Question

In the given circle with centre O, AD = DB = 16 cm and DE = 8 cm. Find the radius of the circle.

Answer 1

Given:

• AD = DB = 16 cm   ...(So AB = 32 cm, chord length)
• DE = 8 cm

Steps:

1. Let the radius of the circle be r = OA.

2. Let the distance from the center O to the midpoint D of chord AB be OD = x.

3. Since AD = 16 cm, by right triangle OAD, use Pythagoras theorem:

r² = x² + 16²

r² = x² + 256

4. Point E lies on the line OD such that segment DE = 8 cm.

5. Because O, D, E are collinear, OE = r can be expressed as:

OE = OD + DE

OE = x + 8

6. Also, the radius squared is:

r² = OE²

= (x + 8)²

= x² + 16x + 64

7. Equate both expressions for r²:

x² + 256 = x² + 16x + 64

Simplifying:

256 = 16x + 64 

256 − 64 = 16x

16x = 192 

x = `192/16`

x = 12

8. Calculate radius:

r = x + 8

= 12 + 8

= 20 cm

The radius of the circle is 20 cm.