## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test

More Exercises

Question 1.
Using trigonometrical tables, find the values of :
(i) sin 48° 52′
(ii) cos 37° 34′
(iii) tan 18° 21′.
Solution:
Using tables, we find that
(i) sin 48° 52′ = .7524 + .0008 = .7532
(ii) cos 37° 34′ = .7934 – .0007 = .7927
(iii) tan 18° 21′ = .3307 + .0010 = .3317.

Question 2.
Use tables to find the acute angle θ, given that
(i) sin θ = 0.5766
(ii) cos θ = 0.2495
(iii) tan θ = 2.4523.
Solution:
Using table, we find that
(i) sin θ = 0.5766 = 0.5764 + 0.0002
= sin (35° 12’+ 1′)
= sin 35° 13′
θ = 35° 13′
(ii) cos θ = 0.2495 = 0.2487 + 0.0008
= cos (75° 36′ – 3′)
= cos 75° 33′
θ = 75° 33′
(iii) tan θ = 2.4523 = 2.4504 + 0.0019
= tan (67° 48′ + 1′)
= tan 67° 49′

Question 3.
If θ is acute and cos θ = 0.53, find the value of tan θ.
Solution:
From the table, we find that
cos θ = 0.53 = .5299 + .0001 = cos 58°
θ = 58°
and tan 58° = 1.6003

Question 4.
Find the value of: sin 22° 11′ + cos 57° 20′ – 2 tan 9° 9′.
Solution:
Using the tables, we find that
sin 22° 11′ = 0.3762 + 0.0014 = 0.3776
cos 57° 20′ = 0.5402 – 0.0005 = 0.5397
tan 9° 9′ = 0.1602 + 0.0009 = 0.1611
∴ sin 22° 11′ + cos 57° 20′ – 2 tan 9° 9′
= 0.3376 + 0.5397 – 0.1611 × 2
= 0.3776 + 0.5397 – 0.3222
= 0.9173 – 0.3222
= .5951.

Question 5.
If θ is acute and sin θ = 0.7547, find the value of: (i) θ (ii) cos θ (iii) 2 cos θ – 3 tan θ.
Solution:
Using the tables, we find that
(i) sin θ = 0.7547 = sin 49°
θ = 49°.
(ii) cos θ = cos 49° = 0.6561.
(iii) tan θ = tan 49° = 1.1504
2 cos θ – 3 tan θ
= 2 × θ.6561 – 3 × 1.1504
= 1.3122 – 3.4512
= – 2.1390

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19

More Exercises

Question 1.
Find the value of the following:
(i) sin 35° 22′
(ii) sin 71° 31′
(iii) sin 65° 20′
(iv) sin 23° 56′.
Solution:
(i) sin 35° 22′
Using the table of natural sines,
we see 35° in the horizontal line and for 18′,
in the vertical column, the value is 0.5779.
Now read 22′ – 18′ = 4′ in the difference column, the value is 10.
Adding 10 in 0.5779 + 10 = 0.5789,
we find sin 35° 22′ = 0.5789.
(ii) sin 71° 31′
Using the table of natural sines, we see 71° in the horizontal line
and for 30′ in the vertical column, the value is 0.9483 and for 31′ – 30′ = 1′,
we see in the mean difference column, the value is 1.
∴ sin 71° 31′ = 0.9483 + 1 = 0.9484.
(iii) sin 65° 20′
Using the table of natural sines, we see 65° in the horizontal line
and for 18′ in the vertical column, the value is .9085 and for 20′ – 18′ = 2′,
we see in the mean difference column. We find 2.
∴ sin 65° 20′ = 0.9085 + 2 = 0.9087 Ans.
(iv) sin 23° 56′
Using the table of natural sines, we see 23° in the horizontal line
and for 54′, we see in vertical column, the value is 0.4051
and for 56′ – 54′ = 2′ in the mean difference. It is 5.
∴ sin 23° 56′ = 0.4051 + 5 = 0.4056

Question 2.
Find the value of the following:
(i) cos 62° 27′
(ii) cos 3° 11′
(iii) cos 86° 40′
(iv) cos 45° 58′.
Solution:
(i) cos 62° 27′
From the table of natural cosines,
we see 62° in the horizontal line and 24′ in the vertical column, the value is .4633
and 27′ – 24′ = 3′ in the mean difference. Its value is 8.
∴ cos 62° 27′ = 0.4633 – 8 = 0.4625 Ans.
(ii) cos 3° 11′
From the table of natural cosines, we see 3° in the horizontal line
and 6′ in the vertical column, its value is 0.9985
and 11′ – 6′ = 5′ in the mean difference, its value is 1.
∴ cos 3° 11′ = 0.9985 – 1 = 0.9984 Ans.
(iii) cos 86° 40′
From the table of natural cosines, we see 86° in the horizontal line
and 36′ in the vertical column, its value is 0.0593
and for 40′ – 36′ = 4′ in the mean difference, it is 12.
cos 86° 40’= 0.0593 – 12 = 0 0581 Ans.
(iv) cos 45° 58′
From the table of natural cosines, we see 45° in the horizontal column
and 54′ in the vertical column, its value is 0.6959
and for 58′ – 54′ = 4′, in the mean difference, it is 8.’
cos 45° 58′ = 0.6959 – 8 = 0.6951

Question 3.
Find the value of the following :
(i) tan 15° 2′
(ii) tan 53° 14′
(iii) tan 82° 18′
(iv) tan 6° 9′.
Solution:
(i) tan 15° 2′
From the table of natural tangents, we see 15° in the horizontal line,
its value is 0.2679 and for 2′, in the mean difference, it is 6.
tan 15° 2′ = 0.2679 + 6 = 0.2685.
(ii) tan 53° 14′
From the table of natural tangents, we see 53° in the horizontal line
and 12′ in the vertical column, its value is 1.3367
and 14′ – 12′ = 2′ in the mean difference, it is 16.
∴ tan 53° 14′ = 1.3367 + 16 = 1 .3383 Ans.
(iii) tan 82° 18′
From the table of natural tangents, we see 82° in the horizontal line
and 18′ in the vertical column, its value is 7.3962.
∴ tan 82° 18’= 7.3962.
(iv) tan 6° 9′
From the table of natural tangents, we see 6° in the horizontal line
and 6′ in the vertical column, its value is .1069
and 9′ – 6′ = 3′, in the mean difference, it is 9.
tan 6°9′ = .1069 + 9 = .1078.

Question 4.
Use tables to find the acute angle θ, given that:
(i) sin θ = – 5789
(ii) sin θ = – 9484
(iii) sin θ = – 2357
(iv) sin θ = – 6371.
Solution:
(i) sin θ = – 5789
From the table of natural sines,
we look for the value (≤ 5789), which must be very close to it,
we find the value .5779 in the column 35° 18′ and in mean difference,
we see .5789 – .5779 = .0010 in the column of 4′.
θ = 35° 18’+ 4’= 35° 22′ Ans.
(ii) sin θ = . 9484
From the table of natural sines,
we look for the value (≤ 9484) which must be very close to it,
we find the value .9483 in the column 71° 30′
and in the mean differences,
we see .9484 – 9483 = 0001, in the column of 1′.
θ = 71° 30′ + 1′ = 71° 31′ Ans.
(iii) sin θ = – 2357
From the table of natural sines,
we look for the value (≤ 2357) which must be very close to it,
we find the value .2351 in the column 13° 36′ and in the mean difference,
we see .2357 – 2351 = .0006, in the column of 2′.
θ = 13° 36′ +2’= 13° 38′ Ans.
(iv) sin θ = .6371
From the table of natural sines,
we look for the value (≤ 6371) which must be very close to it,
we find the value .6361 in the column 39° 30′ and in the mean difference,
we see .6371 – .6361 = .0010 in the column of 4′.
θ = 39° 30′ + 4′ = 39° 34′

Question 5.
Use the tables to find the acute angle θ, given that:
(i) cos θ = .4625
(ii) cos θ = .9906
(iii) cos θ = .6951
(iv) cos θ = .3412.
Solution:
(i) cos θ = .4625
From the table of natural cosines,
we look for the value (≤ .4625) which must be very close to it,
we find the value .4617 in the column of 62° 30′ and in the mean difference,
we see .4625 – .4617 = .0008 which is in column of 3′.
θ = 62° 30′ – 3’= 62° 27′.
(ii) cos θ = .9906
From the table of cosines,
we look for the value (≤ .9906) which must be very close to it,
we find the value of .9905 in the column of 7° 54′ and in mean difference,
we see .9906 – 9905 = .0001 which is in column of 3′.
θ = 70 54′ – 3’= 7° 51′
(iii) cos θ = .6951
From the tables of cosines,
we look for the value (≤ 6951) which must be very close to it,
we find the value .6947 in the column of 46° and in mean difference,
.6951 – .6947 = 0.0004 which in the column of 2′.
θ = 46° – 2′ = 45° 58′ Ans.
(iv) cos θ = .3412
From the table of cosines,
we look for the value of (≤ .3412) which must be very close to it,
we find the value .3404 in the column of 70° 6′ and in the mean difference,
.3412 – 3404 = .0008 which is in the column of 3′.
θ = 70° 6′ – 3′ = 70° 3′

