The diagonals of a cyclic quadrilateral ABCD intersect at P and the area of the triangle APB is 24 square cm. If AB = 8 cm and CD = 5cm, then what is the area of the triangle CPD?

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24cm2

15cm2

12.5cm2

9.375cm2

answer is D.

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Detailed Solution

Given that, the diagonals of a cyclic quadrilateral ABCD intersect at P and the area of the triangle APB is 24 square cm. If AB = 8cm and CD = 5cm, we have to find the area of the triangle CPD.
Diagram for assistance:-

https://lh6.googleusercontent.com/z6NKHfS8-78SK7Agm8dpMTPOo_tlHFKgot3EA9lvvfvL0IZHN1sIYaGg7CmgzcZe4CoKfG8B_tbfpwPL6Lc2t7LlZxzLI_sbiAsLktkgdR2LIly72_iYn7Jk_ccEImmGoxMYo9VC

From the figure, it is clear that by the property of the cyclic quadrilateral, ⇒∠DCA = ∠ABD
⇒∠DCP = ∠ABP {as P is intersection of AC and BD}
Similarly, angle formed by chord BC in same segment ∠CDB = ∠BAC
⇒ ∠CDP = ∠BAP {as P is intersection of AC and BD}
⇒ ∠APB = ∠DPC (Opposite angles)
In ΔAPB and ΔCPD,
∠ABP = ∠DCP
∠BAP = ∠CDP
∠APB = ∠DPC
ΔAPB ≈ ΔCPD
Ratio of similar triangle = (ratio of sides)2