Question

In the accompanying figure CE and DE are equal chords of a circle with centre O. Arc AB is a quarter-circle. Then the ratio of the area of triangle CED to the area of triangle AOB is:

• A. $2:1$
• B. $\sqrt{3}:1$
• C. $4:1$
• D. $3:1$

Answer 1

Answer: (A)

$2:1$

Arc AB is quarter circle.

$\therefore \angle AOB=\frac{\pi }{2}$

$\therefore Area\left(△AOB\right)=\frac{1}{2}\left(OA\right)\left(OB\right)=\frac{1}{2}{r}^{2}OA=OB=OC=OD=radius=r$

CED is an isosceles triangle ($\because CE=DE$)

$\therefore$ Median and perpendicular is same for an isosceles triangle.

$\therefore OE$ is perpendicular to $CD$.

$\therefore Area\left(△CED\right)=\frac{1}{2}\left(OE\right)\left(CD\right)=\frac{1}{2}\left(r\right)\left(2r\right)=r^2\therefore Area\left(△CED\right):Area\left(△AOB\right)=2:1$