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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 joker’s cap is in the form of a right circular cone of base radius 7cm and height 24cm. Find the area of the sheet required to make 10 such caps.

ANSWER:

It is given that

Radius of the cap = 7cm

Height of the cap = 24cm

We know that

The slant height of the cap

l = √ (r2 + h2)

By substituting the values

l = √ (72 + 242)

On further calculation

l = √ (49 + 576) = √ 625

So we get

l = 25cm

We know that

The curved surface area of one cap = πrl

By substituting the values

The curved surface area of one cap = (22/7) × 7 × 25

So we get

The curved surface area of one cap = 550 cm2

So the curved surface area of 10 conical caps = 10 × 550 = 5500 cm2

**Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.**

ANSWER:

It is given that

Radius of the cap = 7cm

Height of the cap = 24cm

We know that

The slant height of the cap

l = √ (r2 + h2)

By substituting the values

l = √ (72 + 242)

On further calculation

l = √ (49 + 576) = √ 625

So we get

l = 25cm

We know that

The curved surface area of one cap = πrl

By substituting the values

The curved surface area of one cap = (22/7) × 7 × 25

So we get

The curved surface area of one cap = 550 cm2

So the curved surface area of 10 conical caps = 10 × 550 = 5500 cm2

**Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.**

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