Question

In a trapezium $ABCD$, $AB\parallel DC$ , $BD\perp AD,AC\perp BC$, If $AD=15,BC=15$ and $AB=25$. Find $Ar\left(▢ABCD\right)$

Answer 1

Given:

In trapezium $ABCD$,

$AB\parallel DC$

$BD\perp AD$

$AC\perp BC$

$AD=15,BC=15$ and $AB=25$

Now, In $\Delta ABD$

According to Pythagoras theorem,
$A{B}^2=A{D}^2+D{B}^2$

$\Rightarrow {25}^2={15}^2+B{D}^2$

$\Rightarrow 625=225+B{D}^2$

$\Rightarrow B{D}^2=625-225$

$\Rightarrow B{D}^2=400$

$\Rightarrow BD=20$
Now,
Area of the $\Delta ABD=\frac{1}{2}\times AD\times BD$

$=\frac{1}{2}\times 15\times 20$

$=15\times 10$

$\Rightarrow$ Area of the $\Delta ABD=150$ sq. units ---(1)

in figure, by construction $DP\perp AB$

$\Rightarrow$ Area of the $\Delta ABD=\frac{1}{2}\times AB\times DP$

$\Rightarrow \frac{1}{2}\times 25\times DP=150$

$\Rightarrow DP=150\times \frac{2}{25}$

$\Rightarrow DP=12$

Thus the height of trapezium is $12$ units.

Now, In $\Delta ADP$
According to Pythagoras theorem,
$A{D}^2=A{P}^2+D{P}^2$

$\Rightarrow {15}^2={12}^2+A{P}^2$

$\Rightarrow 225=144+A{P}^2$

$\Rightarrow A{P}^2=225-144$

$\Rightarrow A{P}^2=81$

$\Rightarrow AP=9$

Since, By RHS congruence, $\Delta APD\cong \Delta BQC$

and $AP=QB$

$QB = 9$

Now, $CD=PQ=25-(9+9)=7$

Therefore, in trapezium $ABCD$,

Lengths of the parallel sides are $AB=25$ units and $CD=7$ units

Distance between the parallel sides $=12$ units
Area of Trapezium $=\frac{1}{2}\times$ (Sum of parallel sides) $\times$ Height

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

$=\frac{1}{2}\times \left(32\right)\times 12$

$=16\times 12$

$=192$

Hence, $A\left(▢ABCD\right)=192$ sq. units.