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

Gujarat Board GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7

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) r = 6 cm
h = l cm
∴ Volume of the right circular cone
= \(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) x \(\frac{22}{7}\) x (6)² x 7
= 264 cm³
(ii) r = 3.5 cm
h = 12 cm
∴ Volume of the right circular cone
= \(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) x \(\frac{22}{7}\) x (3.5)² x 12
= 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) r = 7 cm
l = 25 cm
r² + h² = l²
⇒ (7)² + h² = (25)²
⇒ h² = (25)² – (7)²
⇒ h² = 625 – 49
⇒ h²= 576
⇒ h = \(\sqrt{576}\)
⇒ h = 24 cm
∴ Capacity = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) x \(\frac{22}{7}\) x (7)² x 24
= 1232 cm³ = 1.232 L

(ii) h = 12 cm
l = 13 cm
r² + h² = l²
⇒ r² + (12)² = (13)²
⇒ r² + 144 = 169
⇒ r² = 169 – 144
⇒ r² = 25
⇒ r = \(\sqrt{25}\)
⇒ r = 5 cm

∴ Capacity = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) x \(\frac{22}{7}\) x (5)² x 12
= \(\frac{2200}{7}\) cm³ = \(\frac{2200}{7000}\)L = \(\frac{11}{35}\) L

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 the base of the cone be r cm.
h = 15 cm
Volume = 1570 cm³
⇒ \(\frac{1}{3}\) πr²h x 1570
\(\frac{1}{3}\) x 3.14 x r² x 15 = 1570
⇒ r² = \(\frac{1570 \times 3}{3.14 \times 15}\)
⇒ r² = 100
⇒ r² = 100
⇒ r = \(\sqrt{100}\)
⇒ r = 10 cm
Hence, the radius of the base of the cone is 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 base of the right circular
cone be r cm.
h = 9 cm
Volume = 48π cm³
⇒ \(\frac{1}{3}\)πr²h = 48π
⇒ \(\frac{1}{3}\)πr²h = 48
⇒ \(\frac{1}{3}\) x r² x 9 = 48
⇒ r² = \(\frac{48 \times 3}{9}\)
⇒ r² = 16
⇒ r² = \(\sqrt{16}\) = 4 cm
⇒ 2r = 2(4) = 8 cm
Hence, the diameter of the base of the right circular cone is 8 cm.

Question 5.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
For conical pit
Diameter = 3.5 cm
∴ Radius (r) = \(\frac{3.5}{2}\) m = 1.75 m
Depth (h) = 12 m
∴ Capacity of the conical pit
\(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) x \(\frac{22}{7}\) x (1.75)m² x 12 m³
= 38.5 m³ = 38.5 x 1000L = 38.5 KL

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:
(i) Diameter of the base = 28 cm
∴ Radius of the base (r) = \(\frac{28}{2}\) cm = 14 cm
Let the height of the cone be h cm.
Volume = 9856 cm³
⇒ \(\frac{1}{3}\) πr²h = 9856
⇒ \(\frac{1}{3}\) x \(\frac{22}{7}\) x (14)² x h = 9856
⇒ h = \(\frac{9856 \times 3 \times 7}{22 \times(14)^{2}}\)
⇒ h = 48 cm
Hence, the height of the cone is 48 cm.

(ii) r = 14 cm
h = 48 cm
∴ I = \(\sqrt{r^{2}+h^{2}}\) = \(\sqrt{(14)^{2}+(48)^{2}}\)
= \(\sqrt{196+2304}\) = \(\sqrt{2500}\) = 50 cm
Hence, the slant height of the cone is 50 cm.

(iii) r = 14 cm
l = 50 cm
∴ Curved surface area = πrl
= \(\frac{22}{7}\) x 14 x 50 = 2200 cm²
Hence, the curved surface area of the cone is 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:
The solid obtained will be a right circular cone whose radius of the base is 5 cm and height is 12 cm.

∴ r = 5 cm
h = 12 cm
∴ Volume = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) x (5)² x 12 cm³
= 100 π cm³
Hence, the volume of the solid so obtained is 100 π cm³

Question 8.
If the triangle ABC in 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:
The solid obtained will be a right circular cone whose radius of the base is 12 cm and the height is 5 cm.
∴ r = 12 cm
h = 5 cm

Volume = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) x π x (12)² x 5 cm³= 24071 cm³
The ratio of the volumes of the two solids obtained = 100π : 240 π = 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:
For heap of wheat
Diameter = 10.5 m
∴ Radius (r) = \(\frac{10.5}{2}\) cm = 5.25 m
Height (A) = 3m
∴ Volume = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) x \(\frac{22}{7}\) x (5.25)² x 3
= 86.625 m²

Slant height (l) = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{(5.25)^{2}+(3)^{2}}\)
= \(\sqrt{27.5625+9}\)
= \(\sqrt{36.5629}\)
= 6.05 m (approx.)
∴ Curved surface area = πrl 22
= \(\frac{22}{7}\) x 5.25 x 6.05
= 99.825 m²
Hence, the area of the canvas required is 99.825 m².