NCERT Solutions for Exercise 13.3 Class 9 Maths Chapter 13 - Surface Area and Volumes

Access Surface Area and Volumes Class 9 Chapter 13 Exercise: 13.3

Q1 Diameter of the base of a cone is $\small 10.5 \hspace{1mm}cm$ and its slant height is $\small 10 \hspace{1mm}cm$ . Find its curved surface area.

Answer:

Given,

Base diameter of the cone = $d=10.5\ cm$

Slant height = $l=10\ cm$

We know, Curved surface area of a cone $= \pi r l$

$\therefore$ Required curved surface area of the cone=

$\\ = \frac{22}{7}\times \frac{10.5}{2}\times10 \\ \\ = 165\ cm^2$

Q2 Find the total surface area of a cone, if its slant height is $\small 21\hspace{1mm}m$ and diameter of its base is $\small 24\hspace{1mm}m$ .

Answer:

Given,

Base diameter of the cone = $d=24\ m$

Slant height = $l=21\ cm$

We know, Total surface area of a cone = Curved surface area + Base area

$= \pi r l + \pi r^2 = \pi r (l + r)$

$\therefore$ Required total surface area of the cone=

$\\ = \frac{22}{7}\times\frac{24}{2}\times(21+12) \\ \\ =\frac{22}{7}\times\frac{24}{2}\times33 \\ = 1244.57 \ m^2$

Q3 (i) Curved surface area of a cone is $\small 308\hspace{1mm}cm^2$ and its slant height is 14 cm. Find radius of the base .

Answer:

Given,

The curved surface area of a cone = $\small 308\hspace{1mm}cm^2$

Slant height $= l = 14\ cm$

(i) Let the radius of cone be $r\ cm$

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

$\therefore$ $\\ \pi rl = 308 \\ \\ \Rightarrow \frac{22}{7}\times r\times14 = 308 \\ \Rightarrow r = \frac{308}{44} = 7$

Therefore, the radius of the cone is $7\ cm$

Q3 (ii) Curved surface area of a cone is $\small 308\hspace{1mm}cm^2$ and its slant height is $\small 14\hspace{1mm}cm$ . Find total surface area of the cone.

Answer:

Given,

The curved surface area of a cone = $\small 308\hspace{1mm}cm^2$

Slant height $= l = 14\ cm$

The radius of the cone is $r =$ $7\ cm$

(ii) We know, Total surface area of a cone = Curved surface area + Base area

$= \pi r l + \pi r^2$

$\\ = 308+\frac{22}{7}\times 7^2 \\ = 308+154 = 462\ cm^2$

Therefore, the total surface area of the cone is $462\ cm^2$

Q4 (i) A conical tent is 10 m high and the radius of its base is 24 m. Find slant height of the tent.

Answer:

Given,

Base radius of the conical tent = $r=24\ m$

Height of the conical tent = $h=10\ m$

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

$\\ =\sqrt{10^2+24^2} \\ = \sqrt{676} \\ = 26\ m$

Therefore, the slant height of the conical tent is $26\ m$

Q4 (ii) A conical tent is 10 m high and the radius of its base is 24 m. Find cost of the canvas required to make the tent, if the cost of $\small 1\hspace{1mm}m^2$ canvas is Rs 70.

Answer:

Given,

Base radius of the conical tent = $r=24\ m$

Height of the conical tent = $h=10\ m$

$\therefore$ Slant height = $l=\sqrt{h^2+r^2} = 26\ m$

We know, Curved surface area of a cone $= \pi r l$

$\therefore$ Curved surface area of the tent

$\\ = \frac{22}{7}\times24\times26 \\ \\ =\frac{13728}{7}\ m^2$

Cost of $1\ m^2$ of canvas = $Rs.\ 70$

$\therefore$ Cost of $\frac{13728}{7}\ m^2$ of canvas =

$Rs.\ (\frac{13728}{7}\times70) = Rs.\ 137280$

Therefore, required cost of canvas to make tent is $Rs.\ 137280$

Q5 What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use $\small \pi =3.14$ ).

Answer:

Given,

Base radius of the conical tent = $r =6\ m$

Height of the tent = $h =8\ m$

We know,

Curved surface area of a cone = $\pi rl = \pi r\sqrt{h^2 + r^2}$

$\therefore$ Area of tarpaulin required = Curved surface area of the tent

$\\ =3.14\times6\times \sqrt{8^2+ 6^2} \\ = 3.14\times6\times 10 \\ = 188.4\ m^2$

Now, let the length of the tarpaulin sheet be $x\ m$

Since $20\ cm$ is wasted, effective length = $x - 20 cm = (x - 0.2)\ m$

Breadth of tarpaulin = $3\ m$

$\\ \therefore [(x - 0.2) \times 3] = 188.4 \\ \Rightarrow x - 0.2 = 62.8 \\ \Rightarrow x = 63\ m$

Therefore, the length of the required tarpaulin sheet will be 63 m.

Q6 The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per $\small 100\hspace{1mm}m^2$ .

Answer:

Given, a conical tomb

The base diameter of the cone = $d =14\ m$

Slant height $= l = 25\ m$

We know, Curved surface area of a cone $= \pi r l$

$\\ = \frac{22}{7}\times\frac{14}{2}\times25 \\ \\ = 22\times25 \\ = 550\ m^2$

Now, Cost of whitewashing per $\small 100\hspace{1mm}m^2$ = $\small Rs.\ 210$

$\therefore$ Cost of whitewashing per $\small 550\hspace{1mm}m^2$ = $\small \\ Rs. (\frac{210}{100}\times550 )$

$\small \\ = Rs.\ (21\times55 ) = Rs.\ 1155$

Therefore, the cost of white-washing its curved surface of the tomb is $\small Rs.\ 1155$ .

Q7 A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Answer:

Given, a right circular cone cap (which means no base)

Base radius of the cone = $r=7\ cm$

Height $= h = 24\ cm$

$\therefore l = \sqrt{h^2+r^2}$

We know, Curved surface area of a right circular cone $= \pi r l$

$\therefore$ The curved surface area of a cap =

$\\ = \frac{22}{7}\times7\times\sqrt{24^2+7^2} \\ \\ = 22\times\sqrt{625} \\ = 22\times25\ \\ = 550\ cm^2$

$\therefore$ The curved surface area of 10 caps = $550\times10 = 5500\ cm^2$

Therefore, the area of the sheet required for 10 caps = $5500\ cm^2$

Q8 A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per $\small m^2$ , what will be the cost of painting all these cones? (Use $\small \pi =3.14$ and take $\small \sqrt{1.04}=1.02$ )

Answer:

Given, hollow cone.

The base diameter of the cone = $d = 40\ cm = 0.4\ m$

Height of the cone = $h = 1\ m$

$\therefore$ Slant height = $l = \sqrt{h^2+r^2}$ $= \sqrt{1^2+0.2^2}$

We know, Curved surface area of a cone = $\pi r l = \pi r\sqrt{h^2+r^2}$

$\therefore$ The curved surface area of 1 cone = $3.14\times0.2\times\sqrt{1.04} = 3.14\times0.2\times1.02$

$= 0.64056\ m^2$

$\therefore$ The curved surface area of 50 cones $= (50\times0.64056)\ m^2$

$= 32.028\ m^2$

Now, the cost of painting $\small 1\ m^2$ area = $\small Rs.\ 12$

$\therefore$ Cost of the painting $32.028\ m^2$ area $= Rs.\ (32.028\times12)$

$= Rs.\ 384.336$

Therefore, the cost of painting 50 such hollow cones is $Rs.\ 384.34\ (approx)$