CBSE Class 10 – Page 125

RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B

January 5, 2022June 15, 2018 by Prasanna

RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Question 1.
Solution:
(i) 36, 84
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 1
36 = 2 x 2 x 3 x 3 = 2² x 3²
84 = 2 x 2 x 3 x 7 = 2² x 3 x 7
HCF = 2² x 3 = 2 x 2 x 3 = 12
LCM = 2² x 3² x 7 = 2 x 2 x 3 x 3 x 7 = 252
Now HCF x LCM = 12 x 252 = 3024
and product of number = 36 x 84 = 3024
HCF x LCM = Product of given two numbers.
(ii) 23, 31
23 = 1 x 23
31 = 1 x 31
HCF= 1
and LCM = 23 x 31 = 713
Now HCF x LCM = 1 x 713 = 713
and product of numbers = 23 x 31 = 713
HCF x LCM = Product of given two numbers
(iii) 96, 404
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 2
96 = 2 x 2 x 2 x 2 x 2 x 3 = 2⁵ x 3
404 = 2 x 2 x 101 = 2² x 101
HCF = 2² = 2 x 2 = 4
LCM = 2⁵ x 3 x 101 = 32 x 3 x 101 = 9696
Now HCF x LCM = 4 x 9696 = 38784
and product of two numbers = 96 x 404 = 38784
HCF x LCM = Product of given two numbers
(iv) 144, 198
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 3
144 = 2 x 2 x 2 x 2 x 3 x 3 = 2⁴ x 32
198 = 2 x 3 x 3 x 11 = 2 x 3² x 11
HCF = 2 x 3² = 2 x 3 x 3 = 18
LCM = 2⁴ x 3² x 11 = 16 x 9 x 11 = 1584
and product of given two numbers = 144 x 198 = 28512

and HCF x LCM = 18 x 1584 = 28512
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 5
HCF x LCM = Product of given two numbers
(v) 396, 1080

396 = 2 x 2 x 3 x 3 x 11 = 2² x 3² x 11
1080 = 2 x 2 x 2 x 3 x 3 x 3 x 5 = 2³ x 3³ x 5
HCF = 2² x 3² = 2 x 2 x 3 x 3 = 36
LCM = 2³ x 3³ x 11 x 5 = 2 x 2 x 2 x 3 x 3 x 3 x 5 x 11 = 11880
Now HCF x LCM = 36 x 11880 = 427680
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 7
Product of two numbers = 396 x 1080 = 427680

HCF x LCM = Product of two given numbers.
(vi) 1152, 1664

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 = 2⁷ x 3²
1664 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 13 = 2⁷ x 13
HCF = 2⁷ = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128
LCM = 2⁷ x 3² x 13 = 128 x 9 x 13 = 128 x 117= 14976
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 10
Now HCF x LCM = 128 x 14976= 1916928

and product of given two numbers = 1152 x 1664 = 1916928

HCF x LCM = Product of given two numbers.

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 13

Question 3.
Solution:
HCF of two numbers = 23
LCM =1449
One number = 161
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 17
Second number = 207

Question 4.
Solution:
HCF of two numbers = 145
LCM = 2175
One number = 725
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 18
Second number = 435

Question 5.
Solution:
HCF of two numbers = 18
and product of two numbers = 12960
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 19
LCM of two numbers = 720

Question 6.
Solution:
HCF= 18
LCM = 760
HCF always divides the LCM completely
760 – 18 = 42 and remainder 4
Hence, it is not possible.

Question 7.
Solution:
(a) \(\frac { 69 }{ 92 }\)
HCF of 69 and 92 = 23
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 20

Question 8.
Solution:
Numbers are 428 and 606 and remainder in each case = 6
Now subtracting 6 from each number, we get 438 – 6 = 432
and 606 – 6 = 600
Required number = HCF of 432 and 600 = 24
The largest required number is 24
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 24

Question 9.
Solution:
The numbers are 320 and 457
and remainders are 5 and 7 respectively
320 – 5 = 315 and 457 – 7 = 450
Now the required greatest number of 315 and 450 is their HCF
Now HCF of 315 and 450 = 45
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 25

Question 10.
Solution:
The numbers are given = 35, 56, 91 and the remainder = 7 in each case,
Now the least number = LCM of 35, 56, 91 = 3640
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 26
LCM = 7 x 5 x 8 x 13 = 3640
Required least number = 3640 + 7 = 3647

Question 11.
Solution:
Given numbers are 28 and 32
Remainders are 8 and 12 respectively
28 – 8 = 20
32 – 12 = 20
Now, LCM of 28 and 32 = 224
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 27
LCM = 2 x 2 x 7 x 8 = 224
Least required number = 224 – 20 = 204

Question 12.
Solution:
The given numbers are 468 and 520
Now LCM of 468 and 520 = 4680
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 28
LCM = 2 x 2 x 13 x 9 x 10 = 4680
When number 17 is increase then required number = 4680 – 17 = 4663

Question 13.
Solution:
LCM of 15, 24, 36 = 360
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 29
Required number = 9999 – 279 = 9720

Question 14.
Solution:
Greatest number of 4 digits is 9999
LCM of 4, 7 and 13 = 364
On dividing 9999 by 364, remainder is 171
Greatest number of 4 digits divisible by 4, 7 and 13 = (9999 – 171) = 9828
Hence, required number = (9828 + 3) = 9831

Question 15.
Solution:
LCM of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37
Number to be added = (60 – 37) = 23

Question 16.
Solution:
We can represent any integer number in the form of: pq + r, where ‘p is divisor, ‘q is quotient’, r is remainder*.
So, we can write given numbers from given in formation As :
43 = pq₁ + r …(i)
91 = pq₂ + r …(ii)
And 183 = pq₃ + r …(iii)
Here, we want to find greatest value of ‘p’ were r is same.
So, we subtract eq. (i) from eq. (ii), we get
Pq₂ – Pq₁ = 48
Also, subtract eq. (ii) from eq. (iii), we get
pq₃ – pq₂ = 92
Also, subtract eq. (i) from eq. (iii), we get
Pq₃ – Pq₁ = 140
Now, to find greatest value of ‘p’ we find HCF of 48, 92 and 140 as,
48 = 2 x 2 x 2 x 2 x 3
92 = 2 x 2 x 2 x 2 x 2 x 3
and 140 = 2 x 2 x 5 x 7
So, HCF (48, 92 and 140) = 2 x 2 = 4
Greatest number that will divide 43, 91 and 183 as to leave the same remainder in each case = 4.