Question 6.
Use tables to find the acute angle θ, given that:
(i) tan θ = .2685
(ii) tan θ = 1.7451
(iii) tan θ = 3.1749
(iv) tan θ = .9347
Solution:
(i) tan θ = .2685
From the table of natural tangent,
we look for the value of (≤ .2685) which must be very close to it,
we find the value .2679 in the column of 15° and in the mean difference,
.2685 – .2679 which is in the column of 2′.
θ = 15° +2′ = 15° 2′ Ans.
(ii) tan θ = 1.7451
From the tables of natural tangents,
we look for the value of (≤ 1.7451) which must be very close to it,
we find the value 1.7391 in the column of 60°’ 6′
and in the mean difference 1.7451 + 1.7391 = 0.0060 which is in the column of 5′.
θ = 60° 6’+ 5’= 60° 11’Ans.
(iii) tan θ = 3.1749
From the tables of natural tangents,
we look for the value of (≤ 3.1749) which must be very close to it,
we find the value 3.1716 in the column of 72° 30′
and in the mean difference 3.1749 – 3.1716 = 0.0033 which is in the column of 1′.
θ = 720 30′ + 1′ = 72° 31′ Ans.
(iv) tan θ = .9347
From the tables of natural tangents,
we look for the value of (≤ .9347 which must be very close to it,
we find the value .9325 in the column of 43°
and in the mean difference .9347 – .9325 = 0.0022 which is in the column of 4′.
θ = 43° + 4′ = 43° 4′

Question 7.
Using trigonometric table, find the measure of the angle A when sin A = 0.1822.
Solution:
sin A = 0.1822
From the tables of natural sines,
we look for the value (≤ .1822) which must be very close to it,
we find the value .1822 in column 10° 30′.
A = 10° 30′

Question 8.
Using tables, find the value of 2 sin θ – cos θ when (i) θ = 35° (ii) tan θ = .2679.
Solution:
(i) θ = 35°
2 sin θ – cos θ = 2 sin 35° – cos 35°
= 2 x .5736 – .8192
(From the tables)
= 1.1472 – .8192 = 0.3280.
(ii) tan θ = .2679
From the tables of natural tangents,
we look for the value of ≤ .2679,
we find the value of the column 15°.
θ = 15°
Now, 2 sin θ – cos θ = 2 sin 15° – cos 15°
= 2 (.2588) – .9659 = 5136 – .9659
= -0.4483

Question 9.
If sin x° = 0.67, find the value of
(i) cos x°
(ii) cos x° + tan x°.
Solution:
sin x° = 0.67
From the table of natural sines,
we look for the value of (≤ 0.67) which must be very close to it,
we find the value .6691 in the column 42° and in the mean difference,
the value of 0.6700 – 0.6691 = 0.0009 which is in the column 4′.
θ = 42° + 4′ = 42° 4′
Now
(i) cos x° = cos 42° 4′ = .7431 – .0008
= 0.7423 Ans.
(ii) cos x° + tan x° = cos 42° 4′ + tan 42° 4′
= 0.7423 + .9025
= 1.6448

Question 10.
If θ is acute and cos θ = .7258, find the value of (i) θ (ii) 2 tan θ – sin θ.
Solution:
cos θ = .7258
From the table of cosines,
we look for the value of (≤ .7258) which must be very close to it,
we find the value .7254 in the column of 43° 30′
and in the mean differences the value of .7258 – .7254 = 0.0004
which in the column of 2′.
(i) θ = 43° 30′ – 2’= 43° 28′.
(ii) 2 tan θ – sin θ
= 2 tan43°28′ – sin43°28′
= 2 (.9479) – .6879
= 1.8958 – .6879
= 1.2079

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Ex 19, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

More Exercises

Question 1.
(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.
(ii) If θ is an acute angle and tan θ = $$\\ \frac { 8 }{ 15 }$$, find the value of sec θ + cosec θ.
Solution:
(i) θ is an acute angle.
cosec θ = √5

Question 2.
Evaluate the following:
(i) $$2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right)$$ – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
(ii) $$\frac { { sec }29^{ O } }{ { cosec }61^{ O } }$$ + 2 cot 8° cot 17° cot 45° cot 73°0 cot 82° – 3(sin2 38° + sin2 52°)
(iii) $$\frac { { sin }^{ 2 }{ 22 }^{ O }+{ sin }^{ 2 }{ 68 }^{ O } }{ { cos }^{ 2 }{ 22 }^{ O }+{ cos }^{ 2 }{ 68 }^{ O } }$$ + sin2 63° + cos 63° sin 27°
Solution:
(i) $$2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right)$$ – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

Question 3.
If $$\\ \frac { 4 }{ 3 }$$ (sec2 59° – cot2 31°) – $$\\ \frac { 2 }{ 2 }$$ sin 90° + 3tan2 56° tan2 34° = $$\\ \frac { x }{ 2 }$$, then find the value of x.
Solution:
Given
$$\\ \frac { 4 }{ 3 }$$ (sec2 59° – cot2 31°) – $$\\ \frac { 2 }{ 2 }$$ sin 90° + 3tan2 56° tan2 34° = $$\\ \frac { x }{ 2 }$$

Prove the following (4 to 11) identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

Question 4.
(i) $$\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA$$
(ii) $$\frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA$$
Solution:
(i) $$\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA$$
L.H.S = $$\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA }$$

Question 5.
(i) $$\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta$$
(ii) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.
Solution:
(i) $$\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta$$
L.H.S = $$\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 }$$

Question 6.
(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ
(ii) $$\frac { cot\theta }{ cosec\theta +1 } +\frac { cosec\theta +1 }{ cot\theta } =2sec\theta$$
Solution:
L.H.S. = sin2 θ + cos4 θ
= (1 – cos2 θ + cos4 θ
= 1 – cos2 θ + cos4 θ
= 1 – cos2 θ (1 – cos2 θ)

Question 7.
(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
(ii) $$\frac { 1 }{ sinA+cosA+1 } +\frac { 1 }{ sinA+cosA-1 } =secA+cosecA$$
Solution:
(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
L.H.S = sec4 A (1 – sin4 A) – 2 tan2 A

Question 8.
(i) $$\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1$$
(ii) (sec A – tan A)2 (1 + sin A) = 1 – sin A.
Solution:
(i) $$\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1$$
L.H.S = $$\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta$$

Question 9.
(i) $$\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA$$
(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
(iii) $$\frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }$$
Solution:
(i) $$\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA$$
L.H.S = $$\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA }$$

Question 10.
$$\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }$$
Solution:
$$\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }$$
L.H.S = $$\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA }$$

Question 11.
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
Solution:
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
L.H.S = 2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1

Question 12.
If cot θ + cos θ = m, cot θ – cos θ = n, then prove that (m2 – n2)2 = 16 run.
Solution:
cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii)

Question 13.
If sec θ + tan θ = p, prove that sin θ = $$\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }$$
Solution:
sec θ + tan θ = p,
prove that sin θ = $$\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }$$
$$\frac { 1 }{ cos\theta } +\frac { sin\theta }{ cos\theta } =p$$

Question 14.
If tan A = n tan B and sin A = m sin B, prove that cos2 A = $$\frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 }$$
Solution:
m = $$\\ \frac { sinA }{ sinB }$$
n = $$\\ \frac { tanA }{ tanB }$$

Question 15.
If sec A = $$x+ \frac { 1 }{ 4x }$$, then prove that sec A + tan A = 2x or $$\\ \frac { 1 }{ 2x }$$
Solution:
sec A = $$x+ \frac { 1 }{ 4x }$$
To prove that sec A + tan A = 2x or $$\\ \frac { 1 }{ 2x }$$

Question 16.
When 0° < θ < 90°, solve the following equations:
(i) 2 cos2 θ + sin θ – 2 = 0
(ii) 3 cos θ = 2 sin2 θ
(iii) sec2 θ – 2 tan θ = 0
(iv) tan2 θ = 3 (sec θ – 1).
Solution:
0° < θ < 90°
(i) 2 cos2 θ + sin θ – 2 = 0

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS

More Exercises

Choose the correct answer from the given four options (1 to 12) :

Question 1.
$${ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta }$$ is equal to
(a) 1
(b) -1
(c) sin2 θ
(d) sec2 θ
Solution:
$${ cot }^{ 2 }\theta -\frac { 1 }{ { sin }^{ 2 }\theta }$$
= $$\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta } -\frac { 1 }{ { sin }^{ 2 }\theta }$$

Question 2.
(sec2 θ – 1) (1 – cosec2 θ) is equal to
(a) – 1
(b) 1
(c) 0
(d) 2
Solution:
(sec2 θ – 1) (1 – cosec2 θ)
= $$\left( \frac { 1 }{ { cos }^{ 2 }\theta } -1 \right) \left( 1-\frac { 1 }{ { sin }^{ 2 }\theta } \right)$$

Question 3.
$$\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta }$$ is equal to
(a) 2 sin2 θ
(b) 2 cos2 θ
(c) sin2 θ
(d) cos2 θ
Solution:
$$\frac { { tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta }$$

= $$=\frac { { sin }^{ 2 }\theta }{ 1 } ={ sin }^{ 2 }\theta$$ (c)

Question 4.
(cos θ + sin θ)2 + (cos θ – sin θ)2 is equal to
(a) – 2
(b) 0
(c) 1
(d) 2
Solution:
(cos θ + sin θ)2 + (cos θ – sin θ)2
= cos2 θ + sin2 θ + 2 sin θ cos θ + cos2 θ + sin2 θ – 2 sin θ cos θ
= 2(sin2 θ + cos2 θ)
= 2 × 1 = 2 (d)
(∵ sin2 θ + cos2 θ = 1)

Question 5.
(sec A + tan A) (1 – sin A) is equal to
(a) sec A
(b) sin A
(c) cosec A
(d) cos A
Solution:
(sec A + tan A) (1 – sin A)

