Question

The area of a trapezium with equal non-parallel sides is 168 ${m^2}$. If the lengths of the parallel sides are 36 m and 20 m, find the length of each non-parallel side.

Answer 1

Hint: Start solving the question by constructing perpendiculars from vertices $A$ and $B$. Find the length of perpendiculars using the formula of area of a trapezium, $A = \dfrac{{{\text{sum of parallel sides}}}}{2} \times h$, where $h$ is the length of perpendiculars. After that, use Pythagoras theorem in any of the triangles formed by the perpendicular to find the length of non-parallel sides.


Complete step by step answer:
We will first construct perpendiculars $AE$ on $FC$ and $BD$ as shown in the figure,

Since, $AB\parallel FC$, the height of perpendiculars is same, that is, $AE = BD$
We are given that the area of trapezium is 168 ${m^2}$.
The area of trapezium is, $A = \dfrac{{{\text{sum of parallel sides}}}}{2} \times h$, where $h$is the length of perpendiculars.
Substitute the given values in the formula to find the value of $h$.
$
  168 = \dfrac{{36 + 20}}{2} \times h \\
  168 = \dfrac{{56}}{2} \times h \\
  168 = 28h \\
  h = \dfrac{{168}}{{28}} \\
  h = 6{\text{m}} \\
$
Thus, it implies that the length of the perpendiculars is 6m , $AE = BD = 6{\text{ m}}$
Now, we have to find the length of sides $AF$ and $BC$. It is given that the length of the non-parallel sides is equal. Hence, $AF = BC$.
Apply Pythagoras theorem in $\Delta AEF$.
Pythagoras theorem states that, ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2}{\text{ + }}{\left( {{\text{Base}}} \right)^2}$
In $\Delta AEF$, $A{F^2} = A{E^2} + F{E^2}$
We will find the value of $FE$ because we have to substitute it in the above equation.
Since, $AB\parallel FC$ , $AE \bot FC$ and $AB \bot FC$, the length of $AB$ and $ED$ are equal.
Thus, $FE + ED + DC = 36$,
Consider $\Delta AEF$ and $\Delta BDC$
From both the triangles, we observed that,
$AF = BC\left( {{\text{Hypotenuse}}} \right)$
$AE = BD\left( {{\text{Perpendicular}}} \right)$
$\angle E = \angle D = {90^ \circ }\left( {{\text{Right angle}}} \right)$
Hence, $\Delta AEF \cong \Delta BDC$ by the RHS property of congruence.
Also, $FE$ and $DC$ will be equal as triangles $\Delta AEF$ and $\Delta BDC$ are congruent by RHS property.
$
  2FE + 20 = 36 \\
  2FE = 16 \\
  FE = 8{\text{m}} \\
 $
Now, substituting the values in Pythagoras theorem, we get,
$
  A{F^2} = {6^2} + {8^2} \\
  A{F^2} = 36 + 64 \\
  A{F^2} = 100 \\
  AF = 10{\text{m}} \\
 $
Thus, the length of the non-parallel sides is 10 m

Note: The length of perpendiculars drawn between parallel lines is the same. Also, triangles, $\Delta AEF$ and $\Delta BDC$ are congruent by RHS property as $\angle E$ and $\angle D$ are right angles, $AE = BD = 6{\text{ m}}$ and hypotenuse $AF = BC$ (given in the question). The congruence of these triangles makes the length $FE$ and $DC$ are equal.