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 \[=\frac{1}{2}(20+40)\times 24\] \[=30\times 24=720\,c{{m}^{2}}\]
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