5204bb0c5fb33_C9_SA1_Maths.pdf

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Learnhive Education Pvt. Ltd.www.learnhive.com

Copyright 2013 Learnhive Education Pvt. Ltd.

SUMMATIVE ASSESSMENT

MATHS

CLASS 9

Question paper: C09-MA-SA1-2013-01

Time: 3 Hrs Max Marks: 90

General Instructions:

A) All questions are compulsory.

B) The question paper consists of 34 questions divided into four sections A,B,C and D.a. Section A comprises of 10 questions of 1 mark each

b. Section B comprises of 5 questions of 2 marks eachc. Section C comprises of 10 questions of 3 marks eachd. Section D comprises 10 questions of 4 marks each

C) Question numbers 1 to 10 in section A are multiple choice questions where you are to selectone correct option out of the given four.

D) There is no overall choice. However, internal choice has been provided in 2 questions of threemarks each and 2 questions of four marks each. You have to attempt only one of the

alternatives in all such questions.

E) Use of calculator is not permitted.

F) An additional 15 minutes time has been allotted to read this question paper only


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SECTION A

1. Every point on the number line can be represented as a unique __________.

a. Integerb. Real numberc. Natural numberd. Fraction

2. When you add the answer isa.

b. c. d.

3. The degree of polynomial is :a.

b. 2

c. 1d. It is not a polynomial

4. The remainder when is divided by isa. 1

b. c. d.

5. Two sides of a triangle are of lengths 7 cm and 3.5 cm. The length of the 3rdside cannot bea. 4.1 cm

b. 3.4 cm

c. 3.8 cm

d. 3.6 cm


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6. If one angle of a triangle is equal to the sum of other two angles, then the triangle is

a. Obtuse

b. Equilateral

c. Isoscelesd. Right

7.

the angle POB equals

a. 54b. 48c. 108d. 24

8. If two interior angles on the same side of the transversal intersecting two parallel lines are in

the ratio 2 : 3, then the greater of the two angles is ___________.

a. 108b. 54c. 136

d. 1209. In triangle ABC, if BC = AB and B = 80, then A = _________

a. 40b. 80c. 100d. 50

10.The polynomial whose degree is 0 is

a. b. c. 20

d.


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SECTIONB

11.Find the value of p if is a factor of Given If is a factor of then and

12.Express 0.75 in the form of where p and q are integers and q 0.Let = 0.75 ------ (1)10

= 7.

------- (2) [

1 digit is repeating, multiply both sides by 10]

Subtracting (1) from (2)

9= 6.8 9= 0.75 =

13.Check if is a factor of

Since remainder is not zero (it is - 10) the given expression (3x + 7) is not a factor of

14.In the figure, if PQ = XY, prove that PX = QY.

Given PQ = XY

Subtracting XQ from both sides as this is common part between both.

PQXQ = XYXQ

PX = QY [PQXQ = PX and XYXQ = QY]


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15.In ABC, A = 65, C = 35. Which side of the triangle will be the longest? State why it isso, without actual construction.

A = 65, C = 35B = 180- (A + C) = 180- 100B = 80By property of triangles i.e side opposite to largest angle is greatest, we have AC as the longest

side because it is opposite to B = 80which is the greatest angle of the triangle.

SECTION C

16.The perimeter of an equilateral signal board is 120 cm. Find its area.

Let the perimeter be p = 120 cm

semi-perimeter is s = cmLet the side of the equilateral triangle be a then by Herons formula

Area of = For equilateral Area = Also, 40 cm

Area = = = Area =


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17.An insect starts moving on a graph sheet from point O (0,0). It goes right and stops at (0,3) and

then moves upwards by four divisions and then goes left by 6 divisions. Mark its position on

the axes and name the points.

18.Simplify: [( ) ( )][ ][( ) ( )][ ]

= ( )( ) [ ]= =

19.Rationalise

= Using identity (a + b) (ab) = =

20.Find the value of

=

= = 132


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OR

Represent on the number line.

Draw a number line and mark two points: x=0 and x=1

Draw a perpendicular at x=1 that is as long as the distance from 0 to 1.

Connect x=0 to the top of that perpendicular.

The distance from 0 to the top is sqrt(2).

-----------

Using the 0-point as center and radius = sqrt(2),

draw an arc of a circle from the top of the perpendicular down to the number line.Where the arc touches the number line is the sqrt(2) point.

-----------------------

Draw a perpendicular at the point that is "one" high.

Draw the line segment from x=0 to the top of that perpendicular.

The length of that line segment is sqrt(3).

Using the 0-point as center and radius = sqrt(3),

draw an arc of a circle from the top of the perpendicular to the number line.

Where the arc touches the number line is the sqrt(3) point.

21.Factorise: By finding factors of 12 such that 7 is the sum i.e 4 and 3

=

are the roots of

22.Evaluate using suitable identities.

