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Rs aggarwal solutions

CHAPTERS

3. Factorization of Polynomials

4. Linear Equations In Two Variables

6. Introduction To Euclids Geometry

9. Congruence Of Triangles And Inequalities In A Triangle

11. Areas Of Parallelograms And Triangles

14. Areas Of Triangles And Quadrilaterals

15. Volume And Surface Area Of Solids

A conical tent is 10m high and the radius of its base is 24m. Find the slant height of the tent. If the cost of 1m2 canvas is ₹ 70, find the cost of canvas required to make the tent.

ANSWER:

It is given that

The radius of the conical tent = 24m

Height of conical tent = 10m

We know that

Slant height of conical tent can be written as

l = √ (r2 + h2)

By substituting the values

l = √ (242 + 102)

On further calculation

l = √ (576 + 100) = √ 676

So we get

l = 26m

We know that

The curved surface area of conical tent = πrl

By substituting the values

The curved surface area of conical tent = (22/7) × 24 × 26

So we get

The curved surface area of conical tent = (13728/7) m2

It is given that the cost of 1m2 canvas = ₹ 70

So the cost of (13728/7) m2 canvas = ₹ 70 × (13728/7) = ₹ 137280

**Therefore, the slant height of the tent is 26m and the cost of canvas required to make the tent is ₹ 137280.**

ANSWER:

It is given that

The radius of the conical tent = 24m

Height of conical tent = 10m

We know that

Slant height of conical tent can be written as

l = √ (r2 + h2)

By substituting the values

l = √ (242 + 102)

On further calculation

l = √ (576 + 100) = √ 676

So we get

l = 26m

We know that

The curved surface area of conical tent = πrl

By substituting the values

The curved surface area of conical tent = (22/7) × 24 × 26

So we get

The curved surface area of conical tent = (13728/7) m2

It is given that the cost of 1m2 canvas = ₹ 70

So the cost of (13728/7) m2 canvas = ₹ 70 × (13728/7) = ₹ 137280

**Therefore, the slant height of the tent is 26m and the cost of canvas required to make the tent is ₹ 137280.**

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