EXERCISE 13.7
Q 1, Ex 13.7, Page No 233, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Page No 233:
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
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Radius (r) of cone = 6 cmHeight (h) of cone = 7 cm
Volume of cone$=\frac{1}{3} \pi r^{2} h$
$=\left[\frac{1}{3} \times \frac{22}{7} \times(6)^{2} \times 7\right] \mathrm{cm}^{3}$
$=(12 \times 22) \mathrm{cm}^{3}$
=264 cm3
Therefore, the volume of the cone is 264 cm3.
(ii) Radius (r) of cone = 3.5 cm
Height (h) of cone = 12 cm
Volume of cone$=\frac{1}{3} \pi r^{2} h$
$=\left[\frac{1}{3} \times \frac{22}{7} \times(3.5)^{2} \times 12\right] \mathrm{cm}^{3}$
$=\left(\frac{1}{3} \times 22 \times \frac{1}{2} \times 3.5 \times 12\right) \mathrm{cm}^{3}$
=154 cm3
Therefore, the volume of the cone is 154 cm3.
Q 2, Ex 13.7, Page No 233, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Radius (r) of cone = 7 cmSlant height (l) of cone = 25 cm
Height (h) of cone $=\sqrt{l^{2}-r^{2}}$
$=(\sqrt{25^{2}-7^{2}}) \mathrm{cm}$
=24cm
Volume of cone $=\frac{1}{3} \pi r^{2} h$
$=\left(\frac{1}{3} \times \frac{22}{7} \times(7)^{2} \times 24\right) \mathrm{cm}^{3}$
$=(154 \times 8) \mathrm{cm}^{3}$
=1232cm3
Therefore, capacity of the conical vessel
$=\left(\frac{1232}{1000}\right)$ litres (1 litre=1000cm3)
= 1.232 litres
(ii) Height (h) of cone = 12 cm
Slant height (l) of cone = 13 cm
Radius (r) of cone $=\sqrt{l^{2}-h^{2}}$
$=(\sqrt{13^{2}-12^{2}}) \mathrm{cm}$
=5cm
Volume of cone $=\frac{1}{3} \pi r^{2} h$
$=\left[\frac{1}{3} \times \frac{22}{7} \times(5)^{2} \times 12\right] \mathrm{cm}^{3}$
$=\left(4 \times \frac{22}{7} \times 25\right) \mathrm{cm}^{3}$
$=\left(\frac{2200}{7}\right) \mathrm{cm}^{3}$
Therefore, capacity of the conical vessel
$=\left(\frac{2200}{7000}\right)$ litres (1 litre=1000cm3)
$=\frac{11}{35}$litres
Q 3, Ex. 13.7, Page No 233, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 3:
The height of a cone is 15 cm. If its volume is 1570 cm3, find the diameter of its base. [Use π = 3.14]Answer:
Height (h) of cone = 15 cmLet the radius of the cone be r.
Volume of cone = 1570 cm3
$\frac{1}{3} \pi r^{2} h=1570 \mathrm{cm}^{3}$
$\Rightarrow\left(\frac{1}{3} \times 3.14 \times r^{2} \times 15\right) \mathrm{cm}=1570 \mathrm{cm}^{3}$
$\Rightarrow r^{2}=100 \mathrm{cm}^{2}$
⇒ r = 10 cm
Therefore, the diameter of the base of cone is
10×2=20 cm.
Q 4, Ex. 13.7, Page No 233, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 4:
If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.Answer:
Height (h) of cone = 9 cmLet the radius of the cone be r.