Question 17.
Solution:
Remainder in all the cases is 6, i.e.,
20 – 14 = 6
25 – 19 = 6
35 – 29 = 6
40 – 34 = 6
The difference between divisor and the corresponding remainder is 6.
Required number = (LCM of 20, 25, 35, 46) – 6 = 1400 – 6 = 1394

Question 18.
Solution:
Number of participants in Hindi = 60
Number of participants in English = 84
Number of participants in Mathematics =108
Minimum number of participants in one room = HCF of 60, 84 and 108 = 12
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 30

Question 19.
Solution:
Number of books in English = 336
Number of books in Mathematics = 240
Number of books in Science = 96
Minimum number of books of each topic in a stack = HCF of 336, 240 and 96 = 48
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 32

Question 20.
Solution:
Length of first piece of timber = 42 m
Length of second piece of timber = 49 m
and length of third piece of timber = 63 m
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 33

Question 21.
Solution:
Lengths are given as 7 m, 3 m 85 cm and 12 m 95 cm = 700 cm, 385 cm and 1295 cm
Greatest possible length that can be used to measure exactly = HCF of 700, 385, 1295 = 35 cm
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 35

Question 22.
Solution:
Number of pens =1001
and number of pencils = 910
Maximum number of pens and pencils equally distributed to the students = HCF of 1001 and 910 = 91
Number of students = 91
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 36

Question 23.
Solution:
Length of the room = 15 m 17 cm = 1517 cm
and breadth = 9 m 2 cm = 902 cm
Maximum side of square tile used = HCF of 1517 and 902 = 41 cm
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 37

Question 24.
Solution:
Measures of three rods = 64 cm, 80 cm and 96 cm
Least length of cloth that can be measured an exact number of times
= LCM of 64, 80, 96
= 960 cm
= 9 m 60 cm
= 9.6 m
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 38
LCM = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 = 960

Question 25.
Solution:
Beep made by first devices after every = 60 seconds
Second device after = 62 seconds
Period after next beep together = LCM of 60, 62
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 39
LCM = 2 x 30 x 31 = 1860 = 1860 seconds = 31 minutes
Time started beep together, first time together = 10 a.m.
Time beep together next time = 10 a.m. + 31 minutes = 10 : 31 a.m.

Question 26.
Solution:
The traffic lights of three roads change after
48 sec., 72 sec. and 108 sec. simultaneously
They will change together after a period of = LCM of 48 sec., 72 sec. and 108 sec.
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 40
= 7 minutes, 12 seconds
First time they light together at 8 a.m. i.e., after 8 hr.
Next time they will light together = 8 a.m. + 7 min. 12 sec. = 8 : 07 : 12 hrs.

Question 27.
Solution:
Tolling of 6 bells = 2, 4, 6, 8, 10, 12 minutes
They take time tolling together = LCM of 2, 4, 6, 8, 10, 12 = 120 minutes = 2 hours
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B 41
LCM of 2 x 2 x 2 x 3 x 5 = 120 min. (2 hr)
They will toll together after every 2 hours Total time given = 30 hours
Number of times, there will toll together in 30 hours = \(\frac { 30 }{ 2 }\) = 15 times
Total numbers of times = 15 + 1 (of starting time) = 16 times

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1B are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A

January 5, 2022June 15, 2018 by Prasanna

RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A.

Other Exercises

Question 1.
Solution:
For any two given positive integers a and b, there exist unique whole numbers q and r such that
a = bq + r, when 0 ≤ r < b.
Here, a is called dividend, b as divisor, q as quotient and r as remainder.
Dividend = (Divisor x Quotient) + Remainder.

Question 2.
Solution:
Using Euclid’s divison Lemma
Dividend = (Divisor x Quotient) + Remainder
= (61 x 27) + 32
= 1647 + 32
= 1679
Required number = 1679

Question 3.
Solution:
Let the required divisor = x
Then by Euclid’s division Lemma,
Dividend = (Divisor x Quotient) + remainder
1365 = x x 31 + 32
=> 1365 = 31x + 32
=> 31x= 1365 – 32 = 1333
x = \(\frac { 1331 }{ 31 }\) = 43
Divisor = 43

Question 4.
Solution:
(i) 405 and 2520
HCF of 405 and 2520 = 45
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A 1

Question 5.
Solution:
Let n be an arbitrary positive integer.
On dividing n by 2, let m be the quotient and r be the remainder, then by Euclid’s division lemma
n = 2 x m + r = 2m + r, 0 ≤ r < 2
n = 2m or 2m + 1 for some integer m.
Case 1 : When n = 2m, then n is even
Case 2 : When n = 2m + 1, then n is odd.
Hence, every positive integer is either even or odd.

Question 6.
Solution:
Let n be a given positive odd integer.
On dividing n by 6, let m be the quotient and r be the remainder, then by Euclid’s division Lemma.
n = 6m + r, where 0 ≤ r < 6 => n = 6m + r, where r = 0, 1, 2, 3, 4, 5
=> n = 6m or (6m + 1) or (6m + 2) or (6m + 3) or (6m + 4) or (6m + 5)
But n = 6m, (6m + 2) and (6m + 4) are even.
Thus when n is odd, it will be in the form of (6m + 1) or (6m + 3) or (6m + 5) for some integer m.

Question 7.
Solution:
Let n be an arbitrary odd positive integer.
On dividing by 4, let m be the quotient and r be the remainder.
So by Euclid’s division lemma,
n = 4m + r, where 0 ≤ r < 4
n = 4m or (4m + 1) or (4m + 2) or (4m + 3)
But 4m and (4m + 2) are even integers.
Since n is odd, so n ≠ 4m or n ≠ (4m + 2)
n = (4m + 1) or (4m + 3) for some integer m.
Hence any positive odd integer is of the form (4m + 1) or (4m + 3) for some integer m.

