ICSE Class 10

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

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Ex 18
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Trigonometric Identities Chapter Test

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

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

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
Radius of a hemispherical bowl

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
Let radius = r, then

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

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

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
Let radius (r₁) = 2x
and radius (r₂) = 3x
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

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

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 17.4

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 17.4

More Exercise

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.4
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Ex 17.5
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Mensuration Chapter Test

Take π = \(\\ \frac { 22 }{ 7 } \), unless stated otherwise.

Question 1.
The adjoining figure shows a cuboidal block of wood through which a circular cylinderical hole of the biggest size is drilled. Find the volume of the wood left in the block.

Solution:
Diameter of the biggest hole = 30 cm.
Radius (r) = \(\\ \frac { 30 }{ 2 } \) = 15 cm
and height (h) = 70 cm.

Question 2.
The given figure shows a solid trophy made of shining glass. If one cubic centimetre of glass costs Rs 0.75, find the cost of the glass for making the trophy.

Solution:
Edge of cubical part = 28 cm
and radius of cylindrical part (r) = \(\\ \frac { 28 }{ 2 } \) = 14cm

Question 3.
From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.
Solution:
Edge of a cube = 14 cm
Volume = (side)³ = (14)³ = 14 × 14 × 14 cm³ = 2744 cm³
Now diameter of the cone cut out from it = 14 cm

Question 4.
A cone of maximum volume is curved out of a block of wood of size 20 cm x 10 cm x 10 cm. Find the volume of the remaining wood.
Solution:
Size of wooden block = 20 cm × 10 cm × 10 cm
Maximum diameter of the cone = 10 cm
and height (h) = 20 cm

Question 5.
16 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm x 8 cm x 8 cm and then the box is filled with water. Find the volume of the water filled in the box.
Solution:
Given
Radius of each glass sphere = 2 cm

Question 6.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of the wood in the entire stand, correct to 2 decimal places.

Solution:
Dimensions of cuboid = 15 cm × 10 cm × 3.5 cm
and radius of each conical depression (r) = 0.5 cm
and depth (h) = 1.4 cm
Volume of cuboid = l × b × h

Question 7.
A cuboidal block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.
Solution:
Side of cuboidal = 7 cm
Diameter of hemisphere = 7 cm
and radius (r) = \(\\ \frac { 7 }{ 2 } \) cm

Question 8.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder (as shown in the given figure). If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:
Height of the cylinder = 10 cm
and radius of the base = 3.5 cm
Total surface area
= Curved surface area of cylinder + 2 × Curved surface area of hemisphere

Question 9.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area of the toy.
Solution:
Total height of the toy = 15.5 cm
Radius of the base of the conical part (r) = 3.5 cm

Question 10.
A circus tent is in the shape of a cylinder surmounted by a cone. The diameter of the cylindrical portion is 24 m and its height is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas used to make the tent.
Solution:
Radius of base of cylindrical portion of tent (r) = \(\\ \frac { 24 }{ 2 } \) = 12 m

Question 11.
An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answer to the nearest m².
Solution:
Total height of the tent = 85 m
Height of cylindrical part (h₁) = 50 m

Question 12.
From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and the total surface of the remaining solid.
Solution:
Radius of solid cylinder (r₁) = 7 cm

Question 13.
The given figure shows a wooden toy rocket which is in the shape of a circular cone mounted on a circular cylinder. The total height of the rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted green and the cylindrical portion red, find the area of the rocket painted with each of these colours. Also, find the volume of the wood in the rocket. Use π = 3.14 and give answers correct to 2 decimal places.

Solution:
In the given figure,
The total height of the toy rocket = 26 cm
Diameter of cylindrical portion = 3 cm

Question 14.
The given figure shows a hemisphere of radius 5 cm surmounted by a right circular cone of base radius 5 cm. Find the volume of the solid if the height of the cone is 7 cm. Give your answer correct to two places of decimal.
Solution:
Height of the conical part (h) = 7 cm
and radius of the base (r) = 5 cm

Question 15.
A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is \(\\ \frac { 2 }{ 3 } \) of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal
Solution:
Radius of base of hemisphere = \(\\ \frac { 7 }{ 2 } \) m

Question 16.
A circular hall (big room) has a hemispherical roof. The greatest height is equal to the inner diameter, find the area of the floor, given that the capacity of the hall is 48510 m³.
Solution:
Let h be the greatest height
and r be the radius of the base
Then 2r = h + r ⇒ h = r

Question 17.
A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains \(41 \frac { 19 }{ 21 } \) m3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building
Solution:
Volume of air in dome = \(41 \frac { 19 }{ 21 } \) m³
= \(\\ \frac { 880 }{ 21 } \) m³
Let radius of the dome = r m
Then height (h) = r m

Question 18.
A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and the height of the cylinder are 6 cm and 12 cm respectively. If the slant height of the conical portion is 5 cm, find the total surface area and the volume of the rocket. (Use π = 3.14).
Solution:
Height of the cylindrical part (A) = 12 cm
Diameter = 6 cm
Radius (r) = \(\\ \frac { 6 }{ 2 } \) = 3 cm
Slant height of the conical part (l) = 5 cm

Question 19.
A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid. (Take π = 3.14)
Solution:
Diameter = 3.5 cm
Radius (r) = \(\\ \frac { 3.5 }{ 2 } \) = 1.75 cm
Height of cylindrical part (h₁) = 10 cm
and height of conical part (h₂) = 6 cm

Question 20.
A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.
Solution:
Height of cylindrical part = 13 cm
Radius = 5 cm
Radius of cone (r) = 5 cm
Height of cone (h) = 12 cm

Question 21.
The given figure shows a model of a solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1 : 200, find
(i) the total surface area of the solid in π m².
(ii) the volume of the solid in π litres.

Solution:
(i) In the given figure,
Height of cylindrical portion (h) = 8 cm
Radius (r) = 3 cm
Scale = 1 : 200

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