three common tangents AB, CD and EF.68. (b) DE is parallel to BC So ∠AED = ∠C = 35º
80º
D E
35º
CB
A
Since ∠A = 80º Then ∠ADE = 65º ∠EDB is supplement to ∠ADE. So, ∠EDB = 180º – ∠ADE = 180º – 65º = 115º69. (c) 70. (d) Angles are = (x + 15°),
6x 2x6 and 30
5 3 + ° + °
We know that Sum of the angles of a triange is 180°.
⇒ 6x 2x
x 15 6 30 1805 3
+ ° + + ° + + ° =
⇒ 15x 18x 10x
51 18015
+ ++ =
⇒ 43x
180 51 12915
= − =
⇒ 43x = 129 × 15 x = 45°
Then angle are = (45+ 15°), 6 45
6 and5
× + ° 2 45
303
× + °
= 60°, 60°, 60° So this is an equilateral triangle.71. (b)72. (d) D
C = 180° D°R = 180D D
R = 180D73. (b) Angle described by hour hand of a clock in 12 hrs
= 360° Angle described by hour hand of clock in 1 hr
36030
12= = °
Angle described by hour hand of clock in 1 min 30 1
60 2
°= =
Angle described by hour hand of clock in 10 min 1
10 52× = °
74. (c)
45
75. (b)
Q R
P
TM
L
Assuming right angled triangle be in a circle where PR is diameter PT = QT = TR (Radii of circle) QT = TR ∠TQR = ∠TRQ ∠TQR = ∠RLM Corresponding angles ∠RLM = ∠LRM i.e. (∠TRQ)76. (d) B
D
C A
2 2 2( ) ( ) ( )− − −= +CD BC CA
2 2 2
1 1 1⇒ = +
CD BC CA
2 2 2
2 2 2 2 2 2
1 1
. .
+⇒ = ⇒ =
CA BC AB
CD BC CA CD BC CA
2 22 2 2 2 2
2
.. .= ⇒ =
BC CACD CD AB BC CA
AB
.⇒ CD AB = BC.CA77. (a)
78. (c) Sum of angle of regular polygon ( )2 180− °
=n
n
⇒ 135 n = 180n – 360 ⇒ 45 n = 360
3608
45= =n
Number of diagonals = 8C2 – 8
8 7
– 82
×= = 20
79. (c) The n-sided polygon can be dinded into ‘n’ triangle with O, the Centre of the circule as one veotex for each triange. The altitude of each triangle is r. Let the sides of the polygon be ‘a1’, a2 ... an. (Given a1 = a2 = ... an)
D The area of polygon is nr pr
2 2=
Area of polygon = 1 na r a ra2r pr
2 2 2 2+ + − − − =
So, option (c) is correct.
80. (c) F C
G
DH
O
B
E
A
Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH. Using Pythagorean theorem we have from the above diagram:
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
= + = +
= + = +
= + = +
= + = +
OA AH OH AH AE
OC CG OG EB HD
OB EO BE AH BE
OD HD OH HD AE
Adding these equalities, we get
2 2 2 2 2 2
2 2 2 2 2 2
+ = + + +
+ = + + +
OA OC AH HD AE EB
OB OD AH HD AE EB
From these equalities, we get
2 2 2 2+ = +OA OC OB OD
81. (b) For finding the sum of the interior angles of a polygon is the same, whether the polygon is regular or irregular. So, we would use the formula (n – 2) × 180°, where n is the number of sides in the polygon.
Let one angle be of the polygon be x and other 5 equal angles be y.