Question 8.
Solution:
Let a = n³ – n
=> a = n (n² – 1)
=> a = n (n – 1) (n + 1) [(a² – b²) = (a – b) (a + b)]
=> a = (n – 1 ) n (n + 1)
We know that,
(i) If a number is completely divisible by 2 and 3, then it is also divisible by 6.
(ii) If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
(iii) If one of the factor of any number is an even number, then it is also divisible by 2.
a = (n – 1) n (n + 1) [From Eq. (i)]
Now, sum of the digits
= n – 1 + n + n + 1 = 3n
= Multiple of 3, where n is any positive integer.
and (n – 1) n (n +1) will always be even, as one out of (n – 1) or n or (n + 1) must be even.
Since, conditions (ii) and (iii) is completely satisfy the Eq. (i).
Hence, by condition (i) the number n3 – n is always divisible by 6, where n is any positive integer.
Hence proved.

Question 9.
Solution:
Let x = 2m + 1 and y = 2m + 3 are odd positive integers, for every positive integer m.
Then, x² + y² = (2m + 1)² + (2m + 3)²
= 4m² + 1 + 4 m + 4m² + 9 + 12m [(a + b)² = a² + 2ab + b²]
= 8m² + 16m + 10 = even
= 2(4m² + 8m + 5) or 4(2m² + 4m + 2) + 1
Hence, x² + y² is even for every positive integer m but not divisible by 4.

Question 10.
Solution:
We find HCF (1190, 1145) using the following steps:
RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1A 3
(i) Since 1445 > 1190, we divide 1445 by 1190 to get 1 as quotient and 255 as remainder.
By Euclid’s division lemma, we get
1445 = 1190 x 1 + 255 …(i)
(ii) Since the remainder 255 ≠ 0, we divide 1190 by 255 to get 4 as a quotient and 170 as a remainder.
By Euclid’s division lemma, we get
1190 = 255 x 4 + 170 …(ii)
(iii) Since the remainder 170 ≠ 0, we divide 255 by 170 to get 1 as quotient and 85 as remainder.
By Euclid’s division lemma, we get
255 = 170 x 1 +85 …(iii)
(iv) Since the remainder 85 ≠ 0, we divide 170 by 85 to get 2 as quotient and 0 as remainder.
By Euclid’s division lemma, we get
170 = 85 x 2 + 0 …(iv)
The remainder is now 0, so our procedure steps
HCF (1190, 1445) = 85
Now, from (iii), we get
255 = 170 x 1 + 85
=> 85 = 255 – 170 x 1
= (1445 – 1190) – (1190 – 255) x 4
= (1445 – 1190) – (1190 – 255) x 4
= (1445 – 1190) x 2 + (1445 – 1190) x 4
= 1445 – 1190 x 2 + 1445 x 4 – 1190 x 4
= 1445 x 5 – 1190 x 6
= 1190 x (-6) + 1445 x 5
Hence, m = -6, n = 5

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.2

January 5, 2022June 15, 2018 by Prasanna

RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.2

Question 1.
Express the following rational numbers as decimals:
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.2 Q1.1
Solution:

Question 2.
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.2 Q2.1
Solution:

Question 3.
Look at several examples of rational numbers in the form \(\frac { p }{ q }\) (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations. Can you guess what property q must satisfy?
Solution:
The property is if the denominators have factors 2 or 5 or both, the decimal representation will be terminating e.g.
RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.2 Q3.1

Hope given RD Sharma Class 9 Solutions Chapter 1 Number Systems Ex 1.2 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

January 5, 2022June 15, 2018 by Prasanna

RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

Other Exercises

Mark the correct alternative in each of the following :
Question 1.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) √119 cm
Solution:
(d) Radius of a circle OP = 5 cm OQ = 12 cm, PQ is tangent
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 1
OP ⊥ PQ
In right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (12)² = (5)2 + PQ²
=> 144 = 25 + PQ²
PQ² = 144 – 25 = 119
PQ = √119

Question 2.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
(a) Let PQ be the tangent from Q to the circle with O as centre
PQ = 24 cm
OQ = 25 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 2
Let Radius OQ = r
OQ ⊥ PQ
Now in right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (25)² = r² + (24)²
=> 625 = r² + 576
=> r² = 625 – 576 = 49 = (7)²
r = 7
Radius of the circle = 7 cm

Question 3.
The length of the tangent from a point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is
(a) √7 cm
(b) 7 cm
(c) 5 cm
(d) 25 cm
Solution:
(c) Let AB be the tangent from A to the circle of centre O, then
OB = 3 cm
BA = 4 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 3
OB ⊥ BA
In right ∆OBA,
OA² = OB² + BA² (Pythagoras Theorem) = (3)² + (4)² = 9 + 16 = 25 = (5)²
OA = 5
Distance of A from the centre O = 5 cm

Question 4.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution:
(a) PA and PB are the tangents to the circle from P and ∠APB = 80°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 4
∠AOB = 180° – ∠APB = 180°- 80° = 100°
But OP is the bisector of ∠AOB
∠POA = ∠POB = \(\frac { 1 }{ 2 }\) ∠AOB
=> ∠POA = \(\frac { 1 }{ 2 }\) x 100° = 50°

Question 5.
If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then, ∠PTQ is equal to
(a) 60°
(b) 70°
(c) 80°
(d) 90°
Solution:
(b) TP and TQ are the tangents from T to the circle with centre O and OP, OQ are joined and ∠POQ = 110°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 5
But ∠POQ + ∠PTQ = 180°
=> 110° + ∠PTQ = 180°
=> ∠PTQ = 180° – 110° = 70°

Question 6.
PQ is a tangent to a circle with centre O at the point P. If ∆OPQ is an isosceles triangle, then ∠OQP is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(b) In a circle with centre O, PQ is a tangent to the circle at P and ∆OPQ is an isosceles triangle such that OP = PQ
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 6
OP is radius of the circle
OP ⊥ PQ
OP = PQ
∠POQ = ∠OQP
But ∠POQ = ∠PQO = 90° (OP ⊥ PQ)
∠OQP = ∠POQ = 45°

Question 7.
Two equal circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB =
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(d) Two circles with centres O and O’ touch each other at C externally
A common tangent is drawn which touches the circles at A and B respectively.
Join OA, O’B and O’O which passes through C
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 7
AO = BO’ (radii of the equal circle)
AB || OO’
=> AOO’B is a rectangle
Draw another common tangent through C which intersects AB at D, then DA = DC = DB
ADCO and BDCO’ are squares
AC and BC are the diagonals of equal square
AC = BC
∠DAC = ∠DBC = 45°
∠ACB = 90°

