solutions section a - wordpress.com · · 2012-08-10... 3.4cm, 3.5cm, 7cm (1 mark) ans4. (b) 100...
TRANSCRIPT
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