So, according to the question,
30= +x y ( 2) 180 5− × ° = +n x y
(6 2) 180⇒ − × ° 30 5 180.4 6 30= + + ⇒ = +y y y
720 6 30 6 690 115⇒ = + ⇒ = ⇒ = °y y y
82. (c) 2
3 2
4
1
S = 2 3 4 9
2 2
+ += as r = 2 × 1 = 2m2
Area =
9 9 9 92 3 4
2 2 2 2
9 5 3 12 2 2 2
− − −
= 3
154
m2
46
2
3
2
31
Area = 1
2 32× × = 3 m2
Area = 1
[1 3] 22
+ × = 4
Total area = 2 + 3 + 4 + 3
54
= 9 + 3
54
83. (a) D is diameter of each circle D Side of square = D
A B
D C
Diagonal of square = 2 2D D+ = D 2
Diameter of shaded circle = D 2 D− = D( 2 1)−
Diameter = 2 D D D( 2 1)− = −84. (a)
85. (b) A B 50
ACB 90 90 652 2
∠ ×∠ = − = − =
86. (b) 87. (a) Q ∠DAC & ∠ACD are complimentary ∴ y2 = 9 × 16 ⇒ y = 12 ∴ option (a)88. (a) ∠CAF = 100°. Hence ∠BAC = 80° Also, ∠OCA = (90°–∠ACF) = 90° – 50° = 40° = ∠OAC
(Since the triangle OCA is isosceles) Hence ∠OAB = 40° In isosceles ∆OAB, ∠OBA will also be 40° Hence, ∠BOA = 180 – 40 – 40 = 100°89. (b)
EC
2
B
A
D
2 2
2 2
2
FM
60°
30°
Given BC & EF are each 2 feet. Since area of rectangle is length × width.
To find out BF or CE, Take ∆ABF. It has two equal sides (AB = AF), so the perpendicular from A to line BF divides ABF into two congruent ∆s.
So, each of the two triangles is 30°-60°-90° right angle ∆ with hypotenuse 2.
In ∆ABM cos 30° = BM 3 BMBM 3
AB 2 2⇒ = ⇒ =
So, BF = 2 × BM = 2 3
Area of rectangle = 2 3 × 2 = 4 3
ALITER : Splitting hexagon of area “12a” in its part.
2a
a a
a a
2a
a a
a a
area of hexagon
= 12a = 236 (2)
4×
12a = 6 3
area of rectangle = 8a = 4 3
90. (c)
60° 40°
A
E DB C
In ∆ABC, ∠A + ∠B + ∠C = 180° ∠A + 60° + 40° = 180° ∠A = 180° – 60° – 40° = 80° AD bisects ∠BAC ∴ ∠A = ∠BAD + ∠DAC ∠BAD = ∠DAC = 40° Now, In ∆ABE ∠B + ∠E + ∠BAE = 180° 60° + 90° + ∠ BAE = 180° ∠BAE = 30° ∴ ∠EAD = ∠BAD – ∠BAE = 40° – 30° = 10°
47
91. (c)
G
A
CBD
1
1 1
2
AG = BC (Given) BD = DC (given AD is median) So, GD = BD = DC (Q centroid divides median in 2 : 1 ratio) ∆BGD & ∆GCD are both isosceles ∆. Then ∠BGC = 90°92. (c) Should be XY since you divide XY into millions of
congruent portions, each portion which is the diameter of the semicircle is very small. So the sum of all the arcs should be XY.
93. (a) From exam approach. 21. We can mark our answer as direct result.
94. (a)
E
BA
D C
60°
In , 90 60 150∆ ∠ = + = °DEC DCE
180 15015
2
−∠ = ∠ = = °CDE DEC
Sol. (95-97) :
OP Q
R
S
III
r1 r2
95. (b) In ∆SOQ and ∆ROP ∠Ois common ∠S = ∠R = 90º (tangent at circle) ∴ ∆SOQ ~ ∆ROP
