Question

In a trapezium ABCD with bases AB and CD where $AB=52,BC=12,CD=39$ and $DA=5$. The area of the trapezium ABCD is

• A. $182$
• B. $195$
• C. $218.4$
• D. $260$

Answer 1

The correct option is C $218.4$

Given,
The parallel sides of a trapezium $ABCD$ are$52$ m and $39$ m and the remaining two sides are $12$ m and $5$ m .
$AB=52$cm
$AE+FB=52-39$
=$13$
Let $AE=x$, therefore $FB=(13-x)$
Now,
$△DEA$ is a right angled triangle.
$D{A}^2=A{E}^2+D{E}^2$
$=>D{E}^2=D{A}^2-A{E}^2$
$=>D{E}^2={12}^2-x^2$
$=>D{E}^2=144-x^2$
Similarly,
$△CBF$ is a right angled triangle.
$C{B}^2=C{F}^2+B{F}^2$
$=>C{F}^2=C{B}^2-B{F}^2$
$=>C{F}^2=5^2-(13-x{)}^2$
$=>C{F}^2=25-\left[\right(13{)}^2+\left(x^2\right)-\mathrm{2.13.}x]$
$=>C{F}^2=25-169-x^2+26x$
$=>C{F}^2=-x^2+26x-144$
Since DE = CF,
$144-x^2=-x^2+26x-144$
$=>-26x=-288$
$=>x=11$
$AE=11$ and
$FB=13-11=2$
$\therefore DE=CF=\sqrt{144-{11}^2}$
=$4.8$
Area of trapezum ABCD=Area of $△DEA$+Area of $△CFB$+Area of rectangle ABEF
$=(\frac{1}{2}\times 11\times 4.8+\frac{1}{2}\times 2\times 4.8+39\times 4.8)m^2$
$=(26.4+4.8+187.2)m^2$
$=218.4m^2$