Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > Class 10 Mathematics: CBSE (Official) Marking Scheme with Solutions (2017-18)
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Page 1
Marking Scheme
Mathematics Class X (2017-18)
Section A
S.No. Answer Marks
1. Non terminating repeating decimal expansion. [1]
2. k = ±4 [1]
3. a
11
= –25 [1]
4. (0, 5) [1]
5. 9 : 49 [1]
6. 25 [1]
Section B
7. LCM (p, q) = a
3
b
3
HCF (p, q) = a
2
b
LCM (p, q) × HCF (p, q) = a
5
b
4
= (a
2
b
3
) (a
3
b) = pq
[1/2]
[1/2]
[1]
8. S
n
= 2n
2
+ 3n
S
1
= 5 = a
1
S
2
= a
1
+ a
2
= 14 ? a
2
= 9
d = a
2
– a
1
= 4
a
16
= a
1
+ 15d = 5 + 15(4) = 65
[1/2]
[1/2]
[1/2]
[1/2]
9. For pair of equations kx + 1y = k
2
and 1x + ky = 1
We have:
2
1 1 1
2 2 2
a k b 1 c k
,,
a 1 b k c 1
? ? ?
For infinitely many solutions,
1 1 1
2 2 2
a b c
a b c
??
?
2
k1
k 1 k 1, –1
1k
? ? ? ? ?
...(i)
and
2
3
1k
k 1 k 1
k1
? ? ? ? ? ...(ii) …(ii)
From (i) and (ii), k = 1
[1/2]
[1/2]
[1/2]
[1/2]
10.
Since
p
1,
3
??
??
??
is the mid-point of the line segment joining the points (2, 0) and
2
0,
9
??
??
??
therefore,
2
0
p1
9
p
3 2 3
?
? ? ?
The line 5x + 3y + 2 = 0 passes through the point (–1, 1) as 5(–1) + 3(1) + 2 = 0
[1]
[1]
11.
(i) P(square number) =
8
113
(ii) P(multiple of 7) =
16
113
[1]
[1]
Page 2
Marking Scheme
Mathematics Class X (2017-18)
Section A
S.No. Answer Marks
1. Non terminating repeating decimal expansion. [1]
2. k = ±4 [1]
3. a
11
= –25 [1]
4. (0, 5) [1]
5. 9 : 49 [1]
6. 25 [1]
Section B
7. LCM (p, q) = a
3
b
3
HCF (p, q) = a
2
b
LCM (p, q) × HCF (p, q) = a
5
b
4
= (a
2
b
3
) (a
3
b) = pq
[1/2]
[1/2]
[1]
8. S
n
= 2n
2
+ 3n
S
1
= 5 = a
1
S
2
= a
1
+ a
2
= 14 ? a
2
= 9
d = a
2
– a
1
= 4
a
16
= a
1
+ 15d = 5 + 15(4) = 65
[1/2]
[1/2]
[1/2]
[1/2]
9. For pair of equations kx + 1y = k
2
and 1x + ky = 1
We have:
2
1 1 1
2 2 2
a k b 1 c k
,,
a 1 b k c 1
? ? ?
For infinitely many solutions,
1 1 1
2 2 2
a b c
a b c
??
?
2
k1
k 1 k 1, –1
1k
? ? ? ? ?
...(i)
and
2
3
1k
k 1 k 1
k1
? ? ? ? ? ...(ii) …(ii)
From (i) and (ii), k = 1
[1/2]
[1/2]
[1/2]
[1/2]
10.
Since
p
1,
3
??
??
??
is the mid-point of the line segment joining the points (2, 0) and
2
0,
9
??
??
??
therefore,
2
0
p1
9
p
3 2 3
?
? ? ?
The line 5x + 3y + 2 = 0 passes through the point (–1, 1) as 5(–1) + 3(1) + 2 = 0
[1]
[1]
11.
(i) P(square number) =
8
113
(ii) P(multiple of 7) =
16
113
[1]
[1]
12. Let number of red balls be = x
? P(red ball) =
x
12
If 6 more red balls are added:
The number of red balls = x + 6
P(red ball) =
x6
18
?
Since,
x 6 x
2 x 3
18 12
? ??
? ? ?
??
??
? There are 3 red balls in the bag.
[1/2]
[1]
[1/2]
Section C
13. Let n = 3k, 3k + 1 or 3k + 2.
(i) When n = 3k:
n is divisible by 3.
n + 2 = 3k + 2 ? n + 2 is not divisible by 3.
n + 4 = 3k + 4 = 3(k + 1) + 1 ? n + 4 is not divisible by 3.