1
+⇒ = = = +
RP OP PQ OQ PQ
SQ OQ OQ OQ
4
13
⇒ = +PQ
OQ or
4 11
3 3= − =
PQ
OQ
⇒ PQ = 7 and OQ = 21
∴ Required ratio 7 1
21 3= =
96. (b) PQ = r1 + r2 = 7 As the ratio of radii is 4 : 3. So, the only value which satisfies the radii of
circle II = 3 ALITER : Q Ratio of I & II diameter is 4 : 3 ∴ We now that radius of II will be multiple of 3. ∴ Option (b)97. (c) In ∆SOQ, ⇒ SO2 + SQ2 = OQ2 ⇒ SO2 = 212 – 32 = (21 – 3) (21 + 3) = 18 × 24 = 432
12 3⇒ =SO
98. (b) Inradius of right angled triangle
= 2
+ −AB BC AB
= 9 40 41
2
+ − = 4 cm
99. (b)
A
B
C
D
P
O
45°
45°
OC = OD and OA = OP = OB OP = 1 m ∴ PC = 1 m OC = 2m ∴ AC = OC – OA
= ( )2 1− m
and AC + CP = ( )2 1− + 1
= 2 m
= 1.414 m = 141.4 cm
100. (c) ( ) 1
( ) 2
∆=
ar CMN
ar ABNM
∴ ( ) 1
( ) 3
∆=
∆ar CMN
ar CAB
⇒ 1
3= =
MN CM
AB CA ⇒ CA 3
CM 1=
⇒ CA 3 11
CM 1
−− =
48
⇒ AM 3 1
CM 1
−=
⇒ 1 3 1
23 1
+= =
−CM
MA (QMA = (CA – CM)
101. (d) In ∆ABC, ∠ C + ∠ AOB = 180° ⇒ 180° – 50° – 80°– ∠AOB = 180° ⇒ ∠AOB = 130°102. (a) ∠OCD = 90° ∠OAC = ∠OCA = 30° ∠ACD =∠ACO + ∠OCD = 30° + 90° = 120° ∴ ∠BAC = 180° – 120° = 60° ⇒ ∠BCD = 60° (∠BCD = ∠BAC) ⇒ ∠OCB = ∠OCB – ∠BCD = 90° – 60° = 30°
103. (d) 1 1 1
: : 6 : 4 : 32 3 4
=
6x + 4x + 3x = 52, or 13x = 52x, or x = 4 Required length = 12 cm.
104. (c) Required ratio = ( )
2
21: 2
2=
a
a
105. (b) ∆’s APO and BQO are similar (∠APO = ∠BQO = 90°, tangent is ⊥r to radius ∠AOP = ∠QOB, vertically opposite angles). ∴ AO : OB : : 2 : 1 and OP : OQ : : 2 : 1, AB = 10
⇒ AO = 2
3 × AB =
20
3 and OP =
2
3 × PQ = 16
3
AP2 = OA2 – OP2 (In ∆OAP, ∠APO = 90°)
= 2 22 2
1 1(20 16 ) 144
3 3− = ×
∴12
43
= =AP
106. (d)
C AB
O3 cm1 cm
2 cm
'O
22 cm
AB = 2 23 1 2 2 cm− =
∴ AC = 4 2 cm
107. (b) The sum of the interior angles of a polygon of n sides
is given by the expression (2n – 4) 2
π
⇒ ( )2 – 4 16202 180
× = ×nπ π
( ) 1620 22 – 4 18
180
×= =n
or 2n = 22 or n = 11
108. (a) In ∆ABC, DE | | BC
By applying basic Proportionality theorem, =AD AE
DB EC But
3
5=
AD
DB (Given)
∴ 3
5=
AE
EC or
3
5 3=
+ +AE
EC AE or 3
8=
AE
AC
or 3
5.6 8=
AE ⇒ 8AE = 3 × 5.6
⇒ AE = 3 × 5.6 /8 ∴ AE = 2.1 cm.
109. (a) Let the diagonals of the rhombus be x and y and the its sides be x.
OB
A
C
Dx
y
Now, 2 2
2
2 2 = +
x yx
or 2 2
2 –4 4=
x yx
3x2 = y2
or 1
3=
x
y or y : x 3 :1=
110. (a)
D C
E
BA
60
In , 90 60 150∆ ∠ = °+ ° = °DEC DCE
180 15015
2
−∠ = ∠ = = °CDE DEC
49
111.
