Question

2. The radius of a circle is 6 cm. The perpendicular distance from the centre of
the chord which is 8 cm in length, is
(a) V5 cm (b) 215 cm (c)
27 cm (a fa​

Answer 1

Question :-

The radius of a circle is 6 cm. The perpendicular distance from the centre of

the chord which is 8 cm in length, is

Answer -

( Refer to the attachment )

We know that,

The perpendicular from centre of circle bisects the chord.

So,

➩ AB = BC = 8/2 = 4

Now we have,

➩ OC = 6 cm

➩ BC = 4cm

We need to find length of OB -

Applying Pythagoras theorem -

➩ $\sf (BC)^2 + (OB)^2 = (OC)^2$

➩ $\sf 4^2 + (OB)^2 = 6^2$

➩ $\sf (OB)^2 + 16 = 36$

➩ $\sf (OB)^2 = 20$

➩ $\sf OB = \sqrt{20}$

➩ $\boxed{\sf OB = 2 \sqrt{5}}$

Answer 2

Given :-

• The radius of a circle is 6 cm. The perpendicular distance from the centre of the chord which is 8 cm in length, is

To find :-

$\boxed{\bf{(Refer~to~the~Attachment)}}$

★ We know that ★

The perpendicular from centre of circle bisects the chord.

So,

→ AB = BC = $\frac{8}{2}$ = 4

★ Now we have ★

→ OC = 6 cm

→ BC = 4 cm

We need to find length of OB -

Applying Pythagoras theorem -

→ (BC)² + (OB)² = (OC)²

→ 4² + (OB)² = 6²

→ (OB)² + 16 = 36

→ (OB)² += 20

→ OB = $\sqrt{20}$

→ $\boxed{\bf{OB = 2\sqrt{5}}}$