Question

A night camp was organised for Class X students for two days and their accommodation was

planned in tents. Each tent is made in the form of a frustum of a cone surmounted by a conical

top. The diameter of the base and the top of the frustum are 20 m and 6 m respectively and

height is 24 m. If the total height of the tent is 28 m, find the area of canvas required for making

the tent ​




NO SPAMMING!!!

Answer 1

The area of canvas required for making the tent is  1068.56 m².

Step-by-step explanation:

The diameter of the lower base of the frustum = 20 m

So, radius, R = 20/2 = 10 m

The diameter of the upper base of the frustum = 6 m

So, the radius, r = 6/2 = 3 m

Height of the frustum, h₁ = 24 m

Since the total height of the tent is given as 28 m

∴ The height of the conical top of the tent, h₂ = 28 - 24 = 4 m

Also we know that the radius of the top of the frustum will be equal to the radius of the base of the conical top, so, the radius of the cone, r = 3 m.

Now,

The slant height "l₁" of the frustum is given by,

= √[h₁² + (R - r)²]

= √[24² + (10 - 3)²]

= √[576 + 49]

= √[626]

= 25 m

and

The slant height "l₂" of the conical top is given by,

= √[h₂² + r²]

= √[4² + 3²]

= √[16 + 9]

= √[25]

= 5 m

Therefore,

The area of canvas required for making the tent is given,

= [C.S.A of the conical top] + [C.S.A of the frustum]

= [πrl₂] + [π(R+r)l₁]

= [(22/7) × 3 × 5] + [(22/7) × (10 + 3) × 25]

= 47.14 + 1021.42

= 1068.56 m²

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