B
CM
E
FA
D
2
1
let ar ∆BFE = 1 unit ∴ ar ∆FEC = 2 units ∴ ar ∆FMC = (1 + 2) units ∴ ar of square = (1 + 2 + 3 + 6) units = 12 units Q 1 unit = 108 ∴ 12units = 12 × 108 = 12 × 12 × 9
Q length of diagonal = 2 area of square×
= 2 12 3 36 2× × =
112. (a) m ∠ADC = 90º (Angle subtended by the diameter on a circle is 90°)
D
BA C
∴ ∆ADC is a right angled triangle. ∴ (DB)2 = AB × BC. (DB is the perpendicular to the hypotenuse) = 9 × 4 = 36 ∴ DB = 6113. (b) As F is the mid-point of AD, CF is the median of the
triangle ACD to the side AD. Hence area of the triangle FCD = area of the triangle ACF. Similarly area of triangle BCE = area of triangle ACE. ∴ Area of ABCD = Area of (CDF + CFA + ACE
+ BCE) = 2 Area (CFA + ACE) = 2 × 13 = 26 sq. units.114. (c) Given AB is a circle and BT is a tangent, ∠BAO = 32º Here, ∠OBT = 90º [∵ Tangent is ⊥ to the radius at the point of contact] OA = OB [Radii of the same circle] ∴ ∠OBA = ∠OAB = 32º [Angles opposite to equal side are equal] ∴ ∠OBT = ∠OBA + ∠ABT = 90º or 32º + x = 90º. ∠x = 90º – 32º = 58º . Also, ∠AOB = 180º – ∠OAB – ∠OBA = 180º – 32º – 32º = 116º
Now Y = 1
2AOB
[Angle formed at the center of a circle is double the angle formed in the remaining part of the circle]
= 1
2 × 116º = 58º.
115. (a)
A
B
C
D
M
O
N
Two parallel chords AB & CD & AB = CD = 8 cm Diameter of circle = AD = 10 cm.
∴ radius = AO = OD = 105
2= cm
AM = MB = 2
AB = 4 cm.
∆AOM is right angle ∆, AO
2 = AM 2 + OM
2
52 = 42 + OM 2
OM 2 = 25 – 16 = 9
⇒ OM = 3 cm. Similarly, OM = ON = 3 cm ∴ Distance between parallel chords = MN = OM + ON = 3 + 3 = 6 cm116. (c) ∠A + ∠B + ∠C = ∠180°
⇒ 1
2∠B + 1
2∠C = 90 – 1
2∠A
⇒ 1
2∠B + 1
2∠C + ∠BOC = 180°
A
B C
O
⇒ 90° – ∠A + ∠BOC= 180°
⇒ ∠BOC = 180° – 90 902 2
∠ ∠ ° − = ° +
A A
Note : students are advised to remember this result.117. (c) ∠APB = 42° + 68° = 110° (Exterior angle of a triangle is equal to sum of opposite interior ∠s).
A
R
B
42°
PC
68°
∆APR ≅∆BPR [SAS condition]
∴ ∠RPB = ∠RPA = 110
2
°
= 55° ∴ In ∆BRP, ∠ABC = 90° – 55° = 35°.
50
118. (a) Let AB be the chord of length 14 cm. at a distance of 6 cm from the centre O. Draw OE ⊥ AB. Then, BE = 7 cm and OE = 6 cm. ∴ OB2 = OE2 + BE2 = (62 + 72) = 85. Let CD be the chord at a distance of 2 cm from O. Now, OF = 2 cm, OD2 = OB2 = 85.
A BE
7 7
O
FDC
∴ FD = 9 cm. ∴ CD = 2 × FD = 18 cm.119. (a) Let YX = YZ = r (same radii); OYZ is a straight line
(contact of circles) YX ⊥ AB (Tangent ⊥ to radius); AX = 9, XB = 5 (given) ⇒ AB = 14, OB = OZ = 7(Same radii) OX + 7 – 5 = 2 In triangle OXY, OY = 7 – r; YX + r, OX = 2 ⇒ OY2 = YX2 + OX2 (Pythagoras’ Theorem) (7– r)2 = r2 + 22 49 – 14r + r2 = r2 + 4 14r = 45
r = 3 3
14 cm.
120. (a) R is the midpoint of side. AB and Q is the midpoint of side AC.. (midpoint theorem)
∴ RP = 1
2 AC = 1
2AB
PQ = 1
2AB
QR
A
PCB
Perimeter of PRQ = 2(RP + PQ)
= 2 1 1
2 2 +
AB AB
= 2AB = 2 × 8 = 16 121. (c) The given hexagon is regular hexagon. ∴ AF = AO = FO = OE = FE In DAFE, by sine rule.
a AE
sin 30 sin120=
° °
⇒ AE = 3a.
