Question

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

Answer 1

We have,

ABCD as the cyclic quadrilateral in which the diagonal AC and BD.

intersect each other at point P.

also, given that,

AB$=8$cm,

CD$=5$cm

Now,

In $\Delta$DCA and $\Delta$APB,

We have

$\angle DCP=\angle ABP$

$\angle CDP=\angle PAB$

Hence,

$\Delta DPC\sim \Delta APB$ (by A.A property)

According to the given question,

$\frac{ar\Delta DPC}{ar\Delta APB}={\left(\frac{DC}{AB}\right)}^2$

$\Rightarrow \frac{ar\Delta DPC}{24}={\left(\frac{5}{8}\right)}^2$

$\Rightarrow \frac{ar\Delta DPC}{24}=\frac{25}{64}$

$\Rightarrow ar\Delta DPV=\frac{25\times 24}{64}$

$ar\Delta DPC=9.375c{m}^2$

Hence, this is the answer.