Question

Find the area of the given trapezium

• A. $259.65sq.cm$
• B. $295.65sq.cm$
• C. $395.65sq.cm$
• D. $255.65sq.cm$

Answer 1

The correct option is A $259.65sq.cm$

Consider Trapezium $ABCD$

where $AB||CD$ and $AB>CD$
Given $AB=32cm,CD=20cm,AD=10cm,BC=16cm$
Let's take points $P,Q$ on $AB$ such that $QDCP$ would be a rectangle.
You get $PQ=20cm$ so $AQ+PB=(32-20)=12cm$
Let $AQ=xcm$

$PB = 32-20-x \ cm=12-x \ cm$

We know $DQ=CP$, opposite sides of rectangle $DCPQ$

so using Pythagoras Theorem in $△ADQ$ and $△BCP$, we get

$A{D}^2-A{Q}^2=B{C}^2-P{B}^2$

${10}^2-x^2={16}^2-(12-x{)}^2$

$100-x^2=256-144-x^2+24x$

$100=112+24x$

$24x=-12$

$x=\frac{1}{2}$ ------ since the length of a side can only be positive.

$\therefore ,In△ADQ$,

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

$D{Q}^2=A{D}^2-A{Q}^2$

$D{Q}^2={10}^2-(\frac{1}{2}{)}^2=100-\frac{1}{4}=\frac{399}{4}$

$DQ=\frac{\sqrt{399}}{2}$

$Areaoftrapezium=\frac{1}{2}h(a+b)$

where, $h-height,a,b-parallelsides$

$Areaoftrapezium=\frac{1}{2}\left(\frac{\sqrt{399}}{2}\right)(32+20)$

$Areaoftrapezium=259.65sq.cm$