122. (a) Q area of ∆ABC = a.b.c
4R= 1
2× BC × AD
8.6. BC 1BC 4.8
4R 2⇒ = × ×
24 x10R 5
48⇒ = =
123. (b)
x 3 + x
9 m3 m3 m
Using Pythagoras theorem,
2 281 (3 )+ = +x x
⇒ 2 281 9 6+ = + +x x x 6 72 12m⇒ = ⇒ =x x
Height of wall = 12 + 3 =15 m124. (c) PR = PB + AR – 5 = 20 + 20 – 5 [ 5 cm=∵ AB ] So, perimeter = PR + PQ + QR = 20 + (20 – 5) + 20 + (20 – 10) + 20 + (20 – 12) = 35 + 30 + 28 = 93125. (b) By exam approach 13.
We get : 1 1 1
PQ AB CD= +
1 1 1
PQ 3 1⇒ = +
3PQ
4⇒ =
∴ CD : PQ = 1 : 3/4 = 1 : 0 . 75126. (d) Remember that a perpendicular from the centre to a
chord divides it into two equal parts.
12
16 B'B"
A"D
A'
O
B
A
In ∆OBB’, OB2 = BB’2 + OB’2
51
⇒ 202 = 162 + OB’2
⇒ 2 2' 20 16 12= − =OB
Similarly in ∆OAA’, 2 2' 20 12 16= − =OA
∴ Distance between the two parallel chords = 16 – 12 = 4 cm or 16 + 12 = 28 cm
127. (b) (OS)2 = (OK)2 + (KS)2
25 = OK2 + 16
⇒ OK = 3 and (OS)2 = (OL)2 + (LN)2
25 = (OL)2 + 9 ⇒ OL = 4 cm ∴ KL = OL – OK = 1 cm ∴ Area of rectangle = 1 × 10 = 10 cm2
128. (c) AB = PQ = 26 cm and PO = OQ = 13 CM
( ) ( )2 2= −CO PO PC
P
C
O
D
Q
( ) ( )2 213 5= −CO
CO = 12 cm CD = 2CO = 24 cm Alternatively: Solve by using the formula of tangents.
129. (d) At ∠A = 60°, BC = b = c and at ∠A = 90°, 2 2= =BC b c
∴ 60° < A < 90°, 2= < <BC c a c
130. (c) ∠A : ∠B : ∠C = 4 : 1 : 1 Hence we can suppose ∠A = 4x, ∠B = x, ∠C = x ∴ 4x + x + x = 180 x = 30 ∴ A = 120, B = 30, C = 30
Now, sin
sin
A
B =
3sin120 sin 60 2
1sin 30 sin 302
° °= =
° °3
1=
131. (b) 9 × 180 – 2 × 360 = 180 × 5 = 900°
( )180 2 360
180 4
× − × = − ∵
n
n
132. (b) ∠ADO is a right angle (angle of semicircle) Again when OD is perpendicular on the chord AC and
OD passes through the centre of circle ABC, then it must bisect the chord AC at D.
∴ AD = CD = 6 cm
A
O
D90°
133. (a) OB = OA = radius of circle ⇒ ∠CAO = ∠OBA (angles in alternate segments are equal) Now, if ∠CAO = ∠OBA ∴ ∠OAC = ∠OAB ∴ option (a) is correct134. (c) Value of BC will lie in between 999 and 3003. i.e. AB – AC < BC < AB + AC Hence 999 < BC < 3003. So, the total values possible for BC = 2003.135. (b) Let the adjacent angles of the parallelogram be 4x and
5x. Then, 4x + 5x = 180 ⇒ 9x = 180 ⇒ x = 20 One angle of quadrilateral = 3 × 80° = 240° Again, sum of angles of quadrilateral 4y + 11y + 9y + 240° = 360° 24y = 120° ⇒ y = 5 Hence, the sum of the largest and the smallest angles
of the quadrilateral = 4 × 5 + 240 = 260°
136. (d)
O O�
P
QBA
∠AQP = 2
π (Angle in the semicircle is 90°)
∠BQP = 2
π (Angle in the semicircle is 90°)
∠AQB = ∠AQP + ∠BQP = + 2
π ⇒π or 180°
137. (b) In ∆AOB AO = BO (radii of circles) ∴ ∠ABO = ∠BAO = 30º In ∆BOC BO = CO (radii of circles) in ∆BOC
52
30º
40º
xº
A
B C
O
∴ ∠BCO = ∠OBC = 40º ∠ABC = ∠ABO + ∠OBC ∠ABC = 30° + 40° = 70° 2 × ∠ABC = ∠AOC ⇒x° = 140138. (a)
A B
S
T
R
Q
P
∆ PQB and ∆ PRA are similar triangle by AAA criteria.
