Question

$ABCD$ is trapezium with $\overline{AB}$ parallel to $\overline{CD}$ and the diagonals interested at the point $P$. The area of $△ABP$ is $32c{m}^2$ and the areea of $△CDP$ is $50c{m}^2$. The area of the trapezium is :

• A. $122c{m}^2$
• B. $364c{m}^2$
• C. $324c{m}^2$
• D. $164c{m}^2$

Answer 1

The correct option is A $122c{m}^2$

In the figure attached with this answer, a trapezium ABCD with AB parallel to CD

The diagonals intersect at P

The area of triangle ABP is 72 $c{m}^2$

Let $h_1$ be the height of triangle ABP and $h_2$ be the height of triangle CDP

Area of ABP = $\frac{1}{2}$x AB x$h_1$ = 72

$\frac{1}{2}\times b\times h_1$ = 72

$b\times h_1$= 72 x 2 =144

and of triangle CDP is 50$c{m}^2$

Area of CDP = $\frac{1}{2}\times CD\times h_2$

$\frac{1}{2}\times a\times h_1$ = 50

a x$h_2$ = 50 x 2 =100

Tthe area of trapezium

$\frac{1}{2}(a{h}_2+b{h}_1)=\frac{1}{2}(100+144)$ = 122 $c{m}^2$