SOLUTIONS

SECTION A

Ans1. (c) 3 (1 mark)

Ans2. (d) 11 (1 mark)

Ans3. (c) 3.4cm, 3.5cm, 7cm (1 mark)

Ans4. (b) 100 (1 mark)

Ans5. (d) -1 (1 mark)

Ans6. (c)

2

2

x +9x+14

=x +7x+2x+14

=x(x+7)+2(x+7)

=(x+2)(x+7)

So the dimensions are (x+2) and (x+7).

So, by substituting x=2, we get the dimensions to be 9 and 4. (1 mark)

Ans7. (d)

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

a= 32 cm, b= 30 cm, c=30 cm

s 46 cm=

Area of the triangle = s(s a)(s b)(s c)− − −

2

46 14 16 16

32 161 cm

= × × ×

=

(1 mark)

Ans8. (a)

Here 36

s 182

= = , and the sides are 10, 13,13.

By Heron’s formula, Area = 18x8x5x5 5x3x4 60= = sq cm. (1 mark)

SECTION B

Ans9. a = 2+ 3

⇒ 1

a=

1

2 3+

1mark

2

= 1 2 3

2 3 2 3

−×

+ −

=

( )2

2

2 3

2 3

−

−

= 2 3− 1mark

2

So, a + 1

a= (2+ 3 ) + (2- 3 ) = 4 (1 mark)

Ans10. Let f(z) = 3z3 + 8z2 -1

The possible integral zeros of f(z) are -1 and 1.

f(z) = 3z3 + 8z2 -1

and f(-1) = 3(-1)3 + 8(-1)2 -1 ≠ ≠ 0

⇒ -1 is not a zero of f(z) (1 Mark)

f(1) = 3(1)3 + 8(1)2 -1 ≠ ≠ 0

⇒ 1 is not a zero of f(z) (1 Mark)

Therefore, f(z) has no integral zero.

Ans11. (-2x+5y-3z)2

= (-2x)2+(5y)2+(-3z)2 +2(-2x)(5y)+2(5y)(-3z)+ 2(-2x)(-3z) (1 mark)

= 4x2+25y2+9z2-20xy-30yz+12zx (1 mark)

Ans12.

Since, OD and OE are the bisectors of ∠AOC and ∠BOC respectively.

Therefore, ∠AOD=∠COD and ∠BOE = ∠COE 1mark

2

Also, ∠DOE =900

Now, ∠AOC+ ∠BOC = ∠AOD+∠COD+∠BOE+∠COE

= ∠COD+∠COD+∠COE+∠COE

= 2(∠COD+∠COE)

=2(∠DOE) =2 x 90o =180o (1 mark)

Hence, points A, B and C are collinear. 1mark

2

OR

As RO is perpendicular to PQ.

Therefore, ∠POR = ∠QOR=900

Now, ∠QOS = ∠QOR+∠ROS

∠QOS = 90o +∠ROS ……………(i) 1mark

2

Since, ∠POR = 90o ⇒ ∠POS+∠ROS=90o

∠POS = 90o - ∠ROS ……………….(ii) 1mark

2

Subtracting (i) from (ii), we get,

∠QOS - ∠POS = 90o + ∠ROS – (90o - ∠ROS)

∠QOS - ∠POS =2∠ROS

∠ROS =1

2 (∠QOS - ∠POS) (1 mark)

Ans13.

In ∆PQS and ∆PRT

PQ = PR (given)

∠Q =∠R (given)

And ∠P=∠P (common) (1 mark)

Therefore, ∆PQS is congruent to ∆PRT. (By ASA) 1mark

2

Thus, QS = RT (by CPCT) 1mark

2

Ans14.