(ii) When n = 3k + 1:
n is not divisible by 3.
n + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1) ? n + 2 is divisible by 3.
n + 4 = (3k + 1) + 4 = 3k + 5 = 3(k + 1) + 2 ? n + 4 is not divisible by 3.
(iii) When n = 3k + 2:
n is not divisible by 3.
n + 2 = (3k + 2) + 2 = 3k + 4 = 3(k + 1) + 1 ? n + 2 is not divisible by 3.
n + 4 = (3k + 2) + 4 = 3k + 6 = 3(k + 2) ? n + 4 is divisible by 3.
Hence exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.
[1]
[1]
[1]
14.
Since
55
and
33
? are the two zeroes therefore,
2
5 5 1
x x (3x 5)
3 3 3
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
is a factor of given polynomial.
We divide the given polynomial by 3x
2
– 5.
x
2
+ 2x + 1
For other zeroes, x
2
+ 2x + 1 = 0 ? (x + 1)
2
= 0, x = –1, –1
? Zeroes of the given polynomial are
55
, , –1and –1
33
? .
[1]
[1]
[1]
Page 3
Marking Scheme
Mathematics Class X (2017-18)
Section A
S.No. Answer Marks
1. Non terminating repeating decimal expansion. [1]
2. k = ±4 [1]
3. a
11
= –25 [1]
4. (0, 5) [1]
5. 9 : 49 [1]
6. 25 [1]
Section B
7. LCM (p, q) = a
3
b
3
HCF (p, q) = a
2
b
LCM (p, q) × HCF (p, q) = a
5
b
4
= (a
2
b
3
) (a
3
b) = pq
[1/2]
[1/2]
[1]
8. S
n
= 2n
2
+ 3n
S
1
= 5 = a
1
S
2
= a
1
+ a
2
= 14 ? a
2
= 9
d = a
2
– a
1
= 4
a
16
= a
1
+ 15d = 5 + 15(4) = 65
[1/2]
[1/2]
[1/2]
[1/2]
9. For pair of equations kx + 1y = k
2
and 1x + ky = 1
We have:
2
1 1 1
2 2 2
a k b 1 c k
,,
a 1 b k c 1
? ? ?
For infinitely many solutions,
1 1 1
2 2 2
a b c
a b c
??
?
2
k1
k 1 k 1, –1
1k
? ? ? ? ?
...(i)
and
2
3
1k
k 1 k 1
k1
? ? ? ? ? ...(ii) …(ii)
From (i) and (ii), k = 1
[1/2]
[1/2]
[1/2]
[1/2]
10.
Since
p
1,
3
??
??
??
is the mid-point of the line segment joining the points (2, 0) and
2
0,
9
??
??
??
therefore,
2
0
p1
9
p
3 2 3
?
? ? ?
The line 5x + 3y + 2 = 0 passes through the point (–1, 1) as 5(–1) + 3(1) + 2 = 0
[1]
[1]
11.
(i) P(square number) =
8
113
(ii) P(multiple of 7) =
16
113
[1]
[1]
12. Let number of red balls be = x
? P(red ball) =
x
12
If 6 more red balls are added:
The number of red balls = x + 6
P(red ball) =
x6
18
?
Since,
x 6 x
2 x 3
18 12
? ??
? ? ?
??
??
? There are 3 red balls in the bag.
[1/2]
[1]
[1/2]
Section C
13. Let n = 3k, 3k + 1 or 3k + 2.
(i) When n = 3k:
n is divisible by 3.
n + 2 = 3k + 2 ? n + 2 is not divisible by 3.
n + 4 = 3k + 4 = 3(k + 1) + 1 ? n + 4 is not divisible by 3.
(ii) When n = 3k + 1:
n is not divisible by 3.
n + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1) ? n + 2 is divisible by 3.
n + 4 = (3k + 1) + 4 = 3k + 5 = 3(k + 1) + 2 ? n + 4 is not divisible by 3.
(iii) When n = 3k + 2:
n is not divisible by 3.
n + 2 = (3k + 2) + 2 = 3k + 4 = 3(k + 1) + 1 ? n + 2 is not divisible by 3.
n + 4 = (3k + 2) + 4 = 3k + 6 = 3(k + 2) ? n + 4 is divisible by 3.
Hence exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.
[1]
[1]
[1]
14.
Since
55
and
33
? are the two zeroes therefore,
2
5 5 1
x x (3x 5)
3 3 3
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
is a factor of given polynomial.
