NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.3- Surface Area and Volumes

Q1 (i) Find the volume of the right circular cone with radius 6 cm, height 7 cm

Answer:

Given,

Radius = $r=6cm$

Height = $h=7cm$

We know,

Volume of a right circular cone = $\frac{1}{3}\pi r^{2}h$

$\therefore$ Required volume = $\frac{1}{3}\times \frac{22}{7}\times 6^2\times 7$

$$\begin{gathered} =22\times 2\times 6 \\ =264c{m}^3 \end{gathered}$$

Q1 (ii) Find the volume of the right circular cone with: radius 3.5 cm, height 12 cm

Answer:

Given,

Radius = $r=3.5cm$

Height = $h=12cm$

We know,

Volume of a right circular cone = $\frac{1}{3}\pi r^{2}h$

$\therefore$ Required volume = $\frac{1}{3}\times \frac{22}{7}\times {3.5}^2\times 12$

$$\begin{gathered} =22\times 0.5\times 3.5\times 4 \\ =11\times 14 \\ =154c{m}^3 \end{gathered}$$

Q2 (i) Find the capacity in litres of a conical vessel with radius 7 cm, slant height 25 cm

Answer:

Given,

Radius = $r=7cm$

Slant height = $l=\sqrt{r^2+h^2}=25cm$

Height = $h=\sqrt{l^2-r^2}=\sqrt{{25}^2-7^2}$

$=\sqrt{(25-7)(25+7)}=\sqrt{(18)(32)}$

$=24cm$

We know,
Volume of a right circular cone = $\frac{1}{3}\pi r^{2}h$

$\therefore$ Volume of the vessel= $\frac{1}{3}\times \frac{22}{7}\times 7^2\times 24$

$$\begin{gathered} =22\times 7\times 8 \\ =154\times 8 \\ =1232c{m}^3 \end{gathered}$$

$\therefore$ Required capacity of the vessel =

$=\frac{1232}{1000}=1.232litres$

Q2 (ii) Find the capacity in litres of a conical vessel with height 12 cm, slant height 13 cm

Answer:

Given,

Height = $h=12cm$

Slant height = $l=\sqrt{r^2+h^2}=13cm$

Radius = $r=\sqrt{l^2-h^2}=\sqrt{{13}^2-{12}^2}$

$=\sqrt{(13-12)(13+12)}=\sqrt{(1)(25)}$

$=5cm$

We know,
Volume of a right circular cone = $\frac{1}{3}\pi r^{2}h$

$\therefore$ Volume of the vessel= $\frac{1}{3}\times \frac{22}{7}\times 5^2\times 12$

$$\begin{gathered} =\frac{22}{7}\times 25\times 4 \\ =\frac{2200}{7}c{m}^3 \end{gathered}$$

$\therefore$ Required capacity of the vessel.

$=\frac{2200}{7\times 1000}=\frac{11}{35}litres$

Q3 The height of a cone is 15 cm. If its volume is 1570 cm3 , find the radius of the base. (Use π=3.14 )

Answer:

Given,

Height of the cone = $h=15cm$

Let the radius of the base of the cone be $rcm$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^{2}h$

$\therefore$ $\frac{1}{3}\times 3.14\times r^2\times 15=1570$

$$\begin{gathered} \Rightarrow 3.14\times r^2\times 5=1570 \\ \Rightarrow r^2=\frac{1570}{15.7} \\ \Rightarrow r^2=100 \\ \Rightarrow r=10cm \end{gathered}$$

Q4 If the volume of a right circular cone of height 9 cm is 48πcm3 , find the diameter of its base.

Answer:

Given,

Height of the cone = $h=9cm$

Let the radius of the base of the cone be $rcm$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^{2}h$

$\therefore$ $\frac{1}{3}\times \pi \times r^2\times 9=48\pi$

$$\begin{gathered} \Rightarrow 3r^2=48 \\ \Rightarrow r^2=16 \\ \Rightarrow r=4cm \end{gathered}$$

Therefore, the diameter of the right circular cone is $8cm$

Q5 A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Answer:

Given,

Depth of the conical pit = $h=12m$

The top radius of the conical pit = $r=\frac{3.5}{2}m$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^{2}h$

$\therefore$ The volume of the conical pit =

$=\frac{1}{3}\times \frac{22}{7}\times {\left(\frac{3.5}{2}\right)}^2\times 12$

$$\begin{gathered} =\frac{1}{3}\times \frac{22}{7}\times \frac{3.5\times 3.5}{4}\times 12 \\ =22\times 0.5\times 3.5 \\ =38.5m^3 \end{gathered}$$

Now, $1m^3=1kilolitre$

$\therefore$ The capacity of the pit = $38.5kilolitre$

Q6 (i) The volume of a right circular cone is 9856cm3 . If the diameter of the base is 28 cm, find height of the cone

Answer:

Given, a right circular cone.

