• # 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.

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 $=\frac{1}{2}(20+40)\times 24$ $=30\times 24=720\text{ }c{{m}^{2}}~$