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 = $r =6\ cm$

Height = $h =7\ cm$

We know,

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

$\therefore$ Required volume = $\frac{1}{3}\times\frac{22}{7}\times6^2\times7$

$\\ = 22\times2\times6 \\ = 264\ cm^3$

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

Answer:

Given,

Radius = $r =3.5\ cm$

Height = $h =12\ cm$

We know,

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

$\therefore$ Required volume = $\frac{1}{3}\times\frac{22}{7}\times3.5^2\times12$

$\\ = 22\times0.5\times3.5\times4 \\ = 11\times14 \\ = 154\ cm^3$

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

Answer:

Given,

Radius = $r =7\ cm$

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

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

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

$= 24\ cm$

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

$\therefore$ Volume of the vessel= $\frac{1}{3}\times\frac{22}{7}\times7^2\times24$

$\\ = 22\times7\times8\\ = 154\times8 \\ = 1232\ cm^3$

$\therefore$ Required capacity of the vessel =

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

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

Answer:

Given,

Height = $h =12\ cm$

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

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

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

$= 5\ cm$

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

$\therefore$ Volume of the vessel= $\frac{1}{3}\times\frac{22}{7}\times5^2\times12$

$\\ = \frac{22}{7}\times25\times4\\ = \frac{2200}{7}\ cm^3$

$\therefore$ Required capacity of the vessel.

$ = \frac{2200}{7\times1000} = \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 = $h =15\ cm$

Let the radius of the base of the cone be $r\ cm$

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

$\therefore$ $\frac{1}{3}\times3.14\times r^2\times15 = 1570$

$\\ \Rightarrow 3.14\times r^2\times5 = 1570 \\ \Rightarrow r^2 = \frac{1570}{15.7} \\ \Rightarrow r^2 = 100 \\ \Rightarrow r = 10\ cm$

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 = $h =9\ cm$

Let the radius of the base of the cone be $r\ cm$

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

$\therefore$ $\frac{1}{3}\times\pi\times r^2\times9 = 48\pi$

$\\ \Rightarrow 3r^2 = 48 \\ \Rightarrow r^2 = 16\\ \Rightarrow r = 4\ cm$

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

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 = $h =12\ m$

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 =

$= \frac{1}{3}\times\frac{22}{7}\times \left (\frac{3.5}{2} \right )^2\times12$

$\\ = \frac{1}{3}\times\frac{22}{7}\times \frac{3.5\times 3.5}{4}\times12 \\ \\ = 22\times 0.5\times 3.5 \\ = 38.5\ m^3$

Now, $1\ m^3 = 1\ kilolitre$

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

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 = $r = \frac{28}{2} = 14\ cm$

The volume of the cone = $\small 9856\hspace{1mm}cm^3$

(i) Let the height of the cone be $h\ m$

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

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

$\\ \Rightarrow \frac{1}{3}\times\frac{22}{7}\times14\times14\times h = 9856 \\ \Rightarrow \frac{1}{3}\times22\times2\times14\times h = 9856 \\ \Rightarrow h = \frac{9856\times3}{22\times2\times14} \\ \\ \Rightarrow h =48\ cm$

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

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 = $\small 9856\hspace{1mm}cm^3$

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

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

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

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

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

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 = $r = \frac{28}{2} = 14\ cm$

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

(iii) We know,

The curved surface area of a cone = $\pi r l$

$\therefore$ Required curved surface area= $\frac{22}{7}\times14\times50$

$\\ = 22\times2\times50 \\ = 2200\ cm^2$

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= $h = 12\ cm$

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

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

We know,

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

The required volume of the cone formed = $\frac{1}{3}\times\pi\times5^2\times12$

$\\ = \pi\times25\times4 \\ = 100\pi\ cm^3$

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= $h = 5\ cm$

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

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

We know,

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

The required volume of the cone formed = $\frac{1}{3}\times\frac{22}{7}\times12^2\times5$

$\\ = \pi\times4\times60 \\ = 240\pi\ cm^3$

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 = $h = 3\ m$

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 = $\frac{1}{3}\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2\times3$

$\\ = 22\times\frac{1.5\times10.5}{4} \\ = 86.625\ m^3$

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 = $\pi r l$

The required area of the canvas to cover the heap = $\frac{22}{7}\times\frac{10.5}{2}\times6.05$

$= 99.825\ m^2$