The radius of the base of the cone = $r=\frac{28}{2}=14cm$

The volume of the cone = $9856c{m}^3$

(i) Let the height of the cone be $hm$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^{2}h$

$\therefore$ $\frac{1}{3}\times \frac{22}{7}\times (14{)}^2\times h=9856$

$$\begin{gathered} \Rightarrow \frac{1}{3}\times \frac{22}{7}\times 14\times 14\times h=9856 \\ \Rightarrow \frac{1}{3}\times 22\times 2\times 14\times h=9856 \\ \Rightarrow h=\frac{9856\times 3}{22\times 2\times 14} \\ \Rightarrow h=48cm \end{gathered}$$

Therefore, the height of the cone is $48cm$

Q6 (ii) The volume of a right circular cone is 9856cm3 . If the diameter of the base is 28 cm, find slant height of the cone

Answer:

Given, a right circular cone.

The volume of the cone = $9856c{m}^3$

The radius of the base of the cone = $r=\frac{28}{2}=14cm$

And the height of the cone = $h=48cm$

(ii) We know, Slant height, $l=\sqrt{r^2+h^2}$

$$\begin{gathered} \Rightarrow l=\sqrt{{14}^2+{48}^2} \\ \Rightarrow l=\sqrt{196+2304}=\sqrt{2500} \\ \Rightarrow l=50cm \end{gathered}$$

Therefore, the slant height of the cone is $50cm$ .

Q6 (iii) The volume of a right circular cone is 9856cm3 . If the diameter of the base is 28 cm, find curved surface area of the cone

Answer:

Given, a right circular cone.

The radius of the base of the cone = $r=\frac{28}{2}=14cm$

And Slant height of the cone = $l=50cm$

(iii) We know,

The curved surface area of a cone = $\pi rl$

$\therefore$ Required curved surface area= $\frac{22}{7}\times 14\times 50$

$$\begin{gathered} =22\times 2\times 50 \\ =2200c{m}^2 \end{gathered}$$

Q7 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 a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= $h=12cm$

Base radius of the cone = $r=5cm$

And, Slant height of the cone = $l=13cm$

We know,

The volume of a cone = $\frac{1}{3}\pi r^{2}h$

The required volume of the cone formed = $\frac{1}{3}\times \pi \times 5^2\times 12$

$$\begin{gathered} =\pi \times 25\times 4 \\ =100\pi c{m}^3 \end{gathered}$$

Therefore, the volume of the solid cone obtained is $100π cm^3$

Q8 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 a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= $h=5cm$

Base radius of the cone = $r=12cm$

And, Slant height of the cone = $l=13cm$

We know,

The volume of a cone = $\frac{1}{3}\pi r^{2}h$

The required volume of the cone formed = $\frac{1}{3}\times \frac{22}{7}\times {12}^2\times 5$

$$\begin{gathered} =\pi \times 4\times 60 \\ =240\pi c{m}^3 \end{gathered}$$

Now, Ratio of the volumes of the two solids = $=\frac{100\pi }{240\pi }$

$=\frac{5}{12}$

Therefore, the required ratio is $5:12$

Q9 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.

Answer:

Given,

Height of the conical heap = $h=3m$

Base radius of the cone = $r=\frac{10.5}{2}m$

We know,

The volume of a cone = $\frac{1}{3}\pi r^{2}h$

The required volume of the cone formed = $\frac{1}{3}\times \frac{22}{7}\times {\left(\frac{10.5}{2}\right)}^2\times 3$

$$\begin{gathered} =22\times \frac{1.5\times 10.5}{4} \\ =86.625m^3 \end{gathered}$$

Now,

The slant height of the cone = $l=\sqrt{r^2+h^2}$

$\Rightarrow l=\sqrt{3^2+{5.25}^2}=\sqrt{9+27.5625}\approx 6.05$

We know, the curved surface area of a cone = $\pi rl$

The required area of the canvas to cover the heap = $\frac{22}{7}\times \frac{10.5}{2}\times 6.05$

$=99.825m^2$