In the given figure, AB is a chord that is 3 cm away from the centre O of a circle of radius 5 cm. If OE is the angle bisector of $∠ A O B$ and if it bisects AB at $90^{\circ}$ then what is the length of AB?

Open in App

Solution

Given, OE bisects AB at $90^{\circ}$ , implies $Δ A O E$ is right-angled.
$O A^{2} = O E^{2} + A E^{2}$ {By using Pythagoras' theorem}
i.e., $(5)^{2} = (3)^{2} + A E^{2}$
$\Rightarrow A E = 4 c m$
We have $O E ⊥ A B$ , so $A E = E B$ .
[Perpendicular drawn from center to a chord bisects the chord]
So, $A B = 2 \times A E = 2 \times 4 c m = 8 c m$

Suggest Corrections

Similar questions

In a circle of radius 10 cm given below, chord AB and CD are equal. If OE bisects AB and OF bisects CD and OF = 6 cm, then length EB is ________ .

Chords AB and CD of a circle with centre O intersect at a point E. If OE bisects angle AED, then prove that AB = CD.

Join BYJU'S Learning Program