Question 6.
$$\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A }$$ is equal to
(a) sec2 A
(b) – 1
(c) cot2 A
(d) tan2 A
Solution:
$$\frac { { 1+tan }^{ 2 }A }{ { 1+cot }^{ 2 }A }$$

Question 7.
If sec θ – tan θ = k, then the value of sec θ + tan θ is
(a) $$1-\frac { 1 }{ k }$$
(b) 1 – k
(c) 1 + k
(d) $$\\ \frac { 1 }{ k }$$
Solution:
sec θ – tan θ = k
$$\frac { 1 }{ cos\theta } -\frac { sin\theta }{ cos\theta } =k$$

Question 8.
Which of the following is true for all values of θ (0° < θ < 90°):
(a) cos2 θ – sin2 θ = 1
(b) cosec2 θ – sec2 θ = 1
(c) sec2 θ – tan2 θ = 1
(d) cot2 θ – tan2 θ = 1
Solution:
∴ sec2 θ – tan2 θ = 1 is true for all values of θ as it is an identity.
(0° < θ < 90°) (c)

Question 9.
If θ is an acute angle of a right triangle, then the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is
(a) 0
(b) 2 sin θ cos θ
(c) 1
(d) 2 sin2 θ
Solution:
sin θ cos (90° – θ) + cos θ sin (90° – θ)
= sin θ sin θ + cos θ cos θ
{ ∵ sin(90° – θ) = cosθ, cos (90° – θ) = sin θ }
= sin2 θ + cos2 θ = 1 (c)

Question 10.
The value of cos 65° sin 25° + sin 65° cos 25° is
(a) 0
(b) 1
(b) 2
(d) 4
Solution:
cos 65° sin 25° + sin 65° cos 25°
= cos (90° – 25°) sin 25° + sin (90° – 25°) cos 25°
= sin 25° . sin 25° + cos 25° . cos 25°
= sin2 25° + cos2 25°
( ∵ sin2 θ + cos2 θ = 1)
= 1 (b)

Question 11.
The value of 3 tan2 26° – 3 cosec2 64° is
(a) 0
(b) 3
(c) – 3
(d) – 1
Solution:
3 tan2 26° – 3 cosec2 64°
= 3 tan2 26° – 3 cosec (90° – 26°)
= 3 tan2 26° – 3 sec2 26°

Question 12.
The value of $$\frac { sin({ 90 }^{ O }-\theta )sin\theta }{ tan\theta } -1$$ is
(a) – cot θ
(b) – sin2 θ
(c) – cos2 θ
(d) – cosec2 θ
Solution:
$$\frac { sin({ 90 }^{ O }-\theta )sin\theta }{ tan\theta } -1$$

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18

More Exercises

Question 1.
If A is an acute angle and sin A = $$\\ \frac { 3 }{ 5 }$$ find all other trigonometric ratios of angle A (using trigonometric identities).
Solution:
sin A = $$\\ \frac { 3 }{ 5 }$$
In ∆ABC, ∠B = 90°
AC = 5 and BC = 3

Question 2.
If A is an acute angle and sec A = $$\\ \frac { 17 }{ 8 }$$, find all other trigonometric ratios of angle A (using trigonometric identities).
Solution:
sec A = $$\\ \frac { 17 }{ 8 }$$ (A is an acute angle)
In right ∆ABC
sec A = $$\\ \frac { AC }{ AB }$$ = $$\\ \frac { 17 }{ 8 }$$
AC = 17, AB = 8

Question 3.
Express the ratios cos A, tan A and sec A in terms of sin A.
Solution:
cos A = $$\sqrt { { 1-sin }^{ 2 }A }$$
tan A = $$\frac { SinA }{ CosA } =\frac { sinA }{ \sqrt { { 1-sin }^{ 2 }A } }$$
sec A = $$\frac { 1 }{ cosA } =\frac { 1 }{ \sqrt { { 1-sin }^{ 2 }A } }$$

Question 4.
If tan A = $$\frac { 1 }{ \sqrt { 3 } }$$, find all other trigonometric ratios of angle A.
Solution:
tan A = $$\frac { 1 }{ \sqrt { 3 } }$$
In right ∆ABC,
tan A = $$\\ \frac { BC }{ AB }$$ = $$\frac { 1 }{ \sqrt { 3 } }$$

Question 5.
If 12 cosec θ = 13, find the value of $$\frac { 2sin\theta -3cos\theta }{ 4sin\theta -9cos\theta }$$
Solution:
12 cosec θ = 13
⇒ cosec θ = $$\\ \frac { 13 }{ 12 }$$
In right ∆ABC,
∠A = θ
cosec θ = $$\\ \frac { AC }{ BC }$$ = $$\\ \frac { 13 }{ 12 }$$

Without using trigonometric tables, evaluate the following (6 to 10) :

Question 6.
(i) cos² 26° + cos 64° sin 26° + $$\frac { tan{ 36 }^{ O } }{ { cot54 }^{ O } }$$
(ii) $$\frac { sec{ 17 }^{ O } }{ { cosec73 }^{ O } } +\frac { tan68^{ O } }{ cot22^{ O } }$$ + cos² 44° + cos² 46°
Solution:
Given that
(i) cos² 26° + cos 64° sin 26° + $$\frac { tan{ 36 }^{ O } }{ { cot54 }^{ O } }$$

Question 7.
(i) $$\frac { sin65^{ O } }{ { cos25 }^{ O } } +\frac { cos32^{ O } }{ sin58^{ O } }$$ – sin 28° sec 62° + cosec² 30° (2015)
(ii) $$\frac { sin29^{ O } }{ { cosec61 }^{ O } }$$ + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).
Solution:
given that
(i) $$\frac { sin65^{ O } }{ { cos25 }^{ O } } +\frac { cos32^{ O } }{ sin58^{ O } }$$ – sin 28° sec 62° + cosec² 30°

Question 8.
(i) $$\frac { { sin }35^{ O }{ cos55 }^{ O }+{ cos35 }^{ O }{ sin }55^{ O } }{ { cosec }^{ 2 }{ 10 }^{ O }-{ tan }^{ 2 }{ 80 }^{ O } }$$
(ii) $${ sin }^{ 2 }{ 34 }^{ O }+{ sin }^{ 2 }{ 56 }^{ O }+2tan{ 18 }^{ O }{ tan72 }^{ O }-{ cot }^{ 2 }{ 30 }^{ O }$$
Solution:
Given that
(i) $$\frac { { sin }35^{ O }{ cos55 }^{ O }+{ cos35 }^{ O }{ sin }55^{ O } }{ { cosec }^{ 2 }{ 10 }^{ O }-{ tan }^{ 2 }{ 80 }^{ O } }$$

Question 9.
(i) $${ \left( \frac { { tan25 }^{ O } }{ { cosec }65^{ O } } \right) }^{ 2 }+{ \left( \frac { { cot25 }^{ O } }{ { sec65 }^{ O } } \right) }^{ 2 }+{ 2tan18 }^{ O }{ tan }45^{ O }{ tan72 }^{ O }$$
(ii) $$\left( { cos }^{ 2 }25+{ cos }^{ 2 }65 \right) +cosec\theta sec\left( { 90 }^{ O }-\theta \right) -cot\theta tan\left( { 90 }^{ O }-\theta \right)$$
Solution:
(i) $${ \left( \frac { { tan25 }^{ O } }{ { cosec }65^{ O } } \right) }^{ 2 }+{ \left( \frac { { cot25 }^{ O } }{ { sec65 }^{ O } } \right) }^{ 2 }+{ 2tan18 }^{ O }{ tan }45^{ O }{ tan72 }^{ O }$$

Question 10.
(i) 2(sec² 35° – cot² 55°) – $$\frac { { cos28 }^{ O }cosec{ 62 }^{ O } }{ { tan18 }^{ O }tan{ 36 }^{ O }{ tan30 }^{ O }{ tan54 }^{ O }{ tan72 }^{ O } }$$
(ii) $$\frac { { cosec }^{ 2 }(90-\theta )-{ tan }^{ 2 }\theta }{ 2({ cos }^{ 2 }{ 48 }^{ O }+{ cos }^{ 2 }{ 42 }^{ O }) } -\frac { { 2tan }^{ 2 }{ 30 }^{ O }{ sec }^{ 2 }{ 52 }^{ O }{ sin }^{ 2 }{ 38 }^{ O } }{ { cosec }^{ 2 }{ 70 }^{ O }-{ tan }^{ 2 }{ 20 }^{ O } }$$
Solution:
(i) 2(sec² 35° – cot² 55°) – $$\frac { { cos28 }^{ O }cosec{ 62 }^{ O } }{ { tan18 }^{ O }tan{ 36 }^{ O }{ tan30 }^{ O }{ tan54 }^{ O }{ tan72 }^{ O } }$$

Question 11.
Prove that following:
(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
(ii) $$\frac { tan\theta }{ tan({ 90 }^{ O }-\theta ) } +\frac { sin({ 90 }^{ O }-\theta ) }{ cos\theta } ={ sec }^{ 2 }\theta$$
(iii) $$\frac { cos({ 90 }^{ O }-\theta )cos\theta }{ tan\theta } +{ cos }^{ 2 }({ 90 }^{ O }-\theta )=1$$
(iv) sin (90° – θ) cos (90° – θ) = $$\frac { tan\theta }{ { 1+tan }^{ 2 }\theta }$$
Solution:
(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
L.H.S. = cos θ sin (90° – θ) + sin θ cos (90° – θ)
= cos θ . cos θ + sin θ . sin θ
= cos2 θ + sin2 θ = 1 = R.H.S.