(1 mark)

ABCD is a square. The diagonals of a square are equal and bisect each other at right

angles. (1 mark)

SECTION C

Ans15. x3 - 8y3 - 36xy - 216

= (x)3+ (-2y)3 + (-6)3 - 3.x(-2y)(-6) (1 mark)

= (x-2y-6) (x2+4y2+36+2xy-12y+6x) (1 mark)

= 0 x (x2+4y2+36+2xy-12y+6x) (x = 2y+6 ⇒ x-2y-6=0)

= 0 (1 mark)

OR

2a

bc

+ 2b

ca

+ 2c

ab

=3

LHS ⇒ 2a

bc

+ 2b

ca

+ 2c

ab

= 3 3 3a b c

abc

+ + (1 mark)

= 3abc

abc (if a+b+c=0 then a3+b3+c3 = 3abc) (1 mark)

=3 = RHS (1 mark)

Ans16. Let x= 0.001

Then, x= 0.001001001……….. (i) 1mark

2

Therefore, 1000x = 1.001001001…………… (ii) (1mark)

racting (i) from (ii), we get,

999x=1 ⇒ x=1

999

Hence, 0.001 =1

999

11 marks2

Ans17. (x + 1

x)2 = 3

x2 +2

1

x+ 2(x) x (

1

x) = 3 (1 mark)

x2 +2

1

x+ 2 =3 (1 mark)

or , x2 + 2

1

x = 3 - 2 = 1 (1 mark)

OR

Given, x=1+ 2

And, 1

x=

1

1 2+

1

x =

1

1 2+x1 2

1 2

−

−

=

( )2

2

1 21 2

1 2

−= − +

−

1

1 marks2

Now, 1

xx

−

= 2 +1 – ( 2 -1) = 2 (1 mark)

31

xx

−

= 23 = 8

1mark

2

Ans18.

‘M’ represents 5 on the number line. (3 marks)

OR

Steps of construction:

1) Mark the distance 2.4 units from a fixed point A on the number line to obtain a

point O such that AO=2.4 Units.

2) From O, mark a distance of 1 unit and mark the new point as B.

3) Find the mid point of AB and mark that point as C.

4) Draw a semicircle with centre C and radius AC.

4) Draw a line perpendicular to AB passing through O and intersecting the semicircle

in P.

5) Then OP= 2.4 (1 mark)

(2 marks)

Ans19.

Since interior angles on the same side of transversal are supplementary.

Therefore, ∠EAB+∠RBA = 1800

1

2 ∠EAB+

1

2 ∠RBA =

1

2 x1800 (i) (1 mark)

As AP and BP are bisectors of ∠EAB and ∠RBA respectively

∠PAB =1

2∠EAB and ∠PBA=

1

2 ∠RBA (ii)

From (i) and (ii) , we get,

∠PAB +∠PBA =900 (1 mark)

In ∆APB we have,

∠PAB +∠PBA +∠APB = 180o

90o+∠APB =180o

∠APB =180o-90o = 90o (1 mark)

Ans20. Let the length of the smallest side = x

Then, the other two sides are x+4 and 2x-6.

Perimeter of the triangle = x+x+4+2x-6=50

4x-2 = 50 ⇒ 4x=52 or x=13 (1 mark)

Sides of the triangle are 13,13+4 and 2x13-6 i.e. 13,17 and 20cm (1 mark)

Let a=13, b=17 and c=20

S=a b c+ +

2 =

+ +13 17 20

2= 25

Area of ∆ = s(s a)(s b)(s c)− − −

= 25(25 13)(25 17)(25 20)− − −

= 25 12 8 5× × ×

= 20 3 = 20 x 5.48 = 109.54 cm2 (1 mark)

Ans21.

(1

2 mark)

Construction: Draw the bisector AO of ∠A. 1mark

2

In ∆ABO and ∆ACO

AB=AC (Given) 1mark

2

AO=OA (Common)

∠BAO=∠CAO (By Construction) 1mark

2

∆ABO ≅ ∆ACO (By S.A.S Congruence criteria) 1mark

2

∠B=∠C (By C.P.C.T) (1 mark)

Ans22.

Through O, draw a line POQ parallel to AB

Now PQ ||AB and CD||AB

So, CD||PQ

Therefore AB||PQ and AO is a transversal

We have, ∠AOQ + ∠OAB = 180O (Co interior angles are supplementary)

∠AOQ + 130O = 180O

∠AOQ = 180O – 130O

Therefore ∠AOQ = 50O (1 mark)

Similarly, PQ||CD and OC is a transversal

Therefore ∠QOC + ∠DCO = 180O

∠QOC + 120O = 180O

∠QOC = 180O – 120O

Therefore ∠QOC = 60O (1 mark)

Therefore ∠AOC = ∠AOQ + ∠QOC

= 50O + 60O = 110O (1 mark)

Ans23. From any one point a line can be drawn to each of the other (n-1) points .

