A) \[~300\,c{{m}^{2}}\]
B) \[~180\,c{{m}^{2}}\]
C) \[~126\,c{{m}^{2}}\]
D) \[306\,c{{m}^{2}}\]
Correct Answer: D
Solution :
In right angled\[\Delta \Alpha \Beta C,\]using Pythagoras theorem \[{{(AC)}^{2}}={{(9)}^{2}}+{{(40)}^{2}}=1681\] \[\Rightarrow \]\[AC=41\,cm\] \[\therefore \]Area of \[\Delta \Alpha \Beta C=\frac{1}{2}\times 9\times 40=180\,c{{m}^{2}}\] Now, in \[\Delta \Alpha DC\] \[s=\frac{15+28+41}{2}=42\,cm\] \[\therefore \]Area of \[\Delta \Alpha DC=\sqrt{42\times 27\times 14\times 1}\] \[=126\,c{{m}^{2}}\] Hence, area of quadrilateral ABCD \[=(180+126)c{{m}^{2}}=306\,c{{m}^{2}}\]You need to login to perform this action.
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