∴ AP AR 5
BP BQ 2= =
P divides AB externally in the ratio of 5 : 2139. (a) In ∆OBP. OB = OP (∵ radius)
25º
35º
O
PB
A
∴ ∠OBP = ∠OPB = 35º In ∆ AOP OA = OP (∵ radius) ∴ ∠OAP = ∠OPA = 25º Now, ∠APB = ∠OPA + ∠OPB = 25º + 35º = 60º Hence, ∠AOB = 2∠APB (Angle be substended by are at centre is twice) (= 2 × 60º = 120º
140. (d)
140º
DCB
A
∠ACB + ∠ACD = 180º (linear pair) ∴ ∠ACB = 180º – 140º = 40º
In ∆ABC ∠BAC + ∠ABC + ∠ACB = 180º ∠BAC + 3 ∠BAC + 40º = 180º 4 ∠BAC = 180º – 40º
∠BAC = 14035º
4=
141. (b) In ∆ABC, ∠ACB = 90° ∴ ∠ACB + ∠ACD ⇒ 90º + 50º = 140º As angle mode by triangle in semicircle is equal to 90º. ∴ In quad. ABCD, ∠BAD + ∠BCD = 180º (Sum of opposite angles of a cyclic quad. is equal to
180°) ∠BAD = 180º – 140º = 40º142. (b) ∠PSO is a right angle (angle of semicircle)
S90°
P
O
Again when OS is perpendicular on chord PR and OS passes through the centre of circle PQR, then it must bisect the chord PR at S.
DPS = RS = 17 cm.143. (c)
A
B
O
C D
12
8
x
According to question, AB = 12 AC = 8 AD = 8 + x As, we know that (AB)2 = AC × CD (12)2 = 8 × x
D
144x 18 cm
8= =
DABD is a right angle because ∠B = 90° By Pythagorean theorem, BD2 + AB2 = AD2
BD2 + AB2 = AD2 DBD2 = (18)2 – (12)2
= 324 – 144 = 180 D BD 6 5 cm=
D Radius BD 6 5
3 5 cm2 2
= = =
53
144. (c)
L
M
P Q
R S
U
PQ || RS PR|| QS D PQRS is a || gm ∠LPR = 35° and ∠UST = 70° ∠UST = ∠RSQ (Vertically opposite)
∠RSQ = ∠RPQ (opposite angle of 11 gm) ∠LPR + ∠RPQ + ∠MPQ = 180° 35° + 70° + ∠MPQ = 180° ∠MPQ = 180 – 105 ∠MPQ = 75°145. (b) To calculate the angle between the hour hand and the
minute hand of a clock when the time is 4 : 36 pm, we can say that the angle will be approximately equal to the angle made from 4 : 20 pm to 4 : 36 pm.
Therefore, we need to calculate the angle made by the hands of a clock in 16 minutes.
In 60 minutes, the angle made by the hands of a clock is 360°. So, the angle made by the hands of a clock in
16 minutes will be 360
16 9660
× = ° .
Thus, the angle lies between 72° to 108° i.e. 2 3
.5 5
π π< θ <
146. (b)
A B
C
A
B
C The above two figures shows that (b) is the only correct
answer.147. (b) The sides of a triangle in geometric progression are a,
ar, ar2
Triangle is right angled. Therefore, we use Pythagoras theorem.
