Question

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

• A. 17cm
• B. 15cm
• C. 4cm
• D. 8cm

Answer 1

Answer: (D)

Given, AD = 34 cm and AB = 30 cm

In figure, OM is perpendicular to AB.

Further, the perpendicular from the centre of a circle to a chord bisects the chord.

AM = BM = $\frac{1}{2}AB=15cm$

In right angled triangle OMA, ${OA}^2={OM}^2+{AM}^2$ [BY using Pythagoras theorem ]

${17}^2={OM}^2+{15}^2$

$OM=\sqrt{289-225}$

$OM=8cm$

Taking positive square root , because length is always positive.

Hence the distance of the chord from the centre is $8cm$

option D will be the answer.