Question

In the adjoining figure, points $A,B,C,D$ lie on a circle. $AD=24$ and $BC=12$. What is the ratio of the area of $\triangle CBE$ to that of $\triangle ADE$? 

 

• A. $1:4$
• B. $1:2$
• C. $1:3$
• D. Insufficient data

Hint:

When two chords intersect inside a circle, triangles formed by corresponding chord-pairs are often similar; area ratios then follow from the square of the chord-length ratio.

Answer 1

The Correct Option is A

\(E\) is the intersection of chords \(AB\), \(CD\). Vertical angles at \(E\) are equal, and inscribed angles \(\angle ADE=\angle CBE\), so \(\triangle ADE\sim\triangle CBE\) with ratio \(\dfrac{AD}{CB}=\dfrac{24}{12}=2\).

1. Using \(\text{Area}=\tfrac12(\text{side}_1)(\text{side}_2)\sin\theta\) with the common angle \(\theta\) at \(E\), the area ratio equals the square of the side ratio.
2. \(\dfrac{[CBE]}{[ADE]}=\left(\dfrac{CB}{AD}\right)^2=\left(\dfrac{1}{2}\right)^2=\dfrac14\).

So the ratio of the areas is \(1:4\).