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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 chord of length 30cm is drawn at a distance of 8cm from the centre of a circle. Find out the radius of the circle.

Consider AB as the chord of the circle with O as the centre

Construct OL ⊥ AB

From the figure, we know that OL is the distance from the centre of the chord

It is given that AB = 30cm and OL = 8cm

The perpendicular from the centre of a circle to a chord bisects the chord

So we get

AL = ½ × AB

By substituting the values

AL = ½ × 30

By division

AL = 15cm

Consider △ OLA

Using the Pythagoras theorem it can be written as

OA^2 = OL^2 + AL^2

By substituting the values we get

OA^2 = 8^2 + 15^2

On further calculation

OA^2 = 64 + 225

By addition

OA^2 = 289

By taking the square root

OA = √289

So we get

OA = 17cm

Therefore, the radius of the circle is 17cm.

Consider AB as the chord of the circle with O as the centre

Construct OL ⊥ AB

From the figure, we know that OL is the distance from the centre of the chord

It is given that AB = 30cm and OL = 8cm

The perpendicular from the centre of a circle to a chord bisects the chord

So we get

AL = ½ × AB

By substituting the values

AL = ½ × 30

By division

AL = 15cm

Consider △ OLA

Using the Pythagoras theorem it can be written as

OA^2 = OL^2 + AL^2

By substituting the values we get

OA^2 = 8^2 + 15^2

On further calculation

OA^2 = 64 + 225

By addition

OA^2 = 289

By taking the square root

OA = √289

So we get

OA = 17cm

Therefore, the radius of the circle is 17cm.

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