question_answer

The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are equal, each being 26 cm, find the area of the trapezium.

Answer:

Let ABCD be the trapezium such that AB = 40 cm and CD = 20 cm and AD = BC = 26 cm.                                                                                                                                                              

Now, draw CL || AD

Then, ALCD is a parallelogram

So, AL=CD=20cmandCL=AD=26cm.

In\[\Delta CLB\], we have

CL = CB = 26 cm

Therefore, \[\Delta CLB\]is an isosceles triangle.

Draw altitude CM of \[\Delta CLB.~\]

Since \[\Delta CLB\] is an isosceles triangle.

So, CM is also the median.

Then, LM = \[MB=\frac{1}{2}BL=\frac{1}{2}\text{ }\times 20cm=10cm\]

[as BL = AB -AL = (40 - 20) cm = 20 cm].

Applying Pythagoras theorem in \[\Delta CLM,\]

we have,                     \[C{{L}^{2}}=C{{M}^{2}}+L{{W}^{2}}\]

\[{{26}^{2}}=C{{M}^{2}}+\text{1}{{\text{0}}^{2}}\]

\[C{{M}^{2}}={{26}^{2}}-{{10}^{2}}\]

\[=\left( 26-10 \right)\left( 26+10 \right)\]

\[=16\times 36=576\]

CM = \[\sqrt{576}\] = 24 cm

Hence,   the area of the trapezium\[=\frac{1}{2}\times \] (sum of parallel sides) \[\times \] Height

\[=\frac{1}{2}(20+40)\times 24\]

\[=30\times 24=720\text{ }c{{m}^{2}}~\]