(1 Mark)

So in all, n(n-1) lines can be drawn. But, in counting so , we would have counted

every line twice. (A line from point A to B is same as the line from point B to point A)

(1 Mark)

So, the number of lines can be drawn through n points is lines can be drawn through

n points is ( 1)

2

n n − (1 Mark)

Ans24.

Consider triangles ABD and ACD

We have, AB = AC (given)

AD = AD ...common

∠ BAD=∠ CAD... (AD is angle bisector) ⇒ ABD ACD∆ ≅ ∆ ..(By SAS congruence criterion) (1 Mark)

⇒ ∠ ADB=∠ ADC.... (By CPCT)

BD = CD... (i)

But they form a linear pair.

⇒ ∠ ADB=∠ ADC=90o

In triangle ABD 2 2 2AD +BD = AB (Using Pythagoras theorem) (1 Mark)

By substituting the values given, we get

BD = 4 cm ⇒ BC = 2BD = 8cm.... from (i) (1 Mark)

SECTION D

Ans25. Let p(x) = 2x4+x3-14x2-19x-6 and q(x) = x2+3x+2

Then, q(x) = x2+3x+2 = (x+1) (x+2) (1 mark)

Now, p(-1) = 2(-1)4+(-1)3-14(-1)2-19(-1)-6

= 2-1-14+19-6 = 21-21 =0 (1 mark)

And, p(-2) = 2(-2)4+(-2)3-14(-2)2-19(-2)-6

= 32-8-56+38-6 = 70-70 =0 (1 mark)

(x+1) and (x+2) are the factors of p(x), so p(x) is divisible by (x+1) and (x+2)

Hence, p(x) is divisible by x2+ 3x+2. (1 mark)

OR

Let p(z) = az3+4z2+3z-4 and q(z)= z3-4z+a

When p(z) is divided by z-3 the remainder is given by,

p(3) = a x 33 + 4 x 32 + 3 x 3 - 4 = 27a+36+9-4

= 27a+41 ………………..(i) 1

1 marks2

q(3)= 33 – 4 x 3 + a

= 27-12+a

= 15+a ……………..(ii)

Given that, p(3) = q(3) 1

1 marks2

27a+41 = 15+a

27a - a = -41+15 26a = -26 or a =-1 (1 mark)

Ans26. a) x4 + x4

1-2

= (x2)2 +

( )x2

2

1– 2 x x2 x

x2

1 (1 mark)

= (x2 -x2

1)2

= (x2 -x2

1)(x2 -

x2

1) (1 mark)

b) 2x5 + 432x2y3

=2x5 + 432x2y3

=2x2(x3 + 216y3) (1 mark)

= 2x2[(x)3+ (6y)3]

= 2x2(x+6y) (x2+36y2-6xy) (1 mark)

Ans27.

Since PA is the bisector of ∠QPR

So, ∠QPA=∠RPA …………(i)

In ∆PQM, we have

∠PQM+∠PMQ+∠QPM = 180o

∠PQM+90o+∠QPM = 180o

∠PQM = 90o- ∠QPM ……….(ii) (1 mark)

In ∆PMR, we have

∠PMR+∠PRM+∠RPM = 180o

90o+∠PRM+∠RPM = 180o

∠PRM = 90o - ∠RPM ………….(iii) (1 mark)

Subtracting (iii) from (ii), we get

∠Q-∠<R = (90o-∠QPM)-(90o-∠RPM)

∠Q-∠R = ∠RPM -∠QPM (1 mark)

∠Q-∠R = (∠RPA+∠APM) – (∠QPA-∠APM) ………….(iv)

∠Q-∠R = ∠RPA+∠APM –∠QPA+∠APM (using (i) )

∠Q-∠R = 2∠APM

Hence, ∠APM = 1

2 (∠Q-∠R) (1 mark)

OR

AB=AC ∠ABC=∠ACB

AC=AD ∠ADC =∠ACD 1mark

2

Therefore, ∠ABC + ∠ADC = ∠ACB + ∠ACD

=∠BCD (1 mark)