Prove that following (12 to 30) identities, where the angles involved are acute angles for which the trigonometric ratios as defined:

Question 12.
(i) (sec A + tan A) (1 – sin A) = cos A
(ii) (1 + tan2 A) (1 – sin A) (1 + sin A) = 1.
Solution:
(i) (sec A + tan A) (1 – sin A) = cos A
L.H.S. = (sec A + tan A) (1 – sin A)

Question 13.
(i) tan A + cot A = sec A cosec A
(ii) (1 – cos A)(1 + sec A) = tan A sin A.
Solution:
(i) tan A + cot A = sec A cosec A
L.H.S. = tan A + cot A

Question 14.
(i) $$\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA } =2{ cosec }^{ 2 }A$$
(ii) $$\frac { 1 }{ secA+tanA } +\frac { 1 }{ secA-tanA } =2{ sec }A$$
Solution:
(i) $$\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA } =2{ cosec }^{ 2 }A$$
L.H.S = $$\frac { 1 }{ 1+cosA } +\frac { 1 }{ 1-cosA }$$

Question 15.
(i) $$\frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA }$$
(ii) $$\frac { 1-{ tan }^{ 2 }A }{ { cot }^{ 2 }A-1 } ={ tan }^{ 2 }A$$
(iii) $$\frac { sinA }{ 1+cosA } =cosecA-cotA$$
Solution:
(i) $$\frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA }$$
L.H.S = $$\frac { sinA }{ 1+cosA }$$
(multiplying and dividing by (1 – cosA))

Question 16.
(i) $$\frac { secA-1 }{ secA+1 } =\frac { 1-cosA }{ 1+cosA }$$
(ii) $$\frac { { tan }^{ 2 }\theta }{ { (sec\theta -1) }^{ 2 } } =\frac { 1+cos\theta }{ 1-cos\theta }$$
(iii) $${ (1+tanA) }^{ 2 }+{ (1-tanA) }^{ 2 }=2{ sec }^{ 2 }A$$
(iv) $${ sec }^{ 2 }A+{ cosec }^{ 2 }A={ sec }^{ 2 }A{ .cosec }^{ 2 }A$$
Solution:
(i) $$\frac { secA-1 }{ secA+1 } =\frac { 1-cosA }{ 1+cosA }$$
L.H.S = $$\frac { secA-1 }{ secA+1 }$$

Question 17.
(i) $$\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } =2secA$$
(ii) $$\frac { tanA }{ secA-1 } +\frac { tanA }{ secA+1 } =2cosecA$$
Solution:
(i) $$\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA } =2secA$$
L.H.S = $$\frac { 1+sinA }{ cosA } +\frac { cosA }{ 1+sinA }$$

Question 18.
(i) $$\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } =2{ sec }^{ 2 }A$$
(ii) $$cotA-tanA=\frac { { 2cos }^{ 2 }A-1 }{ sinA-cosA }$$
(iii) $$\frac { cotA-1 }{ 2-{ sec }^{ 2 }A } =\frac { cotA }{ 1+tanA }$$
Solution:
(i) $$\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 } =2{ sec }^{ 2 }A$$
L.H.S = $$\frac { cosecA }{ cosecA-1 } +\frac { cosecA }{ cosecA+1 }$$

Question 19.
(i) $${ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta ={ tan }^{ 2 }\theta { sin }^{ 2 }\theta$$
(ii) $$\frac { cos\theta }{ 1-tan\theta } -\frac { { sin }^{ 2 }\theta }{ cos\theta -sin\theta } =cos\theta +sin\theta$$
Solution:
(i) $${ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta ={ tan }^{ 2 }\theta { sin }^{ 2 }\theta$$
L.H.S = $${ tan }^{ 2 }\theta -{ sin }^{ 2 }\theta$$

Question 20.
(i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ
(ii) 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ = cot4 θ – tan4 θ.
Solution:
(i) cosec4 θ – cosec2 θ = cot4 θ + cot2 θ
L.H.S = cosec4 θ – cosec2 θ

Question 21.
(i) $$\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) } =cot\theta$$
(ii) $$\frac { { tan }^{ 3 }\theta -1 }{ tan\theta -1 } ={ sec }^{ 2 }\theta +tan\theta$$
Solution:
(i) $$\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) } =cot\theta$$
L.H.S = $$\frac { 1+cos\theta -{ sin }^{ 2 }\theta }{ sin\theta (1+cos\theta ) }$$

Question 22.
(i) $$\frac { 1+cosecA }{ cosecA } =\frac { { cos }^{ 2 }A }{ 1-sinA }$$
(ii) $$\sqrt { \frac { 1-cosA }{ 1+cosA } } =\frac { sinA }{ 1+cosA }$$
Solution:
(i) $$\frac { 1+cosecA }{ cosecA } =\frac { { cos }^{ 2 }A }{ 1-sinA }$$
L.H.S = $$\frac { 1+cosecA }{ cosecA }$$

Question 23.
(i) $$\sqrt { \frac { 1+sinA }{ 1-sinA } } =tanA+secA$$
(ii) $$\sqrt { \frac { 1-cosA }{ 1+cosA } } =cosecA-cotA$$
Solution:
(i) $$\sqrt { \frac { 1+sinA }{ 1-sinA } } =tanA+secA$$
L.H.S = $$\sqrt { \frac { 1+sinA }{ 1-sinA } }$$

Question 24.
(i) $$\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA$$
(ii) $$\frac { cotAcotA }{ 1-sinA } =1+cosecA$$
Solution:
(i) $$\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } } =2cosecA$$
L.H.S = $$\sqrt { \frac { secA-1 }{ secA+1 } } +\sqrt { \frac { secA+1 }{ secA-1 } }$$

Question 25.
(i) $$\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } =2(secA+cosecA)$$
(ii) $${ sec }^{ 4 }A-{ tan }^{ 4 }A=1+2{ tan }^{ 2 }A$$
Solution:
(i) $$\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA } =2(secA+cosecA)$$
L.H.S = $$\frac { 1+tanA }{ sinA } +\frac { 1+cotA }{ cosA }$$

Question 26.
(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1
(ii) sec6 A – tan6 A = 1 + 3 tan2 A + 3 tan4 A
Solution:
(i) cosec6 A – cot6 A = 3 cot2 A cosec2 A + 1
L.H.S = cosec6 A – cot6 A

Question 27.
(i) $$\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 } =\frac { 1+cos\theta }{ sin\theta }$$
(ii) $$\frac { sin\theta }{ cot\theta +cosec\theta } =2+\frac { sin\theta }{ cot\theta -cosec\theta }$$
Solution:
(i) $$\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 } =\frac { 1+cos\theta }{ sin\theta }$$
L.H.S = $$\frac { cot\theta -cosec\theta -1 }{ cot\theta -cosec\theta +1 }$$

Question 28.
(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
(ii) (cosecA – sinA)(secA – cosA) sec2A = tanA
(iii) (cosecθ – sinθ)(secθ – cosθ)(tan θ + cotθ) = 1
Solution:
(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
L.H.S = (sinθ + cosθ)(secθ + cosecθ)

Question 29.
(i) $$\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } =2$$
(ii) $$\frac { { tan }^{ 2 }A }{ { 1+tan }^{ 2 }A } +\frac { cot^{ 2 }A }{ 1+{ cot }^{ 2 }A } =1$$
Solution:
(i) $$\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA } =2$$
L.H.S = $$\frac { { sin }^{ 3 }A+{ cos }^{ 3 }A }{ sinA+cosA } +\frac { { sin }^{ 3 }A-{ cos }^{ 3 }A }{ sinA-cosA }$$

Question 30.
(i) $$\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } =\frac { 1 }{ cosA } -\frac { 1 }{ secA-tanA }$$
(ii) $${ (sinA+secA) }^{ 2 }+{ (cosA+cosecA) }^{ 2 }={ (1+secA\quad cosecA) }^{ 2 }$$
(iii) $$\frac { tanA+sinA }{ tanA-sinA } =\frac { secA+1 }{ secA-1 }$$
Solution:
(i) $$\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA } =\frac { 1 }{ cosA } -\frac { 1 }{ secA-tanA }$$
L.H.S = $$\frac { 1 }{ secA+tanA } -\frac { 1 }{ cosA }$$

Question 31.
If sin θ + cos θ = √2 sin (90° – θ), show that cot θ = √2 + 1
Solution:
sin θ + cos θ = √2 sin (90° – θ)
sin θ + cos θ = √2 cos θ
dividing by sin θ

Question 32.
If 7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°, then find the value of θ.
Solution:
7 sin2 θ + 3 cos2 θ = 4, 0° ≤ θ ≤ 90°
3 sin2 θ + 3 cos2 θ + 4 sin2 θ = 4

Question 33.
If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.
Solution:
sec θ + tan θ = m and sec θ – tan θ = n
mn = (sec θ + tan θ) (sec θ – tan θ) = sec2 θ – tan2 θ = 1
(∴ sec2 θ – tan2 θ = 1)
Hence proved.

Question 34.
If x – a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 – y2 = a2 – b2.
Solution:
x – a sec θ + b tan θ and y = a tan θ + b sec θ
To prove that x2 – y2 = a2 – b2.

Question 35.
If x = h + a cos θ and y = k + a sin θ, prove that (x – h)2 + (y – k)2 = a2.
Solution:
x = h + a cos θ and y = k + a sin θ
To prove that (x – h)2 + (y – k)2 = a2.

