Question

In a trapezium ABCD,
seg AB$||$ seg DC
seg BD$\perp$ seg AD,
seg AC$\perp$ seg BC,
If AD$=15$, BC$=15$ and AB$=25$. Find A$(\square$ ABCD$)$.

Answer 1

$AC=\sqrt{{25}^2-{15}^2}$

$AC=\sqrt{400}=20$

Suppose E is a perpendicular from D to AB.

$D{E}^2=A{E}^2+A{D}^2$ and $D{E}^2=E{B}^2+B{D}^2$

$A{E}^2+A{D}^2=E{B}^2+B{D}^2$

$x^2+{15}^2=(25-x{)}^2+{20}^2$

$225=625-50x+400$

$50x=800$

$x=16$

Hence, $D{E}^2=A{E}^2+A{D}^2$

$D{E}^2={16}^2+{15}^2=481$

$DE=\sqrt{481}$