A) \[72\,c{{m}^{2}}\]
B) \[12\sqrt{35}\,c{{m}^{2}}\]
C) \[24\sqrt{35}\,c{{m}^{2}}\]
D) \[150\,c{{m}^{2}}\]
Correct Answer: B
Solution :
As shown in the diagram, area of triangle ABC \[=1/2\]area of the parallelogram ABCD. By Heron's formula, area of triangle \[=\sqrt{S\left( S-a \right)\left( S-b \right)\left( S-c \right)}\] Here \[S=\frac{9+8+13}{2}=15\] \[\therefore \] Area of ABC\[=\sqrt{15(6)(2)(7)}\] \[\sqrt{6\times 7\times 30}=6\sqrt{35}\,\text{sq}\text{.}\,cm.\] Hence, area of ABCD \[=2\times 6\sqrt{35}=12\sqrt{35}\,\text{sq}\text{.m}\]You need to login to perform this action.
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