We divide the given polynomial by 3x
2
– 5.
x
2
+ 2x + 1
For other zeroes, x
2
+ 2x + 1 = 0 ? (x + 1)
2
= 0, x = –1, –1
? Zeroes of the given polynomial are
55
, , –1and –1
33
? .
[1]
[1]
[1]
15. Let the ten’s and the units digit be y and x respectively.
So, the number is 10y + x.
The number when digits are reversed is 10x + y.
Now, 7(10y + x) = 4(10x + y) ? 2y = x …(i)
Also x – y = 3 …(ii)
Solving (1) and (2), we get y = 3 and x = 6.
Hence the number is 36.
[1/2]
[1/2]
[1]
[1/2]
[1/2]
16. Let x-axis divides the line segment joining (–4, –6) and (–1, 7) at the point P in the
ratio 1 : k.
Now, coordinates of point of division
–1– 4k 7 – 6k
P,
k 1 k 1
??
??
??
??
Since P lies on x-axis, therefore
7 6k
0
k1
?
?
?
? 7 – 6k = 0
? k =
7
6
Hence the ratio is
7
1: 6 : 7
6
?
Now, the coordinates of P are
34
,0
13
???
??
??
.
OR
Let the height of parallelogram taking AB as base be h.
Now AB =
2 2 2 2
(7 4) (2 2) 3 4 5 units ? ? ? ? ? ? .
Area ( ? ABC) = ? ?
1 49
4(2 – 9) 7(9 2) 0(–2 – 2) sq units
22
? ? ? ? .
Now,
1 49
AB h
22
? ? ? ?
?
1 49
5h
22
? ? ? ?
? ?h =
49
9.8 units
5
? .
[1/2]
[1]
[1/2]
[1]
[1]
[1]
[1]
17. ? SQN = ? TRM (CPCT as NSQ MTR) ? ? ?
Since, P 1 2 ? ? ? ? ? = P PQR PRQ ? ? ? ? ? (Angle sum property)
? 12 ? ? ? = PQR PRQ ? ? ?
? 2 1 2 PQR ? ? ? (as 1 2 and PQR PRQ) ? ? ? ? ? ? ? ?
1 ? = PQR ?
[1]
[1]
Page 4
Marking Scheme
Mathematics Class X (2017-18)
Section A
S.No. Answer Marks
1. Non terminating repeating decimal expansion. [1]
2. k = ±4 [1]
3. a
11
= –25 [1]
4. (0, 5) [1]
5. 9 : 49 [1]
6. 25 [1]
Section B
7. LCM (p, q) = a
3
b
3
HCF (p, q) = a
2
b
LCM (p, q) × HCF (p, q) = a
5
b
4
= (a
2
b
3
) (a
3
b) = pq
[1/2]
[1/2]
[1]
8. S
n
= 2n
2
+ 3n
S
1
= 5 = a
1
S
2
= a
1
+ a
2
= 14 ? a
2
= 9
d = a
2
– a
1
= 4
a
16
= a
1
+ 15d = 5 + 15(4) = 65
[1/2]
[1/2]
[1/2]
[1/2]
9. For pair of equations kx + 1y = k
2
and 1x + ky = 1
We have:
2
1 1 1
2 2 2
a k b 1 c k
,,
a 1 b k c 1
? ? ?
For infinitely many solutions,
1 1 1
2 2 2
a b c
a b c
??
?
2
k1
k 1 k 1, –1
1k
? ? ? ? ?
...(i)
and
2
3
1k
k 1 k 1
k1
? ? ? ? ? ...(ii) …(ii)
From (i) and (ii), k = 1
[1/2]
[1/2]
[1/2]
[1/2]
10.
Since
p
1,
3
??
??
??
is the mid-point of the line segment joining the points (2, 0) and
2
0,
9
??
??
??
therefore,
2
0
p1
9
p
3 2 3
?
? ? ?
The line 5x + 3y + 2 = 0 passes through the point (–1, 1) as 5(–1) + 3(1) + 2 = 0
[1]
[1]
11.
(i) P(square number) =
8
113
(ii) P(multiple of 7) =
16
113
[1]
[1]
12. Let number of red balls be = x
? P(red ball) =
x
12
If 6 more red balls are added:
The number of red balls = x + 6
P(red ball) =
x6
18
?
Since,
x 6 x
2 x 3
18 12
? ??
? ? ?
??
??
? There are 3 red balls in the bag.
