Question

1.A conical tent was erected by the army
at a base camp with height 3m. and base
diameter 8m. Find the cost of canvas
required for making the tent, if the canvas
cost 70 per 1 sq.m.

Answer 1

$\huge\mathbb \red{ANSWER}$

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$\sf \huge\underline{Question}$

A conical tent was erected by the army at a base camp with height 3m. and base

diameter 8m.

Find the cost of canvas required for making the tent, if the canvas cost 70 per 1 sq.m.

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$\sf \huge\underline{Given}$

$\tt \implies \red{diamter = 3m}$

$\tt \implies \blue{r = \dfrac{d}{2} = \dfrac{8}{2} = 4m}$

$\tt \implies \pink{height \: is \: 3m}$

• do u know slant height...

$\tt \implies \purple{slant \: height = (l) = \sqrt{ {h}^{2} + {r}^{2} }}$

$\tt \implies \purple{ = \sqrt{ {3}^{2} + {4}^{2} = \sqrt{25} = 5m }}$

• now curved surface area of tent and volume of cone

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$\rm{ \boxed{ \underline\blue{ \tt{ \: curved \: area \: of \: tent = \pi \: rl \: }}}}$

$\rm \implies \red{ \frac{22}{7} \times 4 \times 5 = \frac{440}{7} {m}^{2}}$

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• now volume of cone

$\rm{ \boxed{ \underline\blue{ \tt{ \: volume \: of \: cone = \frac{1}{3} \pi \: {r}^{2}h \: }}}}$

$\tt \implies \green{ = \dfrac{1}{3} \times \dfrac{22}{7} \times 4 \times 4 \times 3}$

$\tt \implies \orange{ \dfrac{352}{7} {m}^{2}}$

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Given:

Find the cost of canvas required for making the tent, if the canvas cost 70 per 1 sq.m.

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$\sf \underline{cost \: of \: canvas \: fr \: tent}$

$\sf \implies \blue{csa \times unit}$

$\sf \implies \orange{ = \dfrac{440}{70} \times 70}$

$\sf \implies \purple{ = 4400 \: rupees}$

Answer 2

$hlo$

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Question - conical tent was erected by the army at a base camp with height 3m. and base

diameter 8m.

Find the cost of canvas required for making the tent, if the canvas cost 70 per 1 sq.m.

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Given -

Diameter - 3cm

$r=2d=28=4m</p><p></p><p>\tt \implies \pink{height \: is \: 3m}⟹heightis3m</p><p></p><p>do u know slant height...</p><p></p><p>\tt \implies \purple{slant \: height = (l) = \sqrt{ {h}^{2} + {r}^{2} }}⟹slantheight=(l)=h2+r2</p><p></p><p>\tt \implies \purple{ = \sqrt{ {3}^{2} + {4}^{2} = \sqrt{25} = 5m }}</p><p></p><p>$

$⟹=32+42=25=5m$

• now curved surface area of tent and volume of cone

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Curved area of tent -

$</p><p>722×4×5=7440m2</p><p></p><p>_____________________________________</p><p></p><p>now volume of cone</p><p></p><p>\rm{ \boxed{ \underline\blue{ \tt{ \: volume \: of \: cone = \frac{1}{3} \pi \: {r}^{2}h \: }}}}volumeofcone=31πr2h</p><p></p><p>\tt \implies \green{ = \dfrac{1}{3} \times \dfrac{22}{7} \times 4 \times 4 \times 3}</p><p></p><p>$

$now volume of cone</p><p></p><p>\rm{ \boxed{ \underline\blue{ \tt{ \: volume \: of \: cone = \frac{1}{3} \pi \: {r}^{2}h \: }}}}volumeofcone=31πr2h</p><p></p><p></p><p>$

so answer will be 4450 rupess