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.
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.
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When a diameter is given, triangles formed with it as hypotenuse are right-angled.
Updated On: Aug 6, 2025
- $54\pi$ sq. cm
- $216\pi$ sq. cm
- $162\pi$ sq. cm
- None of these
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The Correct Option is D
Solution and Explanation
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.
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