[1/2]
[1]
[1/2]
Section C
13. Let n = 3k, 3k + 1 or 3k + 2.
(i) When n = 3k:
n is divisible by 3.
n + 2 = 3k + 2 ? n + 2 is not divisible by 3.
n + 4 = 3k + 4 = 3(k + 1) + 1 ? n + 4 is not divisible by 3.
(ii) When n = 3k + 1:
n is not divisible by 3.
n + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1) ? n + 2 is divisible by 3.
n + 4 = (3k + 1) + 4 = 3k + 5 = 3(k + 1) + 2 ? n + 4 is not divisible by 3.
(iii) When n = 3k + 2:
n is not divisible by 3.
n + 2 = (3k + 2) + 2 = 3k + 4 = 3(k + 1) + 1 ? n + 2 is not divisible by 3.
n + 4 = (3k + 2) + 4 = 3k + 6 = 3(k + 2) ? n + 4 is divisible by 3.
Hence exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.
[1]
[1]
[1]
14.
Since
55
and
33
? are the two zeroes therefore,
2
5 5 1
x x (3x 5)
3 3 3
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
is a factor of given polynomial.
We divide the given polynomial by 3x
2
– 5.
x
2
+ 2x + 1
For other zeroes, x
2
+ 2x + 1 = 0 ? (x + 1)
2
= 0, x = –1, –1
? Zeroes of the given polynomial are
55
, , –1and –1
33
? .
[1]
[1]
[1]
15. Let the ten’s and the units digit be y and x respectively.
So, the number is 10y + x.
The number when digits are reversed is 10x + y.
Now, 7(10y + x) = 4(10x + y) ? 2y = x …(i)
Also x – y = 3 …(ii)
Solving (1) and (2), we get y = 3 and x = 6.
Hence the number is 36.
[1/2]
[1/2]
[1]
[1/2]
[1/2]
16. Let x-axis divides the line segment joining (–4, –6) and (–1, 7) at the point P in the
ratio 1 : k.
Now, coordinates of point of division
–1– 4k 7 – 6k
P,
k 1 k 1
??
??
??
??
Since P lies on x-axis, therefore
7 6k
0
k1
?
?
?
? 7 – 6k = 0
? k =
7
6
Hence the ratio is
7
1: 6 : 7
6
?
Now, the coordinates of P are
34
,0
13
???
??
??
.
OR
Let the height of parallelogram taking AB as base be h.
Now AB =
2 2 2 2
(7 4) (2 2) 3 4 5 units ? ? ? ? ? ? .
Area ( ? ABC) = ? ?
1 49
4(2 – 9) 7(9 2) 0(–2 – 2) sq units
22
? ? ? ? .
Now,
1 49
AB h
22
? ? ? ?
?
1 49
5h
22
? ? ? ?
? ?h =
49
9.8 units
5
? .
[1/2]
[1]
[1/2]
[1]
[1]
[1]
[1]
17. ? SQN = ? TRM (CPCT as NSQ MTR) ? ? ?
Since, P 1 2 ? ? ? ? ? = P PQR PRQ ? ? ? ? ? (Angle sum property)
? 12 ? ? ? = PQR PRQ ? ? ?
? 2 1 2 PQR ? ? ? (as 1 2 and PQR PRQ) ? ? ? ? ? ? ? ?
1 ? = PQR ?
[1]
[1]
Also 2 ? = PRQ ?
And SPT ? = QPR ? (common)
PTS ~ PRQ ?? (By AAA similarity criterion)
OR
Construction: Draw AP BC ?
In
2 2 2
ADP, AD AP DP ? ? ?
2 2 2
AD AP (BP BD) ? ? ?
2 2 2 2
AD AP BP BD 2(BP)(BD) ? ? ? ?
2
22
1 BC BC
AD AB BC 2
3 2 3
? ? ? ? ? ?
? ? ?
?? ? ? ? ?
?? ? ? ? ?
? ?
22
7
AD AB BC AB
9
??
22
9AD 7AB ?