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

More Exercise

Question 1.
A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs Rs 300 per m2, find the cost of the tin for making the container.
Solution:
Height of container opened at the top (h) = 1 m = 100 cm
and diameter = 70 cm
∴Radius (r) = $$\\ \frac { 70 }{ 2 }$$ = 35 cm
∴Total surface area = 2πrh + πr2

Question 2.
A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross-section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.
Solution:
Dimensions of the wooden cuboid = 30 cm × 14 cm × 14 cm
Volume = 30 × 14 × 14 = 5880 cm3

Question 3.
Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14.
Solution:
Slant height of a cone (l) = 17 cm
Diameter of base = 30 cm
Radius (r) = $$\\ \frac { 30 }{ 2 }$$ = 15 cm

Question 4.
Find the volume of a cone given that its height is 8 cm and the area of base 156 cm2.
Solution:
Height of a cone = 8 cm
Area of base = 156 cm
.’. Volume = $$\\ \frac { 1 }{ 3 }$$ × area of base × height

Question 5.
The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.
Solution:
Circumference of the edge of bowl = 132 cm

Question 6.
The volume of a hemisphere is $$2425 \frac { 1 }{ 2 }$$ cm2. Find the curved surface area.
Solution:
Volume of a hemisphere = $$2425 \frac { 1 }{ 2 }$$ cm3
= $$\\ \frac { 4851 }{ 2 }$$ cm3

Question 7.
A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy
Solution:
A wooden solid toy is of a shape of a right circular cone
mounted on a hemisphere.
Radius of hemisphere (r) = 4.2 cm
Total height = 10.2 cm

Question 8.
A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. Find the capacity of the capsule.
Solution:
Diameter of cylindrical part = 0.5 cm
Total length of the capsule = 2 cm

Question 9.
A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.
Solution:
Radius of cylinder = $$\\ \frac { 7 }{ 2 }$$cm
and height of cylinder = 19 – 2 × $$\\ \frac { 7 }{ 2 }$$ cm
= 19 – 7 = 12 cm
and radius of hemisphere = $$\\ \frac { 7 }{ 2 }$$ cm

Question 10.
The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm , find its slant height. (Take π = 3.14).
Solution:
Let radius of cone (r) = 5x
then height (h) = 12x

Question 11.
A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.
Solution:
Let height of cone and cylinder = h
Diameter of the base of cone = 3x
Diameter of base of cylinder = 2x

Question 12.
A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.
Solution:
Radius of the cone (r) = 9 cm
Height of the cone (h) = 10 cm
Volume of water filled in cone

Question 13.
An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.
Solution:
Radius of the base of cone = 8 cm

Question 14.
A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of the tent if the canvas used costs Rs 36 per square metre.
Solution:
Diameter of the cylindrical part = 126 m
Radius (r) = $$\\ \frac { 126 }{ 2 }$$ = 63m

Question 15.
The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their
(i) curved surfaces
(ii) volumes.
Solution:
Radius of the base of a cone (r) = 3 cm
Height (h) = 4 cm

Question 16.
A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.
Solution:
Radius of base of a cone (r) = 2. 1 cm
and height (h) = 8.4 cm

Question 17.
How many lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.
Solution:
Dimensions of a solid rectangular lead piece
= 66 cm × 42 cm × 21 cm
.’. Volume = 66 × 42 × 21 cm3
Diameter of a lead shot = 4.2 cm

Question 18.
Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.
Solution:
Radius of a cylinder (r) = 3 cm
Height (h) = 5 cm

Question 19.
A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.
Solution:
Radius of hemisphere = 8 cm
Volume = $$\frac { 2 }{ 3 } \pi { r }^{ 3 }{ cm }^{ 3 }$$

Question 20.
A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of the water in the cylinder.
Solution:
Radius of hemispherical bowl = 6 cm
.’. Volume of the water in the bowl

Question 21.
The diameter of a metallic sphere is 42 cm. It is metled and drawn into a cylindrical wire of 28 cm diameter. Find the length of the wire.
Solution:
Diameter of sphere = 42 cm
Radius of sphere =$$\\ \frac { 42 }{ 2 }$$ = 21 cm

Question 22.
A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Solution:
Radius of sphere = $$\\ \frac { 6 }{ 2 }$$ = 3 cm

Question 23.
A solid sphere of radius 6 cm is metled into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
Solution:
Radius of solid sphere = 6 cm
Volume of solid sphere = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$

Question 24.
A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical vessel, full of water, in such a Way that the whole solid is submerged in water. If the radius of the cylindrical vessel is 5 cm and its height is 10.5 cm, find the volume of water left in the cylindrical vessel.
Solution:
Radius of hemisphere (r) = 3.5 cm
Height of cone (h1) = 4 cm

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS

More Exercise

Choose the correct answer from the given four options (1 to 32) :

Question 1.
In a cylinder, if radius is halved and height is doubled then the volume will be
(a) same
(b) doubled
(c) halved
(d) four times
Solution:
Let radius of cylinder = r
and height = h
then volume = πr²h
If the radius is halved and the height is doubled

Question 2.
In a cylinder, if the radius is doubled &nd height is halved then its curved surface area will be
(a) halved
(b) doubled
(c) same
(d) four times
Solution:
Let radius of a cylinder = r
and height = h
Then curved surface area = 2πrh
Now if radius is doubled and height is halved,
then curved surface area = 2π$$\\ \frac { r }{ 2 }$$ × 2h = 2πrh
which is same (c)

Question 3.
If a well of diameter 8 m has been dug to the depth of 14 m, then the volume of the earth dug out is
(a) 352 m3
(b) 704 m3
(c) 1408 m3
(d) 2816 m3
Solution:
Diameter of a well = 8 m
Radius (r) = $$\\ \frac { 8 }{ 2 }$$ = 4m
Depth (h) = 14 m
Volume of the earth dug put = πr2h
= $$\\ \frac { 22 }{ 7 }$$ × 4 × 4 × 14 m3
= 704 m3 (b)

Question 4.
If two cylinders of the same lateral surface have their radii in the ratio 4 : 9, then the ratio of their heights is
(a) 2 : 3
(b) 3 : 2
(c) 4 : 9
(d) 9 : 4
Solution:
Ratio in two cylinder having same lateral surface area their radii is 4 : 9
Let r1 be the radius of the first and r2 be the second cylinder
and h1, h2 and their heights

Ratio in their heights = 9 : 4 (d)

Question 5.
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
(a) 10 : 17
(b) 20 : 27
(c) 17 : 27
(d) 20 : 37
Solution:
Radii of two cylinder are in the ratio = 2 : 3
Ratio in their height = 5 : 3
Let height of the first cylinder = 5y
and of second = 3y
Now, volume of the first cylinder

Question 6.
The total surface area of a cone whose radius is $$\\ \frac { r }{ 2 }$$ and slant height 2l is
(a) 2πr (l + r)
(b) $$\pi r\left( l+\frac { r }{ 4 } \right)$$
(c) πr(l + r)
(d) 2πrl
Solution:
Radius of a cone = $$\\ \frac { r }{ 2 }$$
and slant height = 2l
total surface area of a cone

Question 7.
If the diameter of the base of cone is 10 cm and its height is 12 cm, then its curved surface area is
(a) 60π cm2
(b) 65π cm2
(c) 90π cm2
(d) 120π cm2
Solution:
Diameter of the base of a cone = 10 cm
Radius (r) = $$\\ \frac { 10 }{ 2 }$$ = 5 cm
and height (h) = 12 cm

Question 8.
If the diameter of the base of a cone is 12 cm and height is 20 cm, then its volume is ,
(a) 240π cm3
(b) 480π cm3
(c) 720π cm3
(d) 960π cm3
Solution:
Diameter of the base of a cone = 12 cm
Radius (r) = $$\\ \frac { 12 }{ 2 }$$ = 6 cm
and height (h) = 20 cm

Question 9.
If the radius of a sphere is 2r, then its volume will be
(a) $$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$
(b) $$4\pi { r }^{ 3 }$$
(c) $$\frac { 8\pi { r }^{ 3 } }{ 3 }$$
(d) $$\frac { 32\pi { r }^{ 3 } }{ 3 }$$
Solution:
Radius of a sphere = 2r

Question 10.
If the diameter of a sphere is 16 cm, then its surface area is
(a) 64π cm2
(b) 256π cm2
(c) 192π cm2
(d) 256 cm2
Solution:
Diameter of a sphere = 16 cm
Radius (r) = $$\\ \frac { 16 }{ 2 }$$ = 8 cm
Surface area = 4π2 = 4π × 8 × 8 cm2 = 256π cm2 (b)

Question 11.
If the radius of a hemisphere is 5 cm, then its volume is
(a) $$\frac { 250 }{ 3 } \pi { \quad cm }^{ 3 }$$
(b) $$\frac { 500 }{ 3 } \pi { \quad cm }^{ 3 }$$
(c) $$75\pi { \quad cm }^{ 3 }$$
(d) $$\frac { 125 }{ 3 } \pi { \quad cm }^{ 3 }$$
Solution:
Radius of a hemisphere (r) = 5 cm

Question 12.
If the ratio of the diameters of the two spheres is 3 : 5, then the ratio of their surface areas is
(a) 3 : 5
(b) 5 : 3
(c) 27 : 125
(d) 9 : 25
Solution:
Ratio in the diameters of two spheres = 3 : 5
Let radius of the first sphere = 3x cm
and radius of the second sphere = 5x cm
Ratio in their surface area

Question 13.
The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 2 : 1
Solution:
Radius of balloon (hemispherical) in the original position = 6 cm
and in increased position = 12 cm
Ratio in their surface areas

Question 14.
The shape of a Gilli, in the game of Gilli- danda, is a combination of

(a) two cylinders
(b) a cone and a cylinders
(c) two cones and a cylinder
(d) two cylinders and a cone
Solution:
The shape of a Gilli is the combination of
two cones and a cylinder (as shown in the figure). (c)

