Question

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 $\angle AOB$ and if it bisects AB at ${90}^{\circ },$ then what is the length of AB?

• A. 6 cm
• B. 3 cm
• C. 8 cm
• D. 5 cm

Answer 1

The correct option is C 8 cm

Given, OE bisects AB at 90${}^o$, implies $△AOE$ is right-angled.
$O{A}^2=O{E}^2+A{E}^2(\text{By using Pythagoras' theorem)}$
i.e., $5^2=3^2+A{E}^2$
$⟹AE=4\text{cm}$
We have OE $\perp$ AB, so AE = EB.
[Perpendicular drawn from center to a chord bisects the chord]
$So,AB=2\times AE=2\times 4\text{cm}=8\text{cm}.$