[1]
[1/2]
[1/2]
[1/2]
[1]
[1/2]
18. Join OC
In ? OPA and ? OCA
OP = OC (radii of same circle)
PA = CA (length of two tangents)
AO = AO (Common)
? ? OPA ? ? OCA (By SSS congruency criterion)
Hence, ? 1 = ? 2 (CPCT)
Similarly ? 3 = ? 4
Now, ? PAB + ? QBA = 180°
? 2 ? 2 + 2 ? 4 = 180°
? ? 2 + ? 4 = 90°
? ? AOB = 90° (Angle sum property)
[1]
[1]
[1]
Page 5
Marking Scheme
Mathematics Class X (2017-18)
Section A
S.No. Answer Marks
1. Non terminating repeating decimal expansion. [1]
2. k = ±4 [1]
3. a
11
= –25 [1]
4. (0, 5) [1]
5. 9 : 49 [1]
6. 25 [1]
Section B
7. LCM (p, q) = a
3
b
3
HCF (p, q) = a
2
b
LCM (p, q) × HCF (p, q) = a
5
b
4
= (a
2
b
3
) (a
3
b) = pq
[1/2]
[1/2]
[1]
8. S
n
= 2n
2
+ 3n
S
1
= 5 = a
1
S
2
= a
1
+ a
2
= 14 ? a
2
= 9
d = a
2
– a
1
= 4
a
16
= a
1
+ 15d = 5 + 15(4) = 65
[1/2]
[1/2]
[1/2]
[1/2]
9. For pair of equations kx + 1y = k
2
and 1x + ky = 1
We have:
2
1 1 1
2 2 2
a k b 1 c k
,,
a 1 b k c 1
? ? ?
For infinitely many solutions,
1 1 1
2 2 2
a b c
a b c
??
?
2
k1
k 1 k 1, –1
1k
? ? ? ? ?
...(i)
and
2
3
1k
k 1 k 1
k1
? ? ? ? ? ...(ii) …(ii)
From (i) and (ii), k = 1
[1/2]
[1/2]
[1/2]
[1/2]
10.
Since
p
1,
3
??
??
??
is the mid-point of the line segment joining the points (2, 0) and
2
0,
9
??
??
??
therefore,
2
0
p1
9
p
3 2 3
?
? ? ?
The line 5x + 3y + 2 = 0 passes through the point (–1, 1) as 5(–1) + 3(1) + 2 = 0
[1]
[1]
11.
(i) P(square number) =
8
113
(ii) P(multiple of 7) =
16
113
[1]
[1]
12. Let number of red balls be = x
? P(red ball) =
x
12
If 6 more red balls are added:
The number of red balls = x + 6
P(red ball) =
x6
18
?
Since,
x 6 x
2 x 3
18 12
? ??
? ? ?
??
??
? There are 3 red balls in the bag.
[1/2]
[1]
[1/2]
Section C
13. Let n = 3k, 3k + 1 or 3k + 2.
(i) When n = 3k:
n is divisible by 3.
n + 2 = 3k + 2 ? n + 2 is not divisible by 3.
n + 4 = 3k + 4 = 3(k + 1) + 1 ? n + 4 is not divisible by 3.
(ii) When n = 3k + 1:
n is not divisible by 3.
n + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1) ? n + 2 is divisible by 3.
n + 4 = (3k + 1) + 4 = 3k + 5 = 3(k + 1) + 2 ? n + 4 is not divisible by 3.
(iii) When n = 3k + 2:
n is not divisible by 3.
n + 2 = (3k + 2) + 2 = 3k + 4 = 3(k + 1) + 1 ? n + 2 is not divisible by 3.
n + 4 = (3k + 2) + 4 = 3k + 6 = 3(k + 2) ? n + 4 is divisible by 3.
Hence exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.
[1]
[1]
[1]
14.
Since
55
and
33
? are the two zeroes therefore,
2
5 5 1
x x (3x 5)
3 3 3
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
is a factor of given polynomial.
We divide the given polynomial by 3x
2
– 5.
x
2
+ 2x + 1
For other zeroes, x
2
+ 2x + 1 = 0 ? (x + 1)
2
= 0, x = –1, –1
? Zeroes of the given polynomial are
55
, , –1and –1
33
? .
[1]
[1]
[1]
15. Let the ten’s and the units digit be y and x respectively.
So, the number is 10y + x.
The number when digits are reversed is 10x + y.
Now, 7(10y + x) = 4(10x + y) ? 2y = x …(i)
Also x – y = 3 …(ii)
Solving (1) and (2), we get y = 3 and x = 6.
Hence the number is 36.
[1/2]
[1/2]
[1]
[1/2]
[1/2]
16. Let x-axis divides the line segment joining (–4, –6) and (–1, 7) at the point P in the
ratio 1 : k.
Now, coordinates of point of division
–1– 4k 7 – 6k
P,
k 1 k 1
??
??
??
??
Since P lies on x-axis, therefore
7 6k
0
k1
?
?
?
? 7 – 6k = 0
? k =
7
6
Hence the ratio is
7
1: 6 : 7
6
?
