Question

Question 1
AD is a diameter of a circle and AB is a chord. If AD = 34cm, AB = 30cm, the distance of AB from the centre of the circle is

(A) 17 cm
(B) 15 cm
(C) 4 cm
(D) 8 cm

Answer 1

(D) 8 cm

Given, AD = 34 cm and AB = 30 cm
Consider the figure given below,


Draw OL $\perp$ AB
Since the perpendicular from the centre of a circle to a chord bisects the chord,
$\therefore AL=LB=\frac{1}{2}AB=15cm$

In right angled $\Delta$ OLA,
$O{A}^2=O{L}^2+A{L}^2$
$\therefore (17{)}^2=O{L}^2+(15{)}^2$
$\Rightarrow 289=O{L}^2+225$
$\Rightarrow 289=O{L}^2+225$
$\Rightarrow O{L}^2=289-225=64$
$\therefore OL=8cm$
[taking positive square root, because lentgh is always positive]
Hence, the distance of the chord from the centre is 8cm.