Question 8.
ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in ∆ABC. The radius of the circle is
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Solution:
(b) In a right ∆ABC, ∠B = 90°
BC = 6 cm, AB = 8 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 8
AC² = AB² + BC² (Pythagoras Theorem) = (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 cm
An incircle is drawn with centre 0 which touches the sides of the triangle ABC at P, Q and R
OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle
OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA
OPBQ is a square
Let r be the radius of the incircle
PB = BQ = r
AR = AP = 8 – r,
CQ = CR = 6 – r
AC = AR + CR
=> 10 = 8 – r + 6 – r
10 = 14 – 2r
=> 2r = 14 – 10 = 4
=> r = 2
Radius of the incircle = 2 cm

Question 9.
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120°, then ∠OPQ is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(c) PQ is a tangent to the circle with centre O, from P, QOR is the diameter and ∠POR = 120°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 9
OQ is radius and PQ is tangent to the circle
OQ ⊥ QP or ∠OQP = 90°
But ∠QOP + ∠POR = 180° (Linear pair)
=> ∠QOP + 120° = 180°
∠QOP = 180° – 120° = 60°
Now in ∆POQ
∠QOP + ∠OQP + ∠OPQ = 180° (Angles of a triangle)
=> 60° + 90° + ∠OPQ = 180°
=> 150° + ∠OPQ = 180°
=> ∠OPQ = 180° – 150° = 30°

Question 10.
If four sides of a quadrilateral ABCD are tangential to a circle, then
(a) AC + AD = BD + CD
(b) AB + CD = BC + AD
(c) AB + CD = AC + BC
(d) AC + AD = BC + DB
Solution:
(b) A circle is inscribed in a quadrilateral ABCD which touches the sides AB, BC, CD and DA at P, Q, R and S respectively then the sum of two opposite sides is equal to the sum of other two opposite sides
AB + CD = BC + AD

Question 11.
The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is
(a) √7 cm
(b) 2√7cm
(c) 10 cm
(d) 5 cm
Solution:
(b) Radius of the circle = 6 cm
and distance of the external point from the centre = 8 cm
Length of tangent = √{(8)² – (6)²}
= √(64 – 36) = √28
= √(4 x 7) = 2√7 cm

Question 12.
AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 12 cm
Solution:
(c) AB and CD are two common tangents to the two circles which touch each other externally at C and intersect AB in D
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 10
CD = 4 cm
DA and DC are tangents to the first circle from D
CD = AD = 4 cm
Similarly DC and DB are tangents to the second circle from D
CD = DB = 4 cm
AB = AD + DB = 4 + 4 = 8 cm

Question 13.
In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then,
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 11
(a) AD = AB + BC + CA
(b) 2AD = AB + BC + CA
(c) 3AD = AB + BC + CA
(d) 4AD = AB + BC + CA
Solution:
(b) AD, AE and BC are the tangents to the circle at D, E and F respectively
D and AE are tangents to the circle from A
AD = AE ……(i)
Similarly, CD = CF and BE = BF ….(ii)
Now AB + AC + BC = AE – BE + AD – CD + CF + BF
= AD – BE + AD – CD + BE + BE
= 2AD [From (i) and (ii)]
or 2 AD = AB + BC + CA

Question 14.
In the figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR =
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 12
(a) 8 cm
(b) 3 cm
(c) 2.5 cm
(d) 5 cm
Solution:
(d) In the figure, 0 is the centre of the circle
QR is tangent to the circle and QOS is a diameter SQ = 6 cm, QR = 4 cm

OQ = \(\frac { 1 }{ 2 }\) QS = \(\frac { 1 }{ 2 }\) x 6 = 3 cm
OQ is radius
OQ ⊥ QR
Now in right ∆OQR
OR² = QR² + QO² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OR = 5 cm

Question 15.
In the figure, the perimeter of ∆ABC is
(a) 30 cm
(b) 60 cm
(c) 45 cm
(d) 15 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 14
Solution:
(a) ∆ABC is circumscribed of circle with centre O
AQ = 4 cm, CP = 5 cm and BR = 6 cm
AQ and AR the tangents to the circle AQ = AR = 4 cm
Similarly BP and BR are tangents,
BP = BR = 6 cm
and CP and CQ are the tangents
CQ = CP = 5 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 15
AB = AR + BR = 4 + 6 = 10 cm
BC = BP + CP = 6 + 5 = 11 cm
AC = AQ + CQ = 4 + 5 = 9 cm
Perimeter of ∆ABC = AB + BC + AC = 10 + 11 + 9 = 30 cm

Question 16.
In the figure, AP is a tangent to the circle with centre O such that OP = 4 cm and ∠OPA = 30°. Then, AP =
(a) 2√2 cm
(b) 2 cm
(c) 2√3 cm
(d) 3√2 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 16
Solution:
(c) In the figure, AP is the tangent to the circle with centre O such that
OP = 4 cm, ∠OPA = 30°
Join OA, let AP = x

Question 17.
AP and AQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ =
(a) 12 cm
(b) 18 cm
(c) 24 cm
(d) 36 cm
Solution:
(c) OP is radius, PA is the tangent
OP ⊥ AP
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 18
Now in right ∆OAP,
OA² = OP² + AP²
(15)² = (9)² + AP²
225 = 81 + AP²
=> AP² = 225 – 81 = 144 = (12)²
AP = 12 cm
But AP = AQ = 12 cm (tangents from A to the circle)
AP + AQ = 12+ 12 = 24 cm

Question 18.
At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Solution:
(d) In the figure, PQ is diameter XPY is tangent to the circle with centre O and radius 5 cm
From P, at a distance of 8 cm AB is a chord drawn parallel to XY
To find the length of AB
Join OA
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 19
XY is tangent and OP is the radius
OP ⊥ XY or PQ ⊥ XY
AB || XY
OQ is ⊥ AB which meets AB at R
Now in right ∆OAR,
OA² = OR² + AR²
(5)² = (3)² + AR²
25 = 9 + AR²
=> AR² = 25 – 9 = 16 = (4)²
AR = 4 cm
But R is mid-point of AB
AB = 2 AR = 2 x 4 = 8 cm

Question 19.
If PT is tahgent drawn froth a point P to a circle touching it at T and O is the centre of the circle, then ∠OPT + ∠POT =
(a) 30°
(b) 60°
(c) 90°
(d) 180°
Solution:
(c) In the figure, PT is the tangent to the circle with centre O.
OP and OT are joined
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 20
PT is tangent and OT is the radius
OT ⊥ PT
Now in right ∆OPT
∠OTP = 90°
∠OPT + ∠POT = 180° – 90° = 90°