Volume of cone = 48π cm3
$\Rightarrow \frac{1}{3} \pi r^{2} h=48 \pi \mathrm{cm}^{3}$
$\Rightarrow\left(\frac{1}{3} \pi r^{2} \times 9\right) \mathrm{cm}=48 \pi \mathrm{cm}^{3}$
$\Rightarrow r^{2}=16 \mathrm{cm}^{2}$
⇒r=4cm
Diameter of base = 2r = 8 cm
Q 5, Ex. 13.7, Page No 233, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
Question 5:
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? $\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of pit $=\left(\frac{3.5}{2}\right) \mathrm{m}=1.75 \mathrm{m}$Height (h) of pit = Depth of pit = 12 m
Volume of pit $=\frac{1}{3} \pi r^{2} h$
$=\left[\frac{1}{3} \times \frac{22}{7} \times(1.75)^{2} \times 12\right] \mathrm{cm}^{3}$
= 38.5 m3
Thus, capacity of the pit = (38.5 × 1) kilolitres = 38.5 kilolitres
Q 6, Ex. 13.7, Page No 233, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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
$\left[\text { Assume } \pi=\frac{22}{7}\right]$
Answer:
(i) Radius of cone $=\left(\frac{28}{2}\right) \mathrm{cm}$=14cm
Let the height of the cone be h.
Volume of cone = 9856 cm3
$\Rightarrow \frac{1}{3} \pi r^{2} h=9856 \mathrm{cm}^{3}$
$\Rightarrow\left[\frac{1}{3} \times \frac{22}{7} \times(14)^{2} \times h\right] \mathrm{cm}^{2}$
$=9856 \mathrm{cm}^{3}$
h = 48 cm
Therefore, the height of the cone is 48 cm.
(ii) Slant height (l) of cone $=\sqrt{r^{2}+h^{2}}$
$=[\sqrt{(14)^{2}+(48)^{2}}] \mathrm{cm}$
$=[\sqrt{196+2304}] \mathrm{cm}$
=50cm
Therefore, the slant height of the cone is 50 cm.
(iii) CSA of cone = πrl
$=\left(\frac{22}{7} \times 14 \times 50\right) \mathrm{cm}^{2}$
= 2200 cm2
Therefore, 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.Answer:
When right-angled ΔABC is revolved about its side 12 cm, a cone with height (h) as 12 cm, radius (r) as 5 cm, and slant height (l) 13 cm will be formed.
Volume of cone $=\frac{1}{3} \pi r^{2} h$
$=\left[\frac{1}{3} \times \pi \times(5)^{2} \times 12\right] \mathrm{cm}^{3}$
= 100π cm3
Therefore, the volume of the cone so formed is 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.Answer:
When right-angled ΔABC is revolved about its side 5 cm, a cone will be formed having radius (r) as 12 cm, height (h) as 5 cm, and slant height (l) as 13 cm.
Volume of cone $=\frac{1}{3} \pi r^{2} h$
$=\left[\frac{1}{3} \times \pi \times(12)^{2} \times 5\right] \mathrm{cm}^{3}$
$=240 \pi \mathrm{cm}^{3}$
Therefore, the volume of the cone so formed is 240π cm3.
Required ratio$=\frac{100 \pi}{240 \pi}$
$=\frac{5}{12}$
=5:12
Q 9, Ex. 13.7, Page No 233, Surface Areas & Volumes - NCERT Class 9th Maths Solutions
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.$\left[\text { Assume } \pi=\frac{22}{7}\right]$Answer:
Radius (r) of heap $=\left(\frac{10.5}{2}\right) \mathrm{m}$=5.25m
Height (h) of heap = 3 m
Volume of heap$=\frac{1}{3} \pi r^{2} h$
$=\left(\frac{1}{3} \times \frac{22}{7} \times(5.25)^{2} \times 3\right) \mathrm{m}^{3}$
$=86.625 \mathrm{m}^{3}$
Therefore, the volume of the heap of wheat is 86.625 m3.
Area of canvas required = CSA of cone
$=\pi r l=\pi r \sqrt{r^{2}+h^{2}}$
$=\left[\frac{22}{7} \times 5.25 \times \sqrt{(5.25)^{2}+(3)^{2}}\right] \mathrm{m}^{2}$
$=99.825 \mathrm{m}^{2}$
Therefore, 99.825 m2 canvas will be required to protect the heap from rain.
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