Question

3. In a circle of radius 17 cm, find the distance of a chord of length 16 cm from the centre​

Answer 1

Given :-

• radius of the circle = 17 cm
• length of the chord = 16 cm

To Find :-

• find the distance of the chord from the centre ?

Solution :-

we know that,

• A Perpendicular drawn from the centre of the circle to the chord ,bisects the chord.Hence,the point will be the mid point of the line segment

(see the refer diagram)

here,

• O is the centre
• AB is chord
• OB is radius
• C is the mid point of chord AB

➪ CB = AB/2

➪ CB = 16/2

➪ CB = 8 cm

now, in right angled triangle OBC

by Pythagoras theorem ;

$\bold{ :↦ \:{ (CB)}^{2} +{ (OC)}^{2} = {(OB)}^{2} } \\ \\ \bold{: ↦\: {8}^{2} + { (OC)}^{2} = {17}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: ↦ \: 64 + { (OC)}^{2} = 289 } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{:↦ \:{ (OC)}^{2} = 289 - 64 } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{:↦ \:{ (OC)}^{2} = 225} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : ↦ \: OC = \sqrt{225} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{:↦ \: \boxed{ \bold{OC = 15 \: cm }}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

hence, the distance of the chord from the centre is 15 cm