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# A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of canvas required for the tent.

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Very important question of mensuration including
very important concept of it.need solution!
Rd sharma class 10 maths exercise-16.2 mensuration.

we have to find area of canvas required for the tent.
given that tent is in the form of a right circular cylinder
surmounted by a cone. The diameter of cylinder is 24 m.
The height of the cylindrical portion is 11 m while the
vertex of the cone is 16 m above the ground.

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1. Given,

The diameter of the cylinder (also the same for cone) = 24 m.

So, its radius (R) = 24/2 = 12 m

The height of the Cylindrical part (H1) = 11m

So, Height of the cone part (H2) = 16 – 11 = 5 m

Now,

Vertex of the cone above the ground = 11 + 5 = 16 m

Curved Surface area of the Cone (S1) = πRL = 22/7 × 12 × L

The slant height (L) is given by,

L = √(R2 + H22) = √(122 + 52) = √169

L = 13 m

So,

Curved Surface Area of Cone (S1) = 22/7 × 12 × 13

And,

Curved Surface Area of Cylinder (S2) = 2πRH1

S2 = 2π(12)(11) m2

Thus, the area of Canvas required for tent

S = S1 + S2 = (22/7 × 12 × 13) + (2 × 22/7 × 12 × 11)

S = 490 + 829.38

S = 1319.8 m2

S = 1320 m2

Therefore, the area of canvas required for the tent is 1320 m2

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