Question

The parallel sides of a trapezium is $77cm$ and $60cm$ and its non-parallel sides are $26cm$, and $25cm$. The area of the trapezium is equal to

• A. ${1600cm}^2$
• B. ${1666cm}^2$
• C. ${1644cm}^2$
• D. ${1500cm}^2$

Answer 1

The correct option is C ${1644cm}^2$

Let ABCD be a trapezium where AB//DC

$\therefore AB=60$cm

$DC=77$cm

$AD=26$cm

$BC=25$cm

Draw $AD\perp DC$ and $BQ\perp DC$

$\therefore PQ=60$cm and $AP=BQ=h$

Let $DP=x$ and $QC=y$

In right $\Delta APD$,

Using Pythagoras theorem, we have

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

$\Rightarrow h^2+x^2=(26{)}^2$

$\Rightarrow h^2={26}^2-x^2$ ………..$\left(1\right)$

In right $\Delta BQC$,

Using Pythagoras theorem, we have

$B{Q}^2+Q{C}^2=B{C}^2$

$\Rightarrow h^2+y^2={25}^2$

$\Rightarrow h^2={25}^2-y^2$ …………..$\left(2\right)$

Equating equations $\left(1\right)$ and $\left(2\right)$, we have

${26}^2-x^2={25}^2-y^2$

$\Rightarrow x^2-y^2={26}^2-{25}^2$

$\Rightarrow x^2-y^2=676-625$

$\Rightarrow x^2-y^2=51$ ……….$\left(3\right)$

Also, we have, $x+60+y=77$

$\Rightarrow x=77-60-y$

$\Rightarrow x=17-y$

Substituting value of x in $\left(3\right)$ we get

$(17-y{)}^2-y^2=51$

${17}^2+y^2-34y-y^2=51$

$289-51=34y$

$y=\frac{238}{34}$

$\therefore y=7$cm

Thus, $x=17-y=17-7=10$cm

Substituting values of x and y in $\left(1\right)$ we get

$h^2={26}^2-(10{)}^2$

$=676-100$

$=576$

$\therefore h=\sqrt{576}$

$\Rightarrow h=24$cm

Area of rectangle ABQP$=AB\times BQ$

$=60\times 24$

$=1440c{m}^2$

Area of $\Delta APD=\frac{1}{2}\times x\times h$

$=\frac{1}{2}\times 10\times 24$

$=120c{m}^2$

Area of $\Delta BQC=\frac{1}{2}\times y\times h$

$=\frac{1}{2}\times 7\times 24$

$=84c{m}^2$

Hence area of the trapezium ABCD$=(1440+120+84)c{m}^2$

$=1644c{m}^2$.