Question 15.
If two solid hemisphere of same base radius r are joined together along with their bases, then the curved surface of this new solid is
(a) 4πr2
(b) 6πr2
(c) 3πr2
(d) 8πr2
Solution:
Radius of two solid hemispheres = r
These are joined together along with the bases
Curved surface area = 2π2 × 2 = 4πr2 (a)

Question 16.
During conversion of a solid from one shape to another, the volume of the new shape will
(a) increase
(b) decrease
(c) remain unaltered
(d) be doubled
Solution:
During the conversion of a solid into another,
the volume of the new shaper will be the same.
i.e. remain unaltered (c)

Question 17.
If a solid of one shape is converted to another, then the surface area of the new solid
(a) remains same
(b) increases
(c) decreases
(d) can’t say
Solution:
If a solid of one shape has conversed into another then
the surface area of the new solid will same or not same
i.e. can’t say. (d)

Question 18.
If a marble of radius 2.1 cm is put into a cylindrical cup full of water of radius 5 cm and height 6 cm, then the volume of water that flows out of the cylindrical cup is
(a) 38.8 cm3
(b) 55.4 cm3
(c) 19.4 cm3
(d) 471.4 cm3
Solution:
Radius of a marble = 2.1 cm
Volume of marble = $$\frac { 4 }{ 3 } \pi { { r }^{ 3 }\quad cm }^{ 3 }$$
= $$\\ \frac { 4 }{ 3 }$$ x $$\\ \frac { 22 }{ 7 }$$ × 2.1 × 2.1 × 2.1 cm3
= 38.88 cm3
= 38.8 cm3 (a)

Question 19.
The volume of the largest right circular cone that can be carved out from a cube of edge 4.2 cm is
(a) 9.7 cm3
(b) 77.6 cm3
(c) 58.2 cm3
(d) 19.4 cm3
Solution:
Edge of cube = 4.2 cm
Radius of largest cone cut out = $$\\ \frac { 42.2 }{ 2 }$$ = 2.1 cm
and height = 4.2 cm
Volume = $$\frac { 1 }{ 3 } \pi { { r }^{ 2 }h }$$
= $$\\ \frac { 1 }{ 3 }$$ x $$\\ \frac { 22 }{ 7 }$$ × 2.1 × 2.1 × 4.2 cm3
= 19.404
= 19.4 cm3 (d)

Question 20.
The volume of the greatest sphere cut off from a circular cylindrical wood of base radius 1 cm and height 6 cm is
(a) 288 π cm3
(b) $$\frac { 4 }{ 3 } \pi$$ cm3
(c) 6 π cm3
(d) 4 π cm3
Solution:
Radius of cylinder (r) = 1 cm
Height (h) = 6 cm
The largest sphere that can be cut off from the cylinder of radius 1 cm

Question 21.
The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is
(a) 3 : 4
(b) 4 : 3
(c) 9 : 16
(d) 16 : 9
Solution:
Ratio in volumes of two spheres = 64 : 27

Question 22.
If a cone, a hemisphere and a cylinder have equal bases and have same height, then the ratio of their volumes is
(a) 1 : 3 : 2
(b) 2 : 3 : 1
(c) 2 : 1 : 3
(d) 1 : 2 : 3
Solution:
If a cone, a hemisphere and a cylinder have equal bases = r (say)
and height = h in each case and r = h

Question 23.
If a sphere and a cube have equal surface areas, then the ratio of the diameter of the sphere to the edge of the cube is
(a) 1 : 2
(b) 2 : 1
(c) √π : √6
(d) √6 : √π
Solution:
A sphere and a cube have equal surface area
Let a be the edge of a cube and r be the radius of the sphere, then

Question 24.
A solid piece of iron in the form of a cuboid of dimensions 49 cm x 33 cm x 24 cm is moulded to form a sphere. The radius of the sphere is
(a) 21 cm
(b) 23 cm
(c) 25 cm
(d) 19 cm
Solution:
Dimension of a cuboid = 49 cm × 33 cm × 24 cm
Volume of a cuboid = 49 × 33 × 24 cm3
⇒ Volume of sphere = Volume of a cuboid
Volume of a sphere = 49 × 33 × 24 cm3

Question 25.
If a solid right circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of a sphere, then the radius of the sphere is
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 12 cm
Solution:
Height of a circular cone (h) = 24 cm
and radius (r) = 6 cm

Question 26.
If a solid circular cylinder of iron whose diameter is 15 cm and height 10 cm is melted and recasted into a sphere, then the radius of the sphere is
(a) 15 cm
(b) 10 cm
(c) 7.5 cm
(d) 5 cm
Solution:
Diameter of a cylinder = 15 cm
Radius = $$\\ \frac { 15 }{ 2 }$$ cm
and height = 10 cm

Question 27.
The number of balls of radius 1 cm that can be made from a sphere of radius 10 cm is
(a) 100
(b) 1000
(c) 10000
(d) 100000
Solution:
Radius of sphere (R) = 10 cm
Volume of sphere = $$\frac { 4 }{ 3 } \pi { R }^{ 3 }=\frac { 4 }{ 3 } \pi { (10) }^{ 3 }{ cm }^{ 3 }$$

Question 28.
A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm
Solution:
The internal diameter of the metallic shell = 4 cm
and external diameter = 8 cm

Question 29.
A cubical icecream brick of edge 22 cm is to be distributed among some children by filling icecream cones of radius 2 cm and height 7 cm upto its brim. The number of children who will get the icecream cones is
(a) 163
(b) 263
(c) 363
(d) 463
Solution:
Edge of a cubical icecream brick = 22 cm
Volume = a3 = (22)3 = 10648 cm3
Radius (r) of ice cream cone (r) = 2 cm
and height (h) = 7 cm

Question 30.
Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is
(a) 4 cm
(b) 3 cm
(b) 2 cm
(d) 6 cm
Solution:
Diameter of cylinder = 2 cm
Radius = $$\\ \frac { 2 }{ 2 }$$ = 1 cm
and height = 16 cm

Question 31.
A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that $$\\ \frac { 1 }{ 8 }$$ space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is
(a) 142296
(b) 142396
(c) 142496
(d) 142596
Solution:
Internal edge of a hollow cube = 22 cm
Volume = (side)3 = (22)3 = 22 × 22 × 22 cm3 = 10648 cm3
Diameter of spherical marble = 0.5 cm = $$\\ \frac { 1 }{ 2 }$$

Question 32.
In the given figure, the bottom of the glass has a hemispherical raised portion. If the glass is filled with orange juice, the quantity of juice which a person will get is
(a) 135 π cm3
(b) 117 π cm3
(c) 99 π cm3
(d) 36 π cm3

Solution:
Radius of base of cylinder (r) = $$\\ \frac { 6 }{ 2 }$$ cm = 3 cm
and height (h)= 15 cm

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5

More Exercise

Question 1.
The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 m, find its radius.
Solution:
Diameter of metallic sphere = 6 cm
Radius(r) = $$\\ \frac { 6 }{ 2 }$$ = 3cm

Question 2.
The radius of a sphere is 9 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire in metres.
Solution:
Radius of sphere = 9 cm
Volume = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi \times { \left( 9 \right) }^{ 3 }{ cm }^{ 3 }$$

h = 97200 cm = 972 m
Length of wire = 972 m

Question 3.
A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.
Solution:
Radius of a solid hemisphere (r) = 8 cm
Volume = $$\frac { 2 }{ 3 } \pi { r }^{ 3 }=\frac { 2 }{ 3 } \pi \times { \left( 8 \right) }^{ 3 }{ cm }^{ 3 }$$

Question 4.
A rectangular water tank of base 11 m x 6 m contains water upto a height of 5 m. if the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.
Solution:
Base of a water tank = 11 m × 6 m
Height of water level in it (h) = 5 m
Volume of water =11 × 6 × 5 = 330 m³
Volume of water in the cylindrical tank

Question 5.
The rain water from a roof of dimensions 22 m x 20 m drains into a cylindrical vessel having diameter of base 2 m and height; 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.
Solution:
Dimensions of roof = 22 m × 20 m
Let rainfall = x m
.’. Volume of water = 22 × 20 × x m³
Volume of water in cylinder = 22 × 20 × x m³
Diameter of its base = 2 m

Question 6.
The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.
Solution:
Diameter of a cylinder = 40 cm
Radius (r) = $$\\ \frac { 40 }{ 2 }$$ = 20 cm
Height(h) = 9 cm

Question 7.
Eight metallic spheres, each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new (single) sphere.
Solution:
Radius of each metallic sphere (r) = 2 cm
Volume of one sphere = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$

Question 8.
A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.
Solution:
Height of disc cylindrical shaped = 2.5 mm
and base radius = 12 cm
Volume of the disc = πr²h

Question 9.
Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.
Solution:
Weight of first sphere = 1 kg
and weight of second sphere = 7 kg
Radius of smaller sphere = 3 cm
Let r be the radius of a larger sphere

R = 3 x 2 = 6 cm
Diameter of big sphere = 2 x 6 = 12 cm

Question 10.
A hollow copper pipe of inner diameter 6 cm and outer diameter 10 cm is melted and changed into a solid circular cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.
Solution:
Inner diameter of a hollow pipe = 6 cm
and outer diameter = 10 cm
Inner radius (r) = $$\\ \frac { 6 }{ 2 }$$ = 3cm

Question 11.
A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm and height is 72 cm, find the uniform thickness of the cylinder.
Solution:
Radius of a solid sphere (r) = 6 cm
Volume = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi \times { \left( 6 \right) }^{ 3 }{ cm }^{ 3 }$$

