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 = 20 cm and CL = 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 =\[\frac{1}{2}\]\[\times \,20\]cm = 10 cm

\[[as\,BL=ABAL=\left( 4020 \right)\text{ }cm=20\text{ }cm].\]

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

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

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

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

\[=\left( 2610 \right)\left( 26+10 \right)\]

\[=16\times 36=576\]

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

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

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

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