Question

$ABCD$ is trapezium in which $AB\parallel DC,AB=78cm,CD=52cm,AD=28cm$ and $BC=30cm$. Find the area of the trapezium.

Answer 1

R.E.F image

AS $AB\left|\right|CD$ and DM and

$CN\perp AB\Rightarrow DM=CN=h$

In $△DMA,D{A}^2=A{M}^2+D{M}^2$

$\Rightarrow {28}^2=(26-X{)}^2+h^2...(1)$

In $△CNB,C{B}^2=C{N}^2+N{B}^2$

$\Rightarrow {30}^2=X^2+H^2...\left(2\right)$

$E{q}^n\left(1\right)...E{q}^2\left(2\right)-58\times 2=(26-x{)}^2-x^2$

$116=x^2-(x-26{)}^2=(2x-26)26$

$\Rightarrow x-13=\frac{116}{26\times 2}=\frac{29}{13}$

By$e{q}^n,h^2={30}^2-\frac{{29}^2}{{13}^2}=900-\frac{841}{169}=895.02$

$\Rightarrow h=29.9cm$

Area trapezium=$\frac{1}{2}$(sum of parallal sides)height=$\frac{1}{2}\left(130\right)\left(29.2\right)$

$\Rightarrow Area=1898c{m}^2$