(a)2 + (ar)2 = (ar2)2
a2 + a2r2 = a2r4
1 + r2 = r4 or r4 – r2 – 1 = 0
D2 1 1 4( 1)
2
− ± − −=r
2 1 5
2
− ±=r
r = 1 5
2
− +
1 5r
2
− −≠ (Because Radius is not negative)
So, common ratio = 5 1
.2
−
148. (c) Let a = b = c then the sides an 2a, 2a and 2a by heroni formula
2 2a 2aS 2a a
2
+= = +
as D = ( )( )( )( )
2 a a 2 a a 2 a
2 a a 2 a 2 a a 2a
+ + −
+ − + −
( )( ) 22a a 2a a a= + −
2 2 2 2(2a a )a a= − =
Now by putting a = b = c in option (c) satisfies the area a2
i.e. 2a(b c) a(2a)a
2 2
+= =
149. (b)
A
ED
CB10
According to the question and by BPT,
=AD AE
AB AC and ∠A is common.
Therefore, DADE i ABC (by SAS similarity) Since we know that the ratio of areas of two similar
triangles is equal to the square of its proportional sides,
therefore, 2
2
.( ) ( )
.( ) ( )
∆=
∆ar ADE DE
ar ABC BC ...(1)
54
Also, we are given that
1.( ) .( )
5∆ = ∆ar ADE ar ABC
Thus, (1)
2 2
2
.( ) ( ) ( )
5 .( ) 100(10)
∆⇒ = =
∆ar ABC DE DE
ar ABC
21 ( )
5 100⇒ =
DE 1002 5
5⇒ = =DE
150. (d) diagonal of square = 6 2
side of square = 6 2
62
=
3 × side of triangle = 6 × 4
side of triangle = 8
area of triangle 2 2 23 3a 8 16 3 cm
4 4= = × =
151. (a) Area of 23ABC
4∆ = �
from mid point theoram
Area of 2
23 3ADE
4 2 16 ∆ = × = �
�
Area of shaded region 2 23 3
4 16= −� �
23 3
16=
�
152. (c) Let length of the side of the square = x units
x
x
x
x
p–x
CFBp
D
A
q
q – x
E
2 x
D Length of the diagonal = 2 x units
DADE ~ D EFC (By AAA)
DAD DE
EF FC
⇒ q x x
x p x
⇒ x2 = pq–px – qx + x2
⇒ x = pq
p q+
BE = 2 pq
p q+ = Length
of the diagonal
153. (c) Given AD be the diameter of circle.
This circle consist of three semi-circles.
DCB
A
So Area of circle = Area of all 3 semi circles ...(i)
AD is the diameter of all circles.
Let AD = d
Then radius = d/2
Given are of circle = 707 m2
From (i) we get
707 = + +
2 2 2d d d2 2 2 2 2 2π π π
707 =
23 d2 2π
⇒ × =
2d 707 22 3π
Area of shaded region =
2d2 2π
= ×
× = 2707 2 707m
2 3 3π
π
Lost of levelling the shaded region = 63 rs/m2
Lost = 707
633
= 21 × 707 = ` 14847
55
154. (c) BOA
C D
3P
x
X
in D APB
AB = 13
2 cm, BP =
12
2 = 6 cm
AP = ? AP2 = AB2 – BP2
AP2 = 2 2 213 12 5
2 2 2 − =
AP = 5
2 in D AQD AQ2 = AD2 – DQ2
= 2 2 213 5 12
2 2 2 − =
AQ = 12
2= 6
5PQ 6
2= −
73.5
2⇒ =
155. (c) BOA
C D
3P
x
X
Given that AB = 4 cm and CD = 10 cm, let the radius of the circle be r cm.
Since the perpendicular from the center of a circle to a chord bisects the chord, therefore AO = OB = 2 cm and CP = PD = 5 cm.
In DAOX, by Pythagoras theorem, we have
2 2 2 2 22 (3 ) 4 9 6+ + = ⇒ + + + =x r x x r ...(1)
Similarly, in DCPX, we have
2 2 2 2 25 25+ = ⇒ + =x r x r ...(2)
Equation (1) and (2) gives
2 24 9 6 25 13 6 25 6+ + + = + ⇒ + = ⇒x x x x x
= 12 ⇒ x = 2
Now, equation (2) gives
2 225 4 29 29+ = ⇒ = ⇒ =r r r cm
156. (d) OP = 2 2AB + PB + OC× PB
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