Question 20.
In the adjacent figure, if AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD =
(a) 5 cm
(b) 4 cm
(c) 6 cm
(d) 7 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 21
Solution:
(d) In the figure, ∆ABC is the circumscribed a circle
AB = 12 cm, BC = 8 cm and AC = 10 cm
Let AD = a, DB = b and EC = c, then
AF = a, BE = b and FC = c

But AB + BC + AC = 12 + 8 + 10 = 30
a + b + b + c + c + a = 30
=> 2 (a + b + c) = 30
a + b + c = 15
Subtracting BC or b + c from this a = 15 – 8 = 7
AD = 7 cm

Question 21.
In the figure, if AP = PB, then
(a) AC = AB
(b) AC = BC
(c) AQ = QC
(d) AB = BC
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 23
Solution:
(b) In the figure, AP = PB
But AP and AQ are the tangent from A to the circle

AP = AQ
Similarly PB = BR
But AP = PB (given)
AQ = BR ….(i)
But CQ and CR the tangents drawn from C to the circle
CQ = CR
Adding in (i)
AQ + CQ = BR + CR
AC = BC

Question 22.
In the figure, if AP = 10 cm, then BP =
(a) √91 cm
(b) √127 cm
(c) √119 cm
(d) √109 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 25
Solution:
(b) In the figure,
OA = 6 cm, OB = 3 cm and AP = 10 cm

OA is radius and AP is the tangent
OA ⊥ AP
Now in right ∆OAP
OP² = AP² + OA² = (10)² + (6)² = 100 + 36 = 136
Similarly BP is tangent and OB is radius
OP² = OB² + BP²
136 = (3)² + BP2
=> 136 = 9 + BP²
=> BP² = 136 – 9 = 127
BP = √127 cm

Question 23.
In the figure, if PR is tangent to the circle at P and Q is the centre of the circle, then ∠POQ =
(a) 110°
(b) 100°
(c) 120°
(d) 90°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 27
Solution:
(c) In the figure, PR is the tangent to the circle at P.
O is the centre of the circle ∠QPR = 60°

OP is the radius and PR is the tangent OPR = 90°
=> ∠OPQ + ∠QPR = 90°
=> ∠OPQ + 60° = 90°
=> ∠OPQ = 90° – 60° = 30°
OP = OQ (radii of the circle)
∠OQP = 30°
In ∆OPQ,
∠OPQ + ∠OQP + ∠POQ = 180°
=> 30° + 30° + ∠PQR = 180°
=> 60° + ∠POQ = 180°
∠POQ = 180° – 60° = 120°

Question 24.
In the figure, if quadrilateral PQRS circumscribes a circle, then PD + QB =
(a) PQ
(b) QR
(c) PR
(d) PS
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 29
Solution:
(a) In the figure, quadrilateral PQRS is circumscribed a circle

PD = PA (tangents from P to the circle)
Similarly QA = QB
PD + QB = PA + QA = PQ

Question 25.
In the figure, two equal circles touch each other at T, if QP = 4.5 cm, then QR =
(a) 9 cm
(b) 18 cm
(c) 15 cm
(d) 13.5 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 31
Solution:
(a) In the figure, two equal circles touch, each other externally at T
QR is the common tangent
QP = 4.5 cm

PQ = PT (tangents from P to the circle)
Similarly PT = PR
PQ = PT = PR
Now QR = PQ + PR = 4.5 + 4.5 = 9 cm

Question 26.
In the figure, APB is a tangent to a circle with centre O at point P. If ∠QPB = 50°, then the measure of ∠POQ is
(a) 100°
(b) 120°
(c) 140°
(d) 150°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 33
Solution:
(a) In the figure, APB is a tangent to the circle with centre O

∠QPB = 50°
OP is radius and APB is a tangent
OP ⊥ AB
=> ∠OPB = 90°
=> ∠OPQ + ∠QPB = 90°
∠OPQ + 50° = 90°
=> ∠OPQ = 90° – 50° = 40°
But OP = OQ
∠OPQ = OQP = 40°
∠POQ = 180°- (40° + 40°) = 180° – 80° = 100°

Question 27.
In the figure, if tangents PA and PB are drawn to a circle such that ∠APB = 30° and chord AC is drawn parallel to the tangent PB, then ∠ABC =
(a) 60°
(b) 90°
(c) 30°
(d) None of these
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 35
Solution:
(c) In the figure, PA and PB are the tangents to the circle with centre O

∠APB = 30°
Chord AC || BP,
AB is joined
PA = PB
∠PAB = ∠PBA
But ∠PAB + ∠PBA = 180° – 30° = 150°
=> ∠BPA + ∠PBA = 150°
=> 2 ∠PBA = 150°
=> ∠PBA = 75°
AC || BC
∠BAC = ∠PBA = 75°
But ∠PBA = ∠ACB = 75° (Angles in the alternate segment)
∠ABC = 180° – (75° + 75°) = 180° – 150° = 30°

Question 28.
In the figure, PR =
(a) 20 cm
(b) 26 cm
(c) 24 cm
(d) 28 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 37
Solution:
(b) In the figure, two circles with centre O and O’ touch each other externally
PQ and RS are the tangents drawn to the circles

OQ and O’S are the radii of these circles and
OQ = 3 cm, PQ = 4 cm O’S = 5 cm and SR = 12 cm
Now in right ∆OQP
OP² = (OQ)² + PQ² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OP = 5 cm
Similarly in right ∆RSO’
(O’R)² = (RS)² + (O’S)² = (12)² + (5)² = 144 + 25 = 169 = (13)²
O’R = 13 cm
Now PR = OP + OO’ + O’R = 5 + (3 + 5) + 13 = 26 cm

Question 29.
Two circles of same radii r and centres O and O’ touch each other at P as shown in figure. If OO’ is produced to meet the circle C (O’, r) at A and AT is a tangent to the circle C (O, r) such that O’Q ⊥ AT. Then AO : AO’ =
(a) \(\frac { 3 }{ 2 }\)
(b) 2
(c) 3
(d) \(\frac { 1 }{ 4 }\)
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 39
Solution:
(c) Two circles of equal radii touch each other externally at P. OO’ produced meets at A

From A, AT is the tangent to the circle (O, r)
O’Q ⊥ AT
Now AO : AO’ = 3r : r
= 3 : 1 = \(\frac { 3 }{ 1 }\)

