Question

In a tropezium ABCD
seg AB ll seg DC
seg BD $\perp$ seg AD
seg AC $\perp$ seg BC
If AD =15 , BC =15 and AB = 25
Find $\square \left(ABCD\right)$

Answer 1

In $△$ADB, using Pythagoras theorem,

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

${15}^2+B{D}^2={25}^2$ $\Rightarrow BD=\sqrt{625-225}=\sqrt{400}$

$BD=20$

ar$△$ADB $=\frac{1}{2}\ast AD\ast BD$

$\Rightarrow \frac{1}{2}\ast AB\ast DE=\frac{1}{2}\ast AD\ast BD$

$\Rightarrow 25\ast DE=15\ast 20$

$DE=\frac{300}{25}=12$

$\therefore$ In $△AED$ , using Pythagoras theorem

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

$AE=\sqrt{{15}^2-{12}^2}=\sqrt{225-144}=\sqrt{81}=9$

Similarly in $△BFC$,

$BF=9$

$\therefore$ $EF=CD=AB-2\ast AE=25-(2\ast 9)=7$

Hence, $ar\left(ABCD\right)=\frac{1}{2}\ast (AB+CD)\ast DE$

$=\frac{1}{2}\ast (25+7)\ast 12$

$=192$ sq. units