∠DBC + ∠BDC = ∠BCD [Because ∠ABC = ∠DBC and ∠ADC = ∠BDC]

∠DBC + ∠BDC + ∠BCD = 2∠BCD

2∠BCD = 180O (sum of the all angles of Triangle is 180o] (1 mark)

∠BCD = 90O

Hence, ∠BCD = 90O (1 mark)

Ans28. Let p(x) = x3 + 13x2 + 32x + 20

The constant term in p(x) is equal to 20 and the factors of 20 are ±1,±2,±4,±5,±10.

Putting x = -2 in p(x), we have

p(-2) = (-2)3 + 13(-2)2 + 32(-2) + 20 = -8 + 52 – 64 + 20 = -72 + 72 = 0

p(-2)=0 1

1 marks2

As p(-2)=0, so (x+2) is a factor of p(x). Now, divide p(x) by (x+2)

x+2/ x3 + 13x2 + 32x + 20\x2+11x+10

x3 + 2x2

(-) (-)

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

11x2+32x+20

11x2+22x

(-) (-)

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

10x +20

10x+20

(-) (-)

-----------

0

---------- 1

1 marks2

p(x) = (x+2)( x2+11x+10)

= (x+2)( x2+10x + x +10)

=(x+2)(x (x + 10) + 1 (x+10))

= (x+2)( x+10)(x+1) (1 mark)

Ans29.

1mark

2

In ∆ABC we have,

∠A+∠B+∠C=1800 ( sum of the angles of a ∆ is 180)

1

2∠A+

1

2∠B+

1

2∠C=

1

2x 1800

1

2(∠A +∠1+∠2) =90o

∠1+∠2 = 90o –1

2∠A ………..(i)

1(1 marks)2

In ∆OBC, we have,

∠1+∠2+∠BOC = 1800 (sum of the angles of a ∆ is 180)

∠BOC = 180o-(∠1+∠2) (1 mark)

∠BOC = 180o-(90o-1

2∠A) ( using (i))

∠BOC = 90o + 1

2 ∠A (1 mark)

Ans30.

Given that, AD=AE

Therefore, ∠ADE=∠AED (angles opposite to equal sides of a triangle are equal)

So, ∠ADB = ∠AEC (remaining angles of linear pair) (1 mark)

In ∆ADB and ∆AEC

AD=AE (given)

∠ADB= ∠AEC (proved above)

BD=CE (given)

Thus, ∆ADB and ∆AEC are congruent. (By SAS) (2 marks)

Therefore, ∠ABC=∠ACB (Corresponding parts of congruent triangles) (1 mark)

Ans31.

Let us join AC.

In ∆ABC

AB < BC (AB is smallest side of quadrilateral ABCD)

∴ ∠2 < ∠1 (angle opposite to smaller side is smaller) ... (1)

In ∆ADC

AD < CD (CD is the largest side of quadrilateral ABCD)

∴ ∠4 < ∠3 (angle opposite to smaller side is smaller) ... (2)

(1 mark)

On adding equations (1) and (2), we have

∠2 + ∠4 < ∠1 + ∠3

⇒ ∠C < ∠A

⇒ ∠A > ∠C

(1 mark)

Let us join BD.

In ∆ABD

AB < AD (AB is smallest side of quadrilateral ABCD)

∴ ∠8 < ∠5 (angle opposite to smaller side is smaller) ... (3)

In ∆BDC

BC < CD (CD is the largest side of quadrilateral ABCD)

∴ ∠7 < ∠6 (angle opposite to smaller side is smaller) ... (4)

(1 mark)

On adding equations (3) and (4), we have

∠8 + ∠7 < ∠5 + ∠6

⇒ ∠D < ∠B

⇒ ∠B > ∠D (1 mark)

Ans32.

Ans33.

Ans34. The given points A(0,4),B(0,0),C(3,0) can be plotted as follows:

(3 marks)

B is at origin.

AB=4 units, BC = 3 units. 1mark

2

Area of ∆AOB = 1

Base Height2

× ×

=1

3 42

× × = 6 square units. 1mark

2