### AD is a diameter of a circle and AB is a chord. If AD = 26 cm, AB = 10 cm, find the perpendicular distance of AB from the centre of the circle.

**Answer:**

12 cm

**Step by Step Explanation:**

- Let us draw the figure as follows:

We are told that the diameter AD = 26 cm and the length of the chord AB = 10 cm.

We are asked to find the perpendicular distance of AB from the center O.

This is the length of the line segment OC. - We can see that △OAC is a right-angled triangle.

We also know that OA is the radius, i.e half of the diameter AD.

Therefore, AO = 13 cm. - We know that the perpendicular to the chord from the center i.e. OC, divides the chord into two equal parts.

Therefore, AC = CB =

=AB 2

= 5 cm10 2 - Since, △OAC is a right angle triangle, OC
^{2}+ AC^{2}= OA^{2}

⇒ OC^{2}+ 5^{2}= 13^{2}

⇒ OC^{2}= 169 - 25

⇒ OC^{2}= 144

⇒ OC = 12 cm - Therefore, perpendicular distance of AB from the centre of the circle is
**12 cm**.