HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.7

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

Question 1.
Find the volume of the right circular cone with :
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm.
Solution:
(i) We have, Radius of cone (r) = 6 cm
Height of the cone (h) = 7 cm
∴ Volume e the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 6^2 \times 7\)
= 264 cm³.
(ii) We have,
Radius of the cone (r) = 3.5 cm
Height of the cone (h) = 12 cm
∴ Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 3.5^2 \times 12\)
= 154 cm³
Hence, volume of the cone (i) 264 cm³, (ii) 154 cm³.

Question 2.
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:
(i) We have, Radius of the conical vessel (r) = 7 cm
Slant height of the conical vessel (l) = 25 cm
Let the height of the conical vessel be h cm.
∴ l² = h² + r²
⇒ 25² = h² + 7²
⇒ h² = 25² – 7²
⇒ h² = (25 + 7) (25 – 7)
⇒ h² = 32 × 18
⇒ h² = 576
⇒ h= \(\sqrt{576}\)
⇒ h = 24 cm
∴ Volume of the conical vessel = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24\)
= 1232 cm³
= \(\frac{1232}{1000}\) litres
= 1.232 litres.

(ii) We have, Slant height of the conical vessel (l) = 13 cm
Height of the conical vessel (h) = 12 cm
Let the radius of the conical vessel be r cm.
∴ l² = h² + r²
⇒ 13² = 12² + r²
⇒ r² = 13² – 12²
⇒ r² = (13 + 12) (13 – 12)
⇒ r² = 25
⇒ r = \(\sqrt{25}\)
⇒ r = 5 cm
∴ Volume of the conical vessel = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times 5^2 \times 12\)
= \(\frac{2200}{7} \mathrm{~cm}^3\)
= \(\frac{2200}{7 \times 1000} \text { litres }\)
= \(\frac{11}{35} \text { litres }\)
Hence, volume of the cone (i) 1.232 litres, (ii) \(\frac{11}{35} \text { litres }\).

Question 3.
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:
Let the radius of base of a cone be r сm.
We have, Height of the cone (h) = 15 cm
volume of the cone = 1570 cm³
⇒ \(\frac{1}{3}\)πr²h = 1570
⇒ \(\frac{1}{3}\) × 3.14 × r² × 15 = 1570
⇒ 15.7 × r² = 1570
⇒ r² = \(\frac{1570}{157}\)
⇒ r² = 100
⇒ r = \(\sqrt{100}\)
⇒ r = 10 cm
Hence, radius of the base of the cone = 10 cm.

Question 4.
If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Solution:
Let the radius of the basse of the cone be r сm.
We have, Height of the cone (h) = 9 cm
Volume of the cone = 48π cm³
⇒ \(\frac{1}{3}\)πr²h = 48π
⇒ \(\frac{1}{3}\)π × r² × 9 = 48π
⇒ 3πr² = 48π
⇒ r² = \(\frac{48 \pi}{3 \pi}\)
⇒ r² = 16
⇒ r = \(\sqrt{16}\)
⇒ r = 4 cm
∴ Diameter of the base of the cone = 2 × 4
= 8 cm
Hence,diameter of the base of cone = 8 cm

Question 5.
A conical pit of top diameter 3.5 m is 12m deep. What is its capacity in kilolitres ?
Solution:
We have,
Depth of conical pit (h) = 12 m
Diameter of conical pit (d) = 3.5 m
∴ Radius of conical pit (r) = \(\frac{3.5}{2}\) = 1.75 m
∴ Capacity of the conical pit = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times(1.75)^2 \times 12\)
= 38.5 m³
= 38.5 kilolitres
(∵ 1 m³ = 1 kilolitres)
Hence,capacity of the conical pit = 38.5 kilolitres.

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find :
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.
Solution:
We have,
The volume of the cone = 9856 cm³
The diameter of the base of the cone (d) = 28 cm
∴ Radius of the base of the cone (r) = \(\frac{28}{2}\) cm = 14 cm
(i) Let the height of the cone be h cm.
Volume of the cone = 9856 cm³
⇒ \(\frac{1}{3}\)πr²h = 9856
⇒ \(\frac{1}{3} \times \frac{22}{7} \times 14^2 \times h\) = 9856
⇒ \(\frac{616}{3}\) × h = 9856
⇒ h = \(\frac{9856 \times 3}{616}\)
⇒ h = 480cm

(ii) Let the slant height of the cone be 1 cm, then
⇒ l² = h² + r²
⇒ l² = 48² + 14²
⇒ l² = 2304 + 196
⇒ l2 = 2500
⇒ l = \(\sqrt{2500}\)
⇒ l = 50 cm.

(iii) Curved surface of the cone = πrl
= \(\frac{22}{7}\) × 14 × 50
= 2200 cm².
Hence, (i) Height of the cone = 48 cm, (ii) Slant height of the cone = 50 cm, (iii) Curved surface area of the cone = 2200 cm².

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
We have,
Sides of the triangle ABC are AB = 12 cm, BC = 5 cm and AC = 13 cm
Since ΔABC is revolved about the side AB(= 12 cm)
∴ We get a cone (AC’C) as shown in the figure.

Then,radius of the base of the cone (r) = 5 cm
and height of the cone (h) = 12 cm
∴ Volume of the cone (solid) = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × π × 5² × 12
= 100π cm³
Hence, volume of the cone so obtained = 100π cm³.

Question 8.
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
Since ΔABC is revolved about BC(= 5 cm).
∴ We get a cone (CAA’) as shown in the figure.

Then,radius of the base of the cone (r) = 12 cm
and height of the cone (h) = 5 cm
Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × π × 12² × 5
= 240π cm³
Ratio of the volumes of the two solids obtained in Q. 7 and Q. 8
= 100π : 240π = \(\frac{100 \pi}{240 \pi}\)
= \(\frac{5}{2}\) = 5 : 12
Hence,volume of the cone (solid) = 240π cm³,
and required ratio of their volumes = 5 : 12.

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Solution:
We have,
Height of the conical heap (h) = 3 m
Diameter of the base of the conical heap (d) = 10.5 m³

∴ Radius of the base of the conical heap (d) = \(\frac{10.5}{2}\) = 5.25 m
∴ Slant height of a conical heap (l) = \(\sqrt{h^2+r^2}\)
= \(\sqrt{3^2+(5.25)^2}\)
= \(\sqrt{9+27.56}\)
= \(\sqrt{36.56}\)
= 6.05 m
∴ Volume of the conical heap = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times(5.25)^2 \times 3\)
= \(\frac{606.375}{7}\)
= 86.625 m³
Required area of canvas = Curved surface
area of the conical heap = πrl
= \(\frac{22}{7}\) × 5.25 × 6.05
= 16.5 × 6.05
= 99.825 m²
Hence,volume of the heap = 86.625 m³ and required area of canvas = 99.825 m².