A) \[245\text{ }c{{m}^{2}}\]
B) \[240\text{ }c{{m}^{2}}\]
C) \[247.5\text{ }c{{m}^{2}}\]
D) \[250\text{ }c{{m}^{2}}\]
Correct Answer: C
Solution :
[c] In the adjoining figure,. In right angled \[\Delta ADC,\]\[AC=41\,cm,\]\[DC=40\,cm\] \[\because \] \[{{(AC)}^{2}}={{(AD)}^{2}}+{{(DC)}^{2}}\] \[\therefore \] \[{{(AD)}^{2}}={{(AD)}^{2}}+{{(DC)}^{2}}\] \[{{(AD)}^{2}}={{(41)}^{2}}-{{(40)}^{2}}=1681-1600\] \[{{(AD)}^{2}}=81\]\[\Rightarrow \]\[AD=\sqrt{81}\] \[\Rightarrow \]\[AD=9\,cm\] (which is height of trapezium ABCD) \[\therefore \]Area of trapezium \[ABCD=\frac{1}{2}\times \] (Sum of parallel lines) \[\times \](Height) \[=\frac{1}{2}\times (AB+CD)\times AD\] \[=\frac{1}{2}\times (15+40)\times 9\] \[=\frac{1}{2}\times 55\times 9=247.5\,c{{m}^{2}}\] |
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