NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise – 13.7)

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NCERT solutions for class 9 Maths


Surface Areas and Volumes


Chapter 13 


Exercise 13.7


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) Radius of the cone (r) =6 cm
Height (h) = 7 cm


Ex 13.7  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)Radius  of conical vessel (r) = 7 cm and slant height (l) =25 cm

Thus, the capacity of the conical vessel = 1.232 litres.

(ii) Height of conical vessel (h) = 12 cm and slant height (l) = 13 cm

∴ Capacity of the conical vessel

Thus, the  capacity of the conical vessel =  \frac { 11 }{ 35 } litres.


Ex 13.7  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:
Volume of the cone = 1570 cm3
Height of the cone (h) = 15 cm
Volume of cone =  \frac { 1 }{ 3 } πr2h
⇒1570= \frac { 1 }{ 3 }  x 3.14 x r² x 15

⇒1570 = 3.14 x 5 x r²

r² = 1570 / 3.14 x 5 = 100

r = 10 cm

Thus, the radius of the base of cone = 10 cm.


Ex 13.7 Question 4.


If the volume of a right circular cone of height 9 cm is 48 cm³, find the diameter of its base.
Solution:
Volume of the cone = 48 cm3
Height of the cone (h) = 9 cm
Volume of cone =  \frac { 1 }{ 3 } πr2h

Diameter = 2 x radius .
∴ Diameter of the base of the cone = (2 x 4)cm = 8 cm


Ex 13.7  Question 5.


A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
Radius  of pit (r) = \frac { 3.5 }{ 2 } m
Depth (h) = 12m
⇒ Volume (capacity) = \frac { 1 }{ 3 } πr2h

Thus, the capacity of the  pit = 38.5 kilolitres.


Ex 13.7  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:
Volume of the cone V = 9856 cm3
Radius of the base (r) = \frac { 28 }{ 2 } = 14 cm

(i) Let the height of the cone be h cm.

Thus, the height of cone = 48 cm.

(ii) Height of cone h = 48 cm and radius r = 14 cm.                    Let the slant height = l cm.
⇒ l2 = r2 +h2
⇒ l2 = 142 + 482 = 196 + 2304 = 2500
∴ l = 50
Thus, the slant height of cone = 50 cm.

(iii) Slant height of cone l = 50cm and radius r =. 14 cm      Curved surface area of a cone = πrl
∴ Curved surface area = \frac { 22 }{ 7 } x 14 x 50 cm2
= 2200 cm2
Thus, the curved surface area of the cone is 2200 cm2.


Ex 13.7  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:


If the triangle is revolved about 12 cm side, a cone will be formed. Therefore, the radius of of cone r = 5 cm            Height(h) = 12cm and slant height l = 13 cm
∴ Volume of the cone so obtained = \frac { 1 }{ 3 }πr2h
\frac { 1 }{ 3 } x π x (5)2 x 12cm3
= 100 π cm3
Thus, the  volume of the cone = 100πcm3.


Ex 13.7 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 the right triangle is revolved about the side 5 cm.
∴ Height of the cone so obtained (h) = 5 cm
Radius of the cone (r) = 12 cm
Volume of solid =  \frac { 1 }{ 3 } πr2h =  \frac { 1 }{ 3 } x π x 12 x 12 x 5 = 240πcm³
Thus, the volume of solid = 240πcm³


Ex 13.7 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:

Thus, the  volume of heap of wheat = 86.625m3
Let, the slant height of heap of. Wheat = l cm

Thus the required area of the canvas to protect wheat = 99.825 m2