SURFACE AREA AND VOLUME
EXERCISE 13.7
EXERCISE 13.7
Q 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) Here, radius of the cone r = 6 cm
Height (h) = 7 cm
Q 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) Here, r = 7 and I = 25 cm
Q 3. The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base.
Solution: Here, height of the cone (h) = 15 cm
Volume of the cone (v) = 1570 cm3
Let the radius of the base be ‘r’ cm.
Q 4. If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.
Solution: Volume of the cone = 48 π cm3
Height of the cone (h) = 9 cm
Let ‘r’ be its base radius.
∵Diameter = 2 × Radius
∴Diameter of the base of the cone = 2 × 4 = 8 cm
Q5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Solution: Here, diameter of the conical pit = 3.5 m
= 38.5 m3
∴ 1000 cm3 = 1 l and 1000000 cm3 = l m3
∴ 1000 × 1000 cm3 = 1000 l = 1 kl
Also 1000 × 1000 cm3 = 1 m3
⇒1 m3 = 1 kl
⇒38.5 m3 = 38.5 kl
Thus, the capacityof the conical pit is 38.5 kl.
Q6. The volume of right circular cone is 9856 cm3. If the diameter of the be 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
Diameter of the base = 28 cm
Thus, the required height is 48 cm.
(ii) To find the slant height
Let the slant height be ‘l’ cm.
∵ (Slant height)2 = (Radius)2 + (Height)2
∴ l2 = 142 + 482 = 196 + 2304 = 2500 = (50)2
⇒ l = 50
Thus, the required height = 50 cm.
(iii) To find the curved surface area
∵The curved surface area of a cone is given by nrl
Thus, the curved surface area of the cone is 2200 cm2.
Q 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: Sides of the right triangle are 5 cm, 12 cm and 13 cm.
∵The right angled triangle is revolved about the 12 cm side.
∴ Its height is 12 cm and base is 5 cm.
Thus, we have Radius of the base of the cone so formed (r) = 5 cm
Height (h) = 12 cm
Slant height = 13 cm
= π × 100 cm3
= 100π cm3
Thus, the required volume of the cone is 100π cm3.
Q 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
Solution: Here the heap of wheat is in the form of a cone such that
Base diameter = 10.5 m
Thus, the required volume = 86.625 m3
Area of the canvas
∵The area of the canvas to cover the heap must be equal to the curved surface area of the conical heap.
