Question

. What is the length of a chord, which is at a distance of 3 cm from the centre of a circle of
radius 5 cm?
(a) 8 cm
(b) 4cm (c) 2cm (d) 1 cm
​

Answer 1

Plz refers to the attachments .

$\bf \red{ \underline{ \underline{given}}}$

• OB= 5 cm (Radius)
• OM = 3 cm (Distance of chord from the centre)

$\bf \red{ \underline{ \underline{to \: find \: out}}}$

Find the length of a chord (AB).

$\bf \red{ \underline{ \underline{solution}}}$

$\orange{In \: the \: right \: triangle \: OMB,}$

$\bold{By \: pythagoras \: therome }$

$\implies \mathsf {OB {}^{2} =OM {}^{2} +BM {}^{2} }$

$\implies \mathsf {5 {}^{2} = 3 {}^{2} + BM {}^{2} }$

$\implies{25 = 9 + BM {}^{2}}$

$\implies{BM {}^{2} = 25 - 9}$

$\implies{BM = \sqrt{16} }$

$\implies{BM = 4}$

Since, the perpendicular from the centre to a chord bisects the chord.

$\therefore \: AB = 2BM =( 2 \times 4)cm \: = \: 8cm$

Option (a) 8 cm

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

The chord is at a distance of 3 cm from the centre of a circle of radius 5 cm.

To Find:

We need to find the length of chord.

Solution:

We can find the length of chord using pythagoras theorem.

We have,

OB^2 = OM^2 + BM^2

5^2 = 3^2 + BM^2

25 = 9 + BM^2

25 - 9 = BM^2

16 = BM^2

$\sqrt{16 } = {BM}^{2}$

OR BM = 4cm.

We know that the perpendicular from the center to a chord bisects the circle, so

AB = 2(BM)

=> AB = 2(4)

=> AB = 8cm

Hence, the length of chord is 8cm.