NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.1 - Surface Area and Volumes

Q1 Diameter of the base of a cone is $10.5cm$ and its slant height is $10cm$ . Find its curved surface area.

Answer:

Given,

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

Slant height = $l=10cm$

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

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

$$\begin{gathered} =\frac{22}{7}\times \frac{10.5}{2}\times 10 \\ =165c{m}^2 \end{gathered}$$

Q2 Find the total surface area of a cone, if its slant height is $21m$ and diameter of its base is $24m$ .

Answer:

Given,

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

Slant height = $l=21cm$

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

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

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

$$\begin{gathered} =\frac{22}{7}\times \frac{24}{2}\times (21+12) \\ =\frac{22}{7}\times \frac{24}{2}\times 33 \\ =1244.57m^2 \end{gathered}$$

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

Answer:

Given,

The curved surface area of a cone = $308c{m}^2$

Slant height $=l=14cm$

(i) Let the radius of cone be $rcm$

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

$\therefore$ $$\begin{gathered} \pi rl=308 \\ \Rightarrow \frac{22}{7}\times r\times 14=308 \\ \Rightarrow r=\frac{308}{44}=7 \end{gathered}$$

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

Q3 (ii) Curved surface area of a cone is $308c{m}^2$ and its slant height is $14cm$ . Find total surface area of the cone.

Answer:

Given,

The curved surface area of a cone = $308c{m}^2$

Slant height $=l=14cm$

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

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

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

$$\begin{gathered} =308+\frac{22}{7}\times 7^2 \\ =308+154=462c{m}^2 \end{gathered}$$

Therefore, the total surface area of the cone is $462c{m}^2$

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 = $r=24m$

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

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

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

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

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 $1m^2$ canvas is Rs 70.

Answer:

Given,

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

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

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

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

$\therefore$ Curved surface area of the tent

$$\begin{gathered} =\frac{22}{7}\times 24\times 26 \\ =\frac{13728}{7}{m}^2 \end{gathered}$$

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

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

$Rs.(\frac{13728}{7}\times 70)=Rs.137280$

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 $\pi =3.14$ ).

Answer:

Given,

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

Height of the tent = $h=8m$

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

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

Now, let the length of the tarpaulin sheet be $xm$

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

Breadth of tarpaulin = $3m$

$$\begin{gathered} \therefore [(x-0.2)\times 3]=188.4 \\ \Rightarrow x-0.2=62.8 \\ \Rightarrow x=63m \end{gathered}$$

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 $100m^2$ .

Answer:

Given, a conical tomb

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

Slant height $=l=25m$

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

$$\begin{gathered} =\frac{22}{7}\times \frac{14}{2}\times 25 \\ =22\times 25 \\ =550m^2 \end{gathered}$$

Now, Cost of whitewashing per $100m^2$ = $Rs.210$

$\therefore$ Cost of whitewashing per $550m^2$ = $Rs.(\frac{210}{100}\times 550)$

$=Rs.(21\times 55)=Rs.1155$

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

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 = $r=7cm$

Height $=h=24cm$

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

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

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

$$\begin{gathered} =\frac{22}{7}\times 7\times \sqrt{{24}^2+7^2} \\ =22\times \sqrt{625} \\ =22\times 25 \\ =550c{m}^2 \end{gathered}$$

$\therefore$ The curved surface area of 10 caps = $550\times 10=5500c{m}^2$

Therefore, the area of the sheet required for 10 caps = $5500c{m}^2$

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 $m^2$ , what will be the cost of painting all these cones? (Use $\pi =3.14$ and take $\sqrt{1.04}=1.02$ )

Answer:

Given, hollow cone.

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

Height of the cone = $h=1m$

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

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

$\therefore$ The curved surface area of 1 cone = $3.14\times 0.2\times \sqrt{1.04}=3.14\times 0.2\times 1.02$

$=0.64056m^2$

$\therefore$ The curved surface area of 50 cones $=(50\times 0.64056)m^2$

$=32.028m^2$

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

$\therefore$ Cost of the painting $32.028m^2$ area $=Rs.(32.028\times 12)$

$=Rs.384.336$

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