Exercise-11.1, Class 9th, Maths, Exercise 11, NCERT

Useful formula used: Curved surface area (CSA) of a cone $=\pi rl$, total surface area $=\pi r(l+r)$
(Where $r$ = radius of base, $l$ = slant height.)

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Q1

Diameter of base = 10.5 cm, slant height $l=10$ cm.
Radius $r=10.5\mathrm{/}2=5.25$

$$\text{CSA}=\pi rl=\frac{22}{7}\times 5.25\times 10=\frac{22\times 52.5}{7}=165{\text{cm}}^2\mathrm{}$$

Answer: $165{\text{cm}}^2\mathrm{}$

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Q2

Slant height $l=21$ m, diameter $=24$ m$\Rightarrow r=\mathrm{12\; m}$.
Total surface area $=\pi r(l+r)$

$$\text{TSA}=\frac{22}{7}\times 12\times (21+12)=\frac{22}{7}\times 12\times 33=\frac{8712}{7}{\text{m}}^2\approx 1244.57{\text{m}}^2\mathrm{}$$

Answer: ${\displaystyle \frac{8712}{7}}{\text{m}}^2\approx 1244.57{\text{m}}^2\mathrm{}$

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Q3

CSA $=308{\text{cm}}^2,l=14$.
(i) Find $r$ from $\pi rl=308$

$$r=\frac{308}{\pi l}=\frac{308}{(22\mathrm{/}7)\times 14}=\frac{308}{308\mathrm{/}7}=7\text{ cm}\mathrm{.}$$

(ii) Total surface area $=\pi r(l+r)=\frac{22}{7}\times 7\times (14+7)=22\times 21=462{\text{cm}}^2\mathrm{}$

Answer: (i) $r=7$. (ii) TSA $=462{\text{cm}}^2\mathrm{}$

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Q4

Tent: height $h=10$ m, base radius $r=24$ m.

(i) Slant height $l=\sqrt{r^2+h^2}=\sqrt{{24}^2+{10}^2}=\sqrt{576+100}=\sqrt{676}=26$

(ii) CSA $=\pi rl=\frac{22}{7}\times 24\times 26$

Cost of canvas @ ₹70 per m${}^2$:

$$\text{CSA}=\frac{22}{7}\times 24\times 26=\frac{13728}{7}{\text{m}}^2,$$

$$\text{Cost}=\text{CSA}\times 70=\frac{13728}{7}\times 70=13728\times 10=\text{₹}\mathrm{137,280}$$

Answers: (i) $l=26$. (ii) Cost $=\text{₹}\mathrm{137,280}$

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Q5

Tarpaulin width = 3 m. Cone: height $h=\mathrm{8\; m}$, radius $r=6$ m.
(Use $\pi =3.14$) Add 20 cm = 0.20 m extra length for stitching/wastage.

Slant height $l=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$

CSA $=\pi rl=3.14\times 6\times 10=188.4{\text{m}}^2\mathrm{}$

Length of tarpaulin required $={\displaystyle \frac{\text{CSA}}{\text{width}}}={\displaystyle \frac{188.4}{3}}=62.8\text{ m}\mathrm{}$

Add 0.20 m wastage ⇒ required length $=62.8+0.20=63.0\text{ m}\mathrm{}$

Answer: $63.0\text{ m}$ of tarpaulin (3 m wide).

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Q6

Slant height $l=25$ m, base diameter $=14$ ⇒ $r=7$.
CSA $=\pi rl=\frac{22}{7}\times 7\times 25=22\times 25=550{\text{m}}^2\mathrm{}$

White-washing cost @ ₹210 per 100 m${}^2$:

$$\text{Cost}=\frac{210}{100}\times 550=2.1\times 550=\text{₹}1155.$$

Answer: Cost $=\text{₹}\mathrm{1,155}$

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Q7

Joker’s cap: right circular cone, $r=7$ cm, $h=24$ cm.
Slant height $l=\sqrt{7^2+{24}^2}=\sqrt{49+576}=\sqrt{625}=25$

Area of sheet for one cap = CSA = $\pi rl=\frac{22}{7}\times 7\times 25=22\times 25=550{\text{cm}}^2\mathrm{}$

For 10 caps: $10\times 550=5500{\text{cm}}^2\mathrm{}$

Answer: $5500{\text{cm}}^2$ of sheet for 10 caps.

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Q8

50 hollow cones, outer paint required on outer curved surface.
Base diameter $=40$ cm ⇒ $r=20$ cm, height $h=1$
$=100$ cm. Use $\pi =3.14$ and $\sqrt{1.04}=1.02$ (given).

Slant height:

$$l=\sqrt{r^2+h^2}=\sqrt{{20}^2+{100}^2}=\sqrt{400+10000}=\sqrt{10400}\mathrm{.}$$

Compute with the approximation: $\sqrt{10400}=10\sqrt{104}=10\times (10\sqrt{1.04})=100\sqrt{1.04}$
Given $\sqrt{1.04}\approx 1.02$$\approx 100\times 1.02=102\text{ cm}\mathrm{}$

CSA per cone $=\pi rl=3.14\times 20\times 102=3.14\times 2040=6405.6{\text{cm}}^2$

Convert to m${}^2$: $6405.6{\text{cm}}^2=0.64056{\text{m}}^2\mathrm{}$

Total area for 50 cones $=50\times 0.64056=32.028{\text{m}}^2\mathrm{}$

Painting cost @ ₹12 per m${}^2$:

$$\text{Cost}=32.028\times 12=\text{₹}384.336\approx \text{₹}384.34$$

Answer: Approx ₹384.34 to paint the outer sides of all 50 cones.