Question

AB is the diameter of the given circle, while points C and D lie on the circumference as shown. If AB is 15 cm, AC is 12 cm and BD is 9 cm, find the area of the quadrilateral ACBD. 

• A. $54\pi$ sq. cm
• B. $216\pi$ sq. cm
• C. $162\pi$ sq. cm
• D. None of these

Hint:

When a diameter is given, triangles formed with it as hypotenuse are right-angled.

Answer 1

The Correct Option is D

ACBD can be split into two right triangles: $\triangle ABC$ and $\triangle ABD$ since AB is diameter. For $\triangle ABC$: AB = 15, AC = 12 $\Rightarrow$ BC = $\sqrt{15^2 - 12^2} = \sqrt{225 - 144} = 9$ cm. Area = $\frac{1}{2} \times AC \times BC = \frac{1}{2} \times 12 \times 9 = 54$ sq. cm. For $\triangle ABD$: AB = 15, BD = 9 $\Rightarrow$ AD = $\sqrt{15^2 - 9^2} = \sqrt{225 - 81} = 12$ cm. Area = $\frac{1}{2} \times AD \times BD = \frac{1}{2} \times 12 \times 9 = 54$ sq. cm. Total area = $54 + 54 = 108$ sq. cm → not matching any $\pi$-based Option, so correct = None of these.