Question 30.
Two concentric circles of radii 3 cm and 5 cm are given. Then length of chord BC which touches the inner circle at P is equal to
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 41
Solution:
(c) In the figure, two concentric circles of radii 3 cm and 5 cm with centre O
Chord BC touches the inner circle at P
Draw a tangent AB to the inner circle
Join OQ and OA

OQ is radius and AQB is the tangent
OQ ⊥ AB and OQ bisects AB
AQ = QB
Similarly, BP = PC or P is mid-point of BC
But BQ and BP are tangents from B
QB = BP = AQ
In right ∆OAQ,
OA² = AQ² + OQ²
(5)² = AQ² + (3)²
=> AQ² = (5)² – (3)²
=> AQ² = 25 – 9 = 16 = (4)²
AQ = 4 cm
BC = 2 BP = 2 BQ = 2 AQ = 2 x 4 = 8 cm

Question 31.
In the figure, there are two concentric, circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. If PR = 7.5 cm, then PS is equal to
(a) 10 cm
(b) 12 cm
(c) 15 cm
(d) 18 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 43
Solution:
(c) In the figure, two concentric circles with centre O
From a point P on the outer circle,
PRT and PQS are the tangents are drawn to the inner circle at R and Q respectively
PR = 7.5 cm
Join OR and OQ

PT is chord and OR is radius
R is mid-point of PT
Similarly Q is mid-point of PS
But PR = PQ (tangents from P)
PT = 2 PR and PS = 2 PQ
PS = 2 PQ = 2 PR = 2 x 7.5 = 15 cm

Question 32.
In the figure, if AB = 8 cm and PE = 3 cm, then AE =
(a) 11 cm
(b) 7 cm
(c) 5 cm
(d) 3 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 45
Solution:
(c) In the figure, AB and AC are the tangents to the circle from A
DE is another tangent drawn from P
AB = 8 cm, PE = 3 cm

AB = AC (tangents drawn from A to the circle)
Similarly PE = EC and DP = DB
Now AE = AC – CE = AB – PE = 8 – 3 = 5 cm

Question 33.
In the figure, PQ and PR are tangents drawn from P to a circle with centre O. If ∠OPQ = 35°, then
(a) a = 30°, b = 60°
(b) a = 35°, b = 55°
(c) a = 40°, b = 50°
(d) a = 45°, b = 45°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 47
Solution:
(b) In the figure, PQ and PR are the tangents drawn from P to the circle with centre O
∠OPQ = 35°
PO is joined

PQ = PR (tangents from P to the circle)
∠OPQ = ∠OPR
=> 35° = a
=> a = 35°
OQ is radius and PQ is tangent
OQ ⊥ PQ
=> ∠OQP = 90°
In ∆OQP,
∠POQ + ∠QPO = 90°
=> b + 35° = 90°
=> b = 90° – 35° = 55°
a = 35°, b = 55°

Question 34.
In the figure, if TP and TQ are tangents drawn from an external point T to a circle with centre O such that ∠TQP = 60°, then
(a) 25°
(b) 30°
(c) 40°
(d) 60°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 49
Solution:
(b) In the figure, TP and TQ are the tangents drawn from T to the circle with centre O
OP, OQ and PQ are joined
∠TQP = 60°
TP = TQ (Tangents from T to the circle)
∠TQP = ∠TPQ = 60°
∠PTQ = 180° – (60° + 60°) = 180° – 120° = 60°
and ∠POQ = 180° – ∠PTQ = 180° – 60° = 120°
But OP = OQ (radii of the same circle)
∠OPQ = ∠OQP
But ∠OPQ + ∠OQP = 180° – 120° = 60°
But ∠OPQ = 30°

Question 35.
In the figure, the sides AB, BC and CA of triangle ABC, touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is [CBSE 2012]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 50
(a) 11 cm
(b) 10 cm
(c) 14 cm
(d) 15 cm
Solution:
(b) In the figure,
PA = 4 cm, BP = 3 cm, AC = 11 cm
AP and AR are the tangents from A to the circle
AP = AR
=> AR = 4 cm
Similarly BP and BQ are tangents
BQ = BP = 3 cm
AC =11 cm
AR + CR = 11 cm
4 + CR =11 cm
CR = 11 – 4 = 7 cm
CQ and CR are tangents to the circle
CQ = CR = 7 cm
Now, BC = BQ + CQ = 3 + 7 = 10 cm

Question 36.
In the figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ∆EDF is [CBSE 2012]
(a) 18 cm
(b) 13.5 cm
(c) 12 cm
(d) 9 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 51
Solution:
(a) In ∆DEF
DF touches the circle at H
and circle touches ED and EF Produced at K and M respectively
EK = 9 cm
EK and EM are the tangents to the circle
EM = EK = 9 cm
Similarly DH and DK are the tangent
DH = DK and FH and FM are tangents
FH = FM
Now, perimeter of ∆DEF
= ED + DF + EF
= ED + DH + FH + EF
= ED + DK + EM + EF
= EK + EM
= 9 + 9
= 18 cm

Question 37.
In the figure DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF, then the radius of the circle is [CBSE 2013]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 52
(a) 3 cm
(b) 5 cm
(c) 4 cm
(d) 6 cm
Solution:
(b) If figure, DE and DF are tangents to the circle drawn from D.
A is the centre of the circle.
DE = 5 cm and DE ⊥ DF
Join AE, AF

DE is the tangent and AE is radius
AE ⊥ DE
Similarly, AF ⊥ DF
But ∠D = 90° (given)
AFDE is a square
AE = DE (side of square)
But DE = 5 cm
AE = 5 cm
Radius of circle is 5 cm

Question 38.
In the figure, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm, then the radius of the circle (in cm) is [CBSE 2013]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 54
(a) 11
(b) 18
(c) 6
(d) 15
Solution:
(a) In the figure, a circle touches the sides of a quadrilateral ABCD
∠B = 90°, OP = OQ = r
AB = 29 cm, AD = 23 cm, DS = 5 cm
∠B = 90°
BA is tangent and OQ is radius
∠OQB = 90°
Similarly OP is radius and BC is tangents
∠OPB = 90°
But ∠B = 90° (given)
PBQO is a square
DS = 5 cm
But DS and DR are tangents to the circles
DR = 5 cm
But AD = 23 cm
AR = 23 – 5= 18 cm
AR = AQ (tangents to the circle from A)
AQ = 18 cm
But AB = 29 cm
BQ = 29 – 18 = 11 cm
OPBQ is a square
OQ = BQ = 11 cm
Radius of the circle = 11 cm

