A) \[206\,\,c{{m}^{2}}\]
B) \[306\,\,c{{m}^{2}}\]
C) \[356\,\,c{{m}^{2}}\]
D) \[380\,\,c{{m}^{2}}\]
Correct Answer: B
Solution :
Applying Pythagoras theorem in\[\Delta ABC,\]we get \[{{9}^{2}}+{{40}^{2}}=A{{C}^{2}}\] \[\Rightarrow \] \[AC=\sqrt{1681}=41\,\,cm\] \[\therefore \]Area of quadrilateral = Area of \[\Delta ABC+\] Area of \[\Delta ADC\] \[=\frac{1}{2}(9\times 40)+\sqrt{42\times 1\times 14\times 27}\] \[=180+14\times 3\times 3=180+126=306\,\,c{{m}^{2}}\]You need to login to perform this action.
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