Now, the coordinates of P are
34
,0
13
???
??
??
.
OR
Let the height of parallelogram taking AB as base be h.
Now AB =
2 2 2 2
(7 4) (2 2) 3 4 5 units ? ? ? ? ? ? .
Area ( ? ABC) = ? ?
1 49
4(2 – 9) 7(9 2) 0(–2 – 2) sq units
22
? ? ? ? .
Now,
1 49
AB h
22
? ? ? ?
?
1 49
5h
22
? ? ? ?
? ?h =
49
9.8 units
5
? .
[1/2]
[1]
[1/2]
[1]
[1]
[1]
[1]
17. ? SQN = ? TRM (CPCT as NSQ MTR) ? ? ?
Since, P 1 2 ? ? ? ? ? = P PQR PRQ ? ? ? ? ? (Angle sum property)
? 12 ? ? ? = PQR PRQ ? ? ?
? 2 1 2 PQR ? ? ? (as 1 2 and PQR PRQ) ? ? ? ? ? ? ? ?
1 ? = PQR ?
[1]
[1]
Also 2 ? = PRQ ?
And SPT ? = QPR ? (common)
PTS ~ PRQ ?? (By AAA similarity criterion)
OR
Construction: Draw AP BC ?
In
2 2 2
ADP, AD AP DP ? ? ?
2 2 2
AD AP (BP BD) ? ? ?
2 2 2 2
AD AP BP BD 2(BP)(BD) ? ? ? ?
2
22
1 BC BC
AD AB BC 2
3 2 3
? ? ? ? ? ?
? ? ?
?? ? ? ? ?
?? ? ? ? ?
? ?
22
7
AD AB BC AB
9
??
22
9AD 7AB ?
[1]
[1/2]
[1/2]
[1/2]
[1]
[1/2]
18. Join OC
In ? OPA and ? OCA
OP = OC (radii of same circle)
PA = CA (length of two tangents)
AO = AO (Common)
? ? OPA ? ? OCA (By SSS congruency criterion)
Hence, ? 1 = ? 2 (CPCT)
Similarly ? 3 = ? 4
Now, ? PAB + ? QBA = 180°
? 2 ? 2 + 2 ? 4 = 180°
? ? 2 + ? 4 = 90°
? ? AOB = 90° (Angle sum property)
[1]
[1]
[1]
19.
2 2 2
2 2 2 2
cosec 63 tan 24 sin 63 cos63 sin 27 sin 27 sec63
cot 66 sec 27 2(cosec 65 tan 25 )
? ? ? ? ? ? ? ? ? ? ? ?
?
? ? ? ? ? ?
=
2 2 2
2 2 2 2
cosec 63 tan 24 sin 63 cos63 cos(90 – 27 ) sin 27 cosec(90 – 63 )
tan (90 66 ) cosec (90 27 ) 2[cosec 65 cot (90 – 25 )]
? ? ? ? ? ? ? ? ? ? ? ? ? ?
?
? ? ? ? ? ? ? ? ? ? ?
=
2 2 2 2
2 2 2 2
cosec 63 tan 24 sin 63 cos 63 sin 27 cosec27°
tan 24 cosec 63 2(cosec 65 – cot 65 )
? ? ? ? ? ? ? ?
?
? ? ? ? ?
=
11
12
2(1)
?
??
OR
sin cos 2 ? ? ? ?
?
? ?
2
2
(sin cos ) 2 ? ? ? ?
?
22
sin cos 2sin cos 2 ? ? ? ? ? ? ?
? 1 2sin cos 2 ? ? ? ?
?
1
sin cos
2
? ? ? ...(i)
we know,
22
sin cos 1 ? ? ? ? ....(ii)
Dividing (ii) by (i) we get
22
sin cos 1
sin cos 1/2
? ? ?
?
??
? tan cot 2 ? ? ? ?
[1]
[1]
[1]
[1/2]
[1]
[1/2]
[1]
20. We know, AC = r
In ? ACB, ?BC
2
= AC
2
+ AB
2
? BC = AC 2
( AB AC) ?
? BC = r2
Required area = ar( ? ACB) + ar(semicircle on BC as diameter) –ar(quadrant
ABPC)
=
2
2
1 1 r 2 1
r r r
2 2 2 4
??
? ? ? ? ? ? ? ?
??
??
=
2 2 2
r r r
–
2 4 4
??
?
=
2
22
r 196
cm 98 cm
22
??
[1]
[1]
[1]
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