Question 39.
In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 55

Question 40.
Two circles touch each other externally at P. AB is a common tangent to the circle touching them at A and B. The value of ∠APB is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(d) We have, AT = TP and TB = TP (Lengths of the tangents from ext. point T to the circles)
∠TAP = ∠TPA = x (say)
and ∠TBP = ∠TPB = y (say)
Also, in triangle APB,
x + x + x + y + y = 180°
=> 2x + 2y = 180°
=> x + y = 90°
=> ∠APB = 90°

Question 41.
In the figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR= 46, then ∠QOR equals
(a) 67°
(b) 134°
(c) 44°
(d) 46°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 56
Solution:
(b) ∠OQP = 90°
[Tangent is ⊥ to the radius through the point of contact]
∠ORP = 90°
∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° [Angle sum property of a quad.]
90° + 46° + 90° + ∠QOR = 360°
∠QOR = 360° – 90° – 46° – 90° = 134°

Question 42.
In the figure, QR is a common tangent to the given circles touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is [CBSE2014]
(a) 3.8
(b) 7.6
(c) 5.7
(d) 1.9
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 57
Solution:
(b) In the figure, QR is common tangent to the two circles touching each other externally at T
Tangent at T meets QR at P
PT = 3.8 cm
PT and PQ are tangents from P
PT = PQ = 3.8 cm
Similarly PT and PR are tangents
PT = PR = 3.8 cm
QR = 3.8 + 3.8 = 7.6 cm

Question 43.
In the figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, then x =
(a) 10
(b) 9
(c) 8
(d) 7 (CBSE 2014)
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 58
Solution:
(b) In the given figure,
ABCD is a quadrilateral circumscribe a circle and its sides AB, BC, CD and DA touch the circle at P, Q, R and S respectively
AB = x cm, BC = 7 cm, CR = 3 cm, AS = 5 cm
CR and CQ are tangents to the circle from C
CR = CQ = 3 cm
BQ = BC – CQ = 7 – 3 = 4 cm
BQ = and BP are tangents from B
BP = BQ = 4 cm
AS and AP are tangents from A
AP = AS = 5 cm
AB = AP + BP = 5 + 4 = 9 cm
x = 9 cm

Question 44.
If angle between two radii of a circle is 130°, the angle between the tangent at the ends of radii is (NCERT Exemplar)
(a) 90°
(b) 50°
(c) 70°
(d) 40°
Solution:
(b) O is the centre of the circle.
Given, ∠POQ = 130°
PT and QT are tangents drawn from external point T to the circle.
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 59
∠OPT = ∠OQT = 90° [Radius is perpendicular to the tangent at point of contact]
In quadrilateral OPTQ,
∠PTQ + ∠OPT + ∠OQT + ∠POQ = 360°
=> ∠PTQ + 90° + 90° + 130° = 360°
=> ∠PTQ = 360° – 310° = 50°
Thus, the angle between the tangents is 50°.

Question 45.
If two tangents inclined at a angle of 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to [NCERT Exemplar]
(a) \(\frac { 3\surd 3 }{ 2 }\) cm
(b) 6 cm
(c) 3 cm
(d) 3√3 cm
Solution:
(d) Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°.
Join OA and OP.
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 60
Also, OP is a bisector of line ∠APC
∠APO = ∠CPO = 30°
Also, OA ⊥ AP
Tangent at any point of a circle is perpendicular to the radius through the point of contact.

Hence, the length of each tangent is 3√3 cm

Question 46.
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is [NCERT Exemplar]
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Solution:
(b) Let O be the centre of two concentric circles C₁ and C₂, whose radii are r₁ = 4 cm and r₂ = 5 cm.
Now, we draw a chord AC of circle C₂, which touches the circle C₁ at B.
Also, join OB, which is perpendicular to AC. [Tangent at any point of circle is perpendicular to radius throughly the point of contact]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 62
Now, in right angled ∆OBC, by using Pythagoras theorem,
OC² = BC² + BO² [(hypotenuse)² = (base)² + (perpendicular)²]
=> 5² = BC² + 4²
=> BC² = 25 – 16 = 9
=> BC = 3 cm
Length of chord AC = 2 BC = 2 x 3 = 6 cm

Question 47.
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is [NCERT Exemplar]
(a) 4 cm
(b) 5 cm
(b) 6 cm
(d) 8 cm
Solution:
(d) First, draw a circle of radius 5 cm having centre O.
A tangent XY is drawn at point A.
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 63
A chord CD is drawn which is parallel to XY and at a distance of 8 cm from A.
Now, ∠OAY = 90°
[Tangent and any point of a circle is perpendicular to the radius through the point of contact]
∠OAY + ∠OED = 180°
[sum of cointerior is 180°]
=> ∆OED = 180°
Also, AE = 8 cm, Join OC
Now, in right angled ∆OBC
OC² = OE² + EC²
=> EC² = OC² – OE² [by Pythagoras theorem]
EC² = 5² – 3² [OC = radius = 5 cm, OE = AE – AO = 8 – 5 = 3 cm]
EC² = 25 – 9 = 16
=> EC = 4 cm
Hence, length of chord CD = 2 CE = 2 x 4 = 8 cm
[Since, perpendicular from centre to the chord bisects the chord]

Question 48.
From a point P which is at a distance 13 cm from the centre O of a circle of radius 5 cm, the pair of tangent PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is [NCERT Exemplar]
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²
Solution:
(a) Firstly, draw a circle of radius 5 cm having centre O.
P is a point at a distance of 13 cm from O.
A pair of tangents PQ and PR are drawn.
Thus, quadrilateral PQOR is formed.
OQ ⊥ QP [since, AP is a tangent line]
In right angled ∆PQO,
OP² = OQ² + QP²
=> 13² = 5² + QP²
=> QP² = 169 – 25 = 144 = 12²
=> QP = 12 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 64
Now, area of ∆OQP = \(\frac { 1 }{ 2 }\) x QP x QO = \(\frac { 1 }{ 2 }\) x 12 x 5 = 30 cm²
Area of quadrilateral QORP = 2 ∆OQP = 2 x 30 = 60 cm²