By using identity =

= 10000001300

99

= 1000000129700

= 100000029701 = 970299


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23.Find the value of x and y. Given XYPQ

Since XY PQ

If AC is the transversal, then y + 100= 180[y and 100are interior angles on the same side of the transversal]y = 180- 100= 80

Also [angles on a straight line]

OR

Prove that the sum of interior angles of a triangle is 180

Since AB EC and BCD is transversal

ABC =

ECD (corresponding angles) and

BAC =

ACE (Alternate angles)

ABC + BAC = ECD + ACE (adding)

ABC + BAC + BCA = ECD + ACE + BCA (adding BCA to both sides)

ABC + BAC + BCA = 180(ECD, ACE and BCA are angles on straight line)Sum of angles of triangle = 180


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24.In PQR, the bisectors of Q and R intersect each other at point O. Prove that QOR =

For PQRP + Q + R = 180P = 180(Q + R) ------(1)

In OQRQOR +

[OQ and OR are angular bisectors of Q and R]Substituting value of Q + R from equation (1)

QOR + 90- = 180QOR = 180- 90+ QOR = 90+

25.Given PQ XY. Find the value of x.

Draw a line AB through O such that AB XY PQ. Consider lines AB and PQ OSQ =

BOS [alternate angles]

BOS = 100Now BOS + 30

+ AOR = 180

(Angles on a straight line)

AOR = 180130


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AOR = 50Consider line AB and XY and OR is transversal then, AOR + = 180(co-interior angles)

= 18050= 130

SECTION - D

26.Show that

Taking LHS and rationalising the denominators

=

=

= = 2 + 3 = 5 = RHS

27.Evaluate: when a = 2 and n = 3

Given

[(a + b) (ab) =

= [

=

=

28.Factorise:

Let

= =


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Now substituting

The roots are 1, -1 and 3

29.Given that the area of the triangles OPQ and ORS are equal, find the ordinate of R.

Area of OPQ = Area of ORS ----- (1)OQ = -3, OS = 4, PQ = 4, RS = y + 2

From (1)

5


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30.In the figure below, P is a point such that AP = AC. Show that AB AP.

In APC AP ACAPC =ACD gles opposite to equl sies re equl i trigle In ABP, APC is exterior angleAPC =BAP + B exterior gle is equl to sum of iterior opposite glesThis means that

APC

B ------(1)APC =ACP

(1) becomesAPC =ACP BSince in a the side opposite to greater angle is greater AB [side opposite to B is AP and side opposite to ACB is A]

OR

Prove AD = FC, if AB = FE, AB BD and BC = ED and FE EC.

In the second ABD and FCE,AB = FE -----(1)

ABD = FEC = 90-----(2)BC = ED

BC + CD = CD + ED [adding the common part CD to both sides]

BD = CE -----(3)

From (1), (2) and (3) we can say that ABD FCE by SASAD = FE (As the two triangles are congruent, the corresponding sides are equal)

31.The altitude and base of a triangular field is in the ration 6 : 5. If its cost is Rs. 49,57,2 at the

rate of Rs. 36,720 per hectare and 1 hectare = 10,000 sq. m, find dimensions of the field (in

metres).

Given : Cost of field = Rs. 49,57,200

Rate of field = Rs. 36,720 / hectare


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Area = = 135 hectaresArea in m= 135 10000

= 1350000 mLet the base be 5x and altitude be 6x

A r e se ltitue

x x

Are x

x m

Bse is m and altitude is m

32.BD is bisector of ABC. From a point P on BD, perpendiculars PE and PF are drawn to AB

and BC respectively. Prove that

a. Triangle BEP is congruent to triangle BFP

In BEP BFP

1 = 2 (BD bisects ABC)PEB = PFB = 90If 2 out 3 angles of both triangles are equal the 3 rdangle has to be equal (sum of angles of is 180)3 = 4 and side BP is commonBy ASA, BEP BFP

b. PE = PFSince BEP BFPPE = PF (corresponding sides of congruent triangles are equal)


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OR

Consider the square ABCD and the equilateral triangle DEC as shown. Prove that

a. ADE = BCE = 30In square ABCD ADC = BCD = 90In equilateral DEC , EDC = ECD = 60ADE = ADC - EDC

= 9060= 30BCE = BCD - ECD = 9060= 30

ADE = BCE = 30b. ADE BCEIn ADE and BCE, ADE = BCEAD = BC [sides of same square]

DE = CE [sides of equilateral triangle]

ADE BCE [SAS]

33.Two parallel lines are cut by a transversal

a. If the measure of co-interior angles are 2aand (3a10), fix value of a.When 2 parallel lines are cut by a transversal, the sum of con-interior angles is 180 .


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b. If corresponding angles are (2b + 15)and (3b7) find value of b.Corresponding angles are equal

b = 23

34.Find the value of k if the polynomials and when divided by(x4) leave the same remainder.

Let the polynomials be

By remainder theorem, = 1stremainder and = 2ndremainder1stremainder = 2ndremainder {they have the same remainder when divided by } =

= ------(1)

= = ------(2) (1) = (2)

35.Prove that if two lines intersect each other, then the vertically opposite angles are equal.

Consider the two lines AB and PQ intersecting at O.

To prove 1 = 2 and 3 = 4


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1 + 3 = 180(linear pair)1 = 180- 3 ----- (1)

Now take the linear pair of 2 and 3

2 + 3 = 1802 = 180- 3 ----- (2)From (1) and (2)

1 = 2 = 180- 3 1 = 2Similarly 2 and 4 are a linear pair

2 + 4 = 1804 = 180- 2 or 2 = 180- 4 ----- (3)From (2) and (3) 3 = 4


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