Question

In a trapezium $ABCD$ with $AB$ parallel to $CD,$ the diagonals intersect at $P.$ The area of $△ABP$ is $72{cm}^2$ and of $\Delta CDP$ is $50{cm}^2.$ Find the area of the trapezium.

Answer 1

Assuming the Trapezium to be isosceleus

Consider $△ABP$

Area of $△=\frac{1}{2}\times PA\times PB=72\text{cm}$

Since $PA=PB⟹PA=12\text{cm}$

Consider $△CDP$

Area of $△=\frac{1}{2}\times PC\times PD=50\text{cm}$

Since $PC=PD⟹PC=10\text{cm}$

Consider $△APC$

Area of $△=\frac{1}{2}\times PC\times PA=\frac{1}{2}\times 10\times 12=60\text{cm}$

Consider $△BPD$

Area of $△=\frac{1}{2}\times PD\times PB=\frac{1}{2}\times 10\times 12=60\text{cm}$

Area of trapezium $72+60+60+50=240{\text{cm}}^2$