Question 12.
A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal of the tube is $$\\ \frac { 1 }{ 2 }$$ cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone correct to one decimal place.
Solution:
Internal radius of a hollow metallic cylindrical tube (r) = 3 cm
and height (h) = 21 cm

Question 13.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm. Find the height of the cone. (2002)
Solution:
Internal diameter of a hollow sphere = 4 cm
and external diameter = 8 cm
Internal radius (r) = 2 cm
and external radius (R) = 4 cm
Volume of hollow sphere

Question 14.
A well with inner diameter 6 m is dug 22 m deep. Soil taken out of it has been spread evenly all round it to a width of 5 m to form an embankment. Find the height of the embankment.
Solution:
Inner diameter of a well = 6 m
Depth (h) = 22 m

Question 15.
A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.
Solution:
Internal diameter of cylindrical can = 21 cm
Radius (R) = $$\\ \frac { 21 }{ 2 }$$ cm

Question 16.
There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water go down when the sphere is removed?
Solution:
Radius of the cylindrical jar (R) = 8 cm
Height of water level (h) = 14 cm
Volume of water = πR²h

Question 17.
A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid cone submerged? (2002)
Solution:
Height of conical vessel (h) = 20 cm
and diameter = 16.8 cm

Question 18.
A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder?
Solution:
Radius of a solid metallic cylindrical (r) = 14 cm
and height (h) = 12 cm

Question 19.
How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm x 11 cm x 12 cm?
Solution:
Diameter of a shot = 3 cm
Radius (r) = $$\\ \frac { 3 }{ 2 }$$ cm
Volume of one shot = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$

Question 20.
How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm?
Solution:
Diameter of lead shot = 4 cm
Radius (r) = $$\\ \frac { 4 }{ 2 }$$ = 2 cm and
volume =$$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$

Question 21.
Find the number of metallic circular discs with 1.5 cm base diameter and height 0.2 cm to be melted to form a circular cylinder of height 10 cm and diameter 4.5 cm.
Solution:
Radius of the circular disc (r) = 0.75 cm
Height of circular disc (h) = 0.2 cm
Radius of cylinder (R) = 2.25 cm
Height of cylinder (H) = 10 cm
Now,

Question 22.
A solid metal cylinder of radius 14 cm and height 21 cm is melted down and recast into spheres of radius 3.5 cm. Calculate the number of spheres that can be made.
Solution:
Radius of a solid metallic cylinder (r) = 14 cm
and height (h) = 21 cm
Volume of cylinder = πr²h

Question 23.
A metallic sphere of radius 10.5 cm is melted and then recast into small cenes, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained. (2005)
Solution:
Radius of a metallic sphere (r) = 10.5 cm
Volume = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$
= $$\\ \frac { 4 }{ 3 }$$ × π × 10.5 × 10.5 × 10.5 = 1543.5π cm³

Question 24.
A certain number of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the number of cones. (2016)
Solution:
Radius of each cone (r) = 2 cm
and height (h) = 3 cm

Question 25.
A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, $$\\ \frac { 2 }{ 5 }$$ of the water flows out. Find the number of lead shots dropped into the vessel. (2003)
Solution:
Radius of the top of the inverted conical vessel (R) = 2.5 cm
and height (h)= 11 cm

Question 26.
The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained? (2007)
Solution:
Surface area of a metallic sphere = 616 cm²

Question 27.
The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate
(i) the radius of the solid sphere.
(ii) the number of cones recast. (Use π = 3.14).
Solution:
Surface area of a solid metallic sphere = 1256 cm²

Question 28.
Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboid pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?
Solution:
Speed of water flow = 15 km/h
Diameter of pipe = 14 cm

Question 29.
A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate :
(i) the total surface area of the can in contact with water when the sphere is in it.
(ii) the depth of the water in the can before the sphere was put into the can. Given your answer as proper fractions.
Solution:
Radius of a cylindrical can = 3.5 cm
Radius of the sphere = 3.5 cm
and height of water level in the can = 3.5 × 2 = 7 cm

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3

More Exercise

Question 1.
Find the surface area of a sphere of radius :
(i) 14 cm
(ii) 10.5 cm
Solution:
(i) Radius (r) = 14 cm
Surface area = $$4\pi { r }^{ 2 }=4\times \frac { 22 }{ 7 } \times 14\times 14$$ cm2
= 2964 cm2

Question 2.
Find the volume of a sphere of radius :
(i) 0.63 m
(ii) 11.2 cm
Solution:
(i) Radius (r) = 0.63 m
Volume = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$

Question 3.
Find the surface area of a sphere of diameter: (i) 21 cm (ii) 3.5 cm
Solution:
(i) Diameter = 21 cm
Radius (r) = $$\\ \frac { 21 }{ 2 }$$ cm

Question 4.
A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm3, find the mass of the shot-put.
Solution:
Radius of the metallic shot-put = 4.9 cm
Volume = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$

Question 5.
Find the diameter of a sphere whose surface area is 154 cm2.
Solution:
Surface area of a sphere = 154 cm2

Question 6.
Find:
(i) the curved surface area.
(ii) the total surface area of a hemisphere of radius 21 cm.
Solution:
Radius of a hemisphere = 21 cm
(i) Curved surface area = 2πr2

Question 7.
A hemispherical brass bowl has inner- diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.
Solution:
The inner diameter of hemispherical bowl = 10.5 cm

Question 8.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface areas of the balloon in two cases.
Solution:
Original radius of balloon = 7 cm
Radius after filling the air in it = 14 cm
The surface area of balloon, the original position

Question 9.
A sphere and a cube have the same surface. Show that the ratio of the volume of the sphere to that of the cube is √6 : √π
Solution:
Let the edge of a cube = a
Surface area = 6a2
and surface area of sphere = 6a2

Question 10.
(a) If the ratio of the radii of two sphere is 3 : 7, find :
(i) the ratio of their volumes.
(ii) the ratio of their surface areas.
(b) If the ratio of the volumes of the two sphere is 125 : 64, find the ratio of their surface areas.
Solution:
(a) Ratio in radii of two spheres = 3 : 7
Let radius of the first sphere = 3x
and radius of the second sphere = 7x

Question 11.
A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between.
Solution:
Side of a cube = 4 cm
Volume (side)³ = 4 × 4 × 4 = 64 cm³
Diameter of sphere contained by this cube is d = 4 cm

Question 12.
Find the volume of a sphere whose surface area is 154 cm².
Solution:
Given that
Surface area of a sphere = 154 cm²

Question 13.
If the volume of a sphere is $$179 \frac { 2 }{ 3 }$$ cm³, find its radius and the surface area.
Solution:
Given that
Volume of a sphere = $$179 \frac { 2 }{ 3 }$$ cm³
= $$\\ \frac { 539 }{ 3 }$$ cm³

Question 14.
A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Solution:
Radius of a hemispherical bowl (r) = 3.5 cm
= $$\\ \frac { 7 }{ 2 }$$ cm

Question 15.
The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill to its capacity. Find the volume of water pumped into the tank.
Solution:
Internal diameter of a hemispherical tank (r) = 14 m
Radius of the tank = $$\\ \frac { 14 }{ 2 }$$ = 7 m
Water stored in it = 50 kilolitres of water

Question 16.
The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14.
Solution:
Surface area of a solid sphere = 1256 cm²
By cutting it into two hemisphere,
Curved surface area of each hemisphere

Question 17.
Write whether the following statements are true or false. Justify your answer :
(i) The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.
(ii) The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals the volume of a hemisphere of radius r.
(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1 : 2 : 3.
Solution:
(i) The volume of a sphere is equal to the two third of the volume of a cylinder
whose height and diameter are equal to the diameter of the sphere.

(ii) The volume of the longest right circular cone that can be filled in a cube
whose edge is 2r equal to the volume of a hemisphere of radius r

(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height.
The ratio of their volumes is 1 : 2 : 3

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2

More Exercise

Take π = $$\\ \frac { 22 }{ 7 }$$ unless stated otherwise.

Question 1.
Write whether the following statements are true or false. Justify your answer.
(i) If the radius of a right circular cone is halved and its height is doubled, the volume will remain unchanged.
(ii) A cylinder and a right circular cone are having the same base radius and same height. The volume of the cylinder is three times the volume of the cone.
(iii) In a right circular cone, height, radius and slant height are always the sides of a right triangle.
Solution:
(i) If the radius of a right circular cone is halved and its height is doubled,
then the volume will remain unchanged
It is wrong as

(ii) A cylinder and a right circular cone are having the same base radius
and same height the volume of the cylinder is three times the volume of cone – It is true as
Volume of cylinder = $$\pi { r }^{ 2 }h=3\times \frac { 1 }{ 3 } \pi { r }^{ 2 }h$$ = 3(volume of cone)
(iii) In a right circular cone, height, radius and slant height are always the sides of a right triangle
It is true as in a cone and in a right-angled triangle.
Hypotenuse (slant x height) = r2 + h2
and cone is formed by revolving the right triangle about the perpendicular.