Question 49.
If PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°
Solution:
(a) Given, PA and PB are tangent lines.
PA = PB [Since, the length of tangents drawn from an ∠PBA = ∠PAB = θ [say]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 65
In ∆PAB,
∠P + ∠A + ∠B = 180°
[since, sum of angles of a triangle = 180°
50°+ θ + θ = 180°
2θ = 180° – 50° = 130°
θ = 65°
Also, OA ⊥ PA
[Since, tangent at any point of a circle is perpendicular to the radius through the point of contact]
∠PAO = 90°
=> ∠PAB + ∠BAO = 90°
=> 65° + ∠BAO = 90°
=> ∠BAO = 90° – 65° = 25°

Question 50.
The pair of tangents AP and AQ drawn from an external point to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm. The radius of the circle is [NCERT Exemplar]
(a) 10 cm
(b) 7.5 cm
(c) 5 cm
(d) 2.5 cm
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 66

Question 51.
In the figure, if ∠AOB = 125°, then ∠COD is equal to [NCERT Exemplar]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 67
(a) 45°
(b) 35°
(c) 55°
(d) 62\(\frac { 1 }{ 2 }\)°
Solution:
(c) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
∠AOB + ∠COD = 180°
=> ∠COD = 180° – ∠AOB = 180° – 125° = 55°

Question 52.
In the figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to [NCERT Exemplar]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 68
(a) 20°
(b) 40°
(c) 35°
(d) 45°
Solution:
(b) Given, AB || PR
∠ABQ = ∠BQR = 70° [alternate angles]
Also QD is perpendicular to AB and QD bisects AB.
In ∆QDA and ∆QDB
∠QDA = ∠QDB [each 90°]
AD = BD
QD = QD [common side]
∆ADQ = ∆BDQ [by SAS similarity criterion]
Then, ∠QAD = ∠QBD …(i) [c.p,c.t.]
Also, ∠ABQ = ∠BQR [alternate interior angle]
∠ABQ = 70° [∠BQR = 70°]
Hence, ∠QAB = 70° [from Eq. (i)]
Now, in ∆ABQ,
∠A + ∠B + ∠Q = 180°
=> ∠Q = 180° – (70° + 70°) = 40°

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

January 5, 2022June 15, 2018 by Prasanna

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Other Exercises

Mark the correct alternative in each of the following :

Question 1.
If 6 θ is an acute angle such that cos θ =
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 1
Solution:

Question 2.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 4
Solution:

Question 3.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 6

Solution:

Question 4.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 9
Solution:

Question 5.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 12
Solution:

Question 6.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 14
Solution:

Question 7.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 16
Solution:

Question 8.
If 6 is an acute angle such that tan2 6 = 8 \(\frac { 8 }{ 7 }\), then the value of
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 18
Solution:

Question 9.
If 3 cos θ = 5 sin 6, then the value of
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 21
Solution:

Question 10.
If tan2 45° – cos2 30° = x sin 45° cos 45°, then x =
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 26
Solution:

Question 11.
The value of cos217° – sin2 73° is
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 28
Solution:
cos2 17° – sin2 73° = cos2 (90° – 73°) – sin2 73°
= sin2 73° – sin2 73° = 0 (c)

Question 12.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 29
Solution:

Question 13.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 31
Solution:

Question 14.
If A and B are complementary angles then
(a) A – sin B
(b) cos A = cos B
(c) A = tan B
(d) sec A = cosec B
Solution:
∵ A and B are complementary angles
∴ A + B = 90°
⇒ A – 90° – B
sec A = sec (90° – B) = cosec B (d)

Question 15.
If x sin (90° – θ) cot (90° – 6) – cos (90° – θ), then x =
(a) 0
(b) 1
(c) -1
(d) 2,
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 33

Question 16.
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 34
Solution:

Question 17.
If angles A, B, C of a AABC form an increasing AP, then sin B =
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 36
Solution:

Question 18.
If 6 is an acute angle such that sec2 θ =
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 38
Solution:

Question 19.
The value of tan 1° tan 2° tan 3s tan 89° is
(a) 1
(b) -1
(c) 0
(d) None of these
Solution:
tan 1° tan 2° tan 3° tan 44° tan 45° tan 46° tan 89°
= tan 1° tan 2° tan 3° tan 44° tan 45° tan
(90° – 44°) tan (90° – 43°) tan (90° – 1°)’
= tan 1° tan 2° tan 3° tan 44° tan 45° cot
44° cot 43°….. cot 1°
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 40

Question 20.
The value of cos 1″ cos 2° cos 3° cos 180° is
(a) 1
(b) 0
(c) -1
(d) None of these
Solution:
None of these, because we deal here only an angle O<θ ≤ 90° (d)

Question 21.
The value of tan 10° tan 15° tan 75° tan 80° is
(a) -1
(b) 0
(c) 1
(d) None of these
Solution:
tan 10° tan 15° tan 75° tan 80°
= tan 10° tan 15° tan (90° – 15°) tan (90° – 10°) .
= tan 10° tan 15° cot 15° cot 10°
= tan 10° cot 10° tan 15° cot 15°
{∵ tan θ cot θ = 1}
= 1×1 = 1 (C)

Question 22.
The value of
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 41
Solution:

Question 23.
If 6 and 20 – 45° arc acute angles such that sin 0 = cos (20 – 45°), then tan 0 is equal to
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 44
Solution:

Question 24.
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2sin 3θ-√3 tan 3θ is equal to
(a) 1
(b) 0
(c) -1
(d) 1 + √3
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 46

Question 25.
If A + B = 90°, then
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 48

Question 26.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 49
Solution:

Question 27.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 52
Solution:

Question 28.
sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 54

Question 29.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 55
Solution:

Question 30.
If A, B and C are interior angles of a
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 57
Solution:

Question 31.
If cos θ = \(\frac { 2 }{ 3 }\), then 2 sec2 θ + 2 tan2 θ-7 is equal to
(a) 1
(b) 0
(c) 3
(d) 4
Solution:

Question 32.
tan 5° x tan 30° x 4 tan 85° is equal to
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 59
Solution:

Question 33.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 61
(a)-2
(b) 2
(c) 1
(d) 0
Solution:

Question 34.
In the figure, the value of cos Φ is
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 63
Solution:

Question 35.
In the figure, AD = 4 cm, BD = 3 cm and CB = 12 cm, find cot θ.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 65
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.