Question 2.
Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
Solution:
10 Slant height of a cone (l) = 10 cm
and radius of the base = 7 cm
Curved surface area = πrl
= $$\\ \frac { 22 }{ 7 }$$ × 7 × 10 = 220 cm2

Question 3.
Diameter of the base of a cone is 10.5 cm and slant height is 10 cm. Find its curved surface area.
Solution:
The diameter of the base of a cone = 10.5cm
Its radius (r) = $$\\ \frac { 10.5 }{ 7 }$$ = 5.25 cm
and slant height (l) = 10 cm
Curved surface area = πrl
= $$\\ \frac { 22 }{ 7 }$$ × 5.25 × 10 cm2
= 165.0 cm2

Question 4.
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find ,
(ii)total surface area of the cone.
Solution:
Curved surface area of a cone = 308 cm2
Slant height = 14 cm

Question 5.
Find the volume of the right circular cone with
(i) radius 6 cm and height 7 cm
(ii) radius 3.5 cm and height 12 cm.
Solution:
(i) Radius of cone (r) = 6 cm
and height (h) = 7 cm
Volume = $$\frac { 1 }{ 3 } \pi { r }^{ 2 }h$$

Question 6.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
Solution:
and slant height (l) = 25 cm

Question 7.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?
Solution:
Diameter of top of conical pit = 3.5 m
Radius (r) = $$\\ \frac { 3.5 }{ 2 }$$ = 1.75 m
and depth (h) = 12m

Question 8.
If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.
Solution:
Volume of a right circular cone = 48π cm3
Height (h) = 9 cm

Question 9.
The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π = 3.14)
Solution:
Height of cone (h) = 15 cm
Volume = 1570 cm3

Question 10.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100 m2.
Solution:
Slant height of conical tomb (l) = 25 m
and base diameter = 14 m
Radius(r) = $$\\ \frac { 14 }{ 2 }$$ = 7m

Question 11.
A conical tent is 10 m high and the radius of its base is 24 m. Find :
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.
Solution:
Height of a conical tent (h) = 10 m
and radius (r) = 24 m

Question 12.
A Jocker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the cloth required to make 10 such caps.
Solution:
Base radius of a conical cap = 7 cm
and height (h) = 24 cm

= 550 x 10
= 5500 cm2

Question 13.
(a) The ratio of the base radii of two right circular cones of the same height is 3 : 4. Find the ratio of their volumes.
(b) The ratio of the heights of two right circular cones is 5 : 2 and that of their base radii is 2 : 5. Find the ratio of their volumes.
(c) The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find:
(i) the ratio of their volumes.
(ii) the ratio of their lateral surface areas.
Solution:
(i) The ratio in base radii of two right circular cones of the same height = 3 : 4
Let h be the height and radius of first cone = 3x and
Radius of second cone = 4x

Question 14.
Find what length of canvas 2 m in width is required to make a conical tent 20 m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of Rs 80 per metre.
Solution:
Diameter of the base of the conical tent = 20 m
Radius (r) = $$\\ \frac { 20 }{ 2 }$$ = 10 m
and slant height (h) = 42 m

Question 15.
The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.
Solution:
Perimeter of the base of a cone = 44 cm

Question 16.
The volume of a right circular cone is 9856 cm3 and the area of its base is 616 cm2. Find
(i) the slant height of the cone.
(ii) total surface area of the cone.
Solution:
Volume of a circular cone = 9856 cm3
Area of the base = 616 cm2

Question 17.
A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed. (Take π = 3.14)
Solution:
Sides of a right triangle are 6 cm and 8 cm
It is revolved around 8 cm side
Height (h) = 8 cm
Slant height (l) = 10 cm

Question 18.
The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be $$\\ \frac { 1 }{ 27 }$$ of the volume of the given cone, at what height above the base is the section cut?
Solution:
Height of a cone (H) = 30 cm
A small cone is cut off from the top of the cone given

h = 10 cm
∴ A line parallel to base at a distance of 30 – 10 = 20 cm is drawn.

Question 19.
A semi-circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find
(i) the radius of the cone.
(ii) the (lateral) surface area of the cone.
Solution:
Radius of a semi-circular lamina = 35 cm
By folding it a cone is formed whose slant height (l) = r = 35
and half circumference = circumference of the top of the cone

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1

More Exercise

Take π = $$\\ \frac { 22 }{ 7 }$$ unless stated otherwise.

Question 1.
Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of π.
Solution:
Radius of the cylinder (r) = 5 cm
Height (h) = 10 cm
Total surface area = 2πr (h + r)
= 2π x 5(10 + 5) cm2
= 10 x 15π
= 150π cm2

Question 2.
An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1.2m. Neglecting the thickness of its walls, calculate
(i) its outer lateral surface area,
(ii) its capacity in litres.
Solution:
Diameter of cylindrical geyser = 35 cm
Radius (r) = $$\\ \frac { 35 }{ 2 }$$ cm
Height = 1.2 m = 120 cm

Question 3.
A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. If the glass is filled with milk upto a height of 12 cm, find how many litres of milk is needed to serve 1600 students.
Solution:
Number of students = 1600
Diameter of cylindrical glasses = 7 cm
Radius (r) = $$\\ \frac { 7 }{ 2 }$$ cm

Question 4.
In the given figure, a rectangular tin foil of size 22 cm by 16 cm is wrapped around to form a cylinder of height 16 cm. Find the volume of the cylinder.

Solution:
Length of rectangular tin foil (l) = 22 cm
and breadth (b) = 16 cm
By folding lengthwise, the radius of the cylinder

Question 5.
(i) How many cubic metres of soil must be dug out to make a well 20 metres deep and 2 metres in diameter?
(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of Rs 50 per m2, find the cost of plastering.
Solution:
(i) Depth of well (h) = 20 m
and diameter = 2 m

Question 6.
A roadroller (in the shape of a cylinder) has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must make in order to level a playground of size 120 m by 44 m.
Solution:
Diameter of a road roller = 0.7 m
Radius (r) = $$\\ \frac { 0.7 }{ 2 }$$ = 0.35 m
and width (h) = 1.2 m

Question 7.
If the volume of a cylinder of height 7 cm is 448 π cm3, find its lateral surface area and total surface area.
Solution:
Volume of a cylinder = 448 π cm3
Height (h) = 7 cm

Question 8.
A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m3.
Solution:
Height of a wooden pole (h) = 7 m
Diameter = 20 cm

Question 9.
The area of the curved surface of a cylinder is 4400 cm2, and the circumference of its base is 110 cm. Find
(i) the height of the cylinder.
(ii) the volume of the cylinder.
Solution:
Area of the curved surface of a cylinder = 4400 cm2
Circumference of base = 110 cm

Question 10.
A cylinder has a diameter of 20 cm. The area of curved surface is 1000 cm2. Find
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place. (Take π = 3.14)
Solution:
Diameter of a cylinder = 20 cm
Radius (r) = $$\\ \frac { 20 }{ 2 }$$ = 10 cm
Curved surface area = 1000 cm2

Question 11.
The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?
Answer correct to the nearest. 100 words.
Solution:
Height of cylindrical barrel of a pen (h) = 7 cm
Diameter = 5 mm

Question 12.
Find the ratio between the total surface area of a cylinder to its curved surface area given that its height and radius are 7.5 cm and 3.5 cm.
Solution:
Radius of a cylinder (r) = 3.5 cm
and height (h) = 7.5 cm
Total surface area = 2πr(r + h)

Question 13.
The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder?
Solution:
Let the radius of the base of a right circular cylinder = r
and height (h) = h
Volume = πr2h

Question 14.
(i) The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm2. Find the height and the volume of the cylinder.
(ii) The total surface area of a cylinder is 352 cm2. If its height is 10 cm, then find the diameter of the base.
Solution:
Sum of radius and height of a cylinder = 37 cm
Total surface area = 1628 cm2
Let r be radius and h be height, then r × h = 37
and 2πr(r + h) = 1628

Question 15.
The ratio between the curved surface and the total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm2.
Solution:
Ratio between curved surface area and total surface area = 1 : 2
Total surface area = 616 cm2

Question 16.
Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.
Solution:
Volume of two cylinders is the same
Diameter of both cylinder are in the ratio = 3 : 4

Question 17.
A rectangular sheet of tin foil of size 30 cm x 18 cm can be rolled to form a cylinder in two ways along length and along breadth. Find the ratio of volumes of the two cylinders thus formed.
Solution:
Size of the sheet = 30 cm × 18 cm
(i) By rolling lengthwise,
The circumference of the cylinder = 2πr = 30

Question 18.
A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the volume of the metal.
Solution:
Internal diameter of a metal tube = 11.2 cm
and radius (r) = $$\\ \frac { 11.2 }{ 2 }$$ = 5.6 cm
Length (h) = 21 cm
Thickness of metal = 0.4 cm
External radius (R) = 5.6 + 0.4 = 6.0 cm

Question 19.
The given figure shows a metal pipe 77 cm long. The inner diameter of a cross-section is 4 cm and the outer one is 4.4 cm. Find its
(i) inner curved surface area
(ii) outer curved surface area
(iii) total surface area.

Solution:
In the given figure,
Length of metal pipe (h) = 77 cm
Inner diameter = 4 cm

Question 20.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution:
Diameter of the pencil = 7 mm
Radius (R) = $$\\ \frac { 7 }{ 2 }$$ mm = $$\\ \frac { 7 }{ 20 }$$ cm
Diameter of graphite (lead) = 1 mm

Question 21.
A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Solution:
(i) Base of the tin of rectangular base = 5 cm × 4 cm
Height = 15 cm
Volume = lbh = 5 × 4 × 15 = 300 cm³
(ii) Base diameter of cylindrical plastic cylinder = 7 cm

Question 22.
A cylindrical roller made of iron is 2 m long. Its inner diameter is 35 cm and the thickness is 7 cm all round. Find the weight of the roller in kg, if 1 cm³ of iron weighs 8 g.
Solution:
Length of cylindrical roller (h) = 2 m = 200 cm
Diameter = 35 cm
Inner radius = $$\\ \frac { 35 }{ 2 }$$ cm
Thickness = 7 cm

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1, drop a comment below and we will get back to you at the earliest.