A) \[~94\text{ }c{{m}^{2}}\]
B) \[~86\text{ }c{{m}^{2}}\]
C) \[~72\text{ }c{{m}^{2}}\]
D) \[~65\text{ }c{{m}^{2}}\]
Correct Answer: A
Solution :
Area of \[|{{|}^{gm}}ABCD\] \[=ar(\Delta ABL)+ar(\Delta DCL)+ar(\Delta ADL)\] \[=15+32+ar(\Delta \Alpha DL)\] \[=47+\frac{1}{2}ar\,of\,|{{|}^{gm}}\,ABCD\] [Since,\[\text{ }\!\!|\!\!\text{ }{{\text{ }\!\!|\!\!\text{ }}^{\text{gm}}}\text{ABCD}\]and \[\Delta ADL\]are on the same base AD and between the same parallels AD and BC.] \[\Rightarrow \]Area of \[|{{|}^{gm}}ABCD-\frac{1}{2}ar\]of \[\text{ }\!\!|\!\!\text{ }{{\text{ }\!\!|\!\!\text{ }}^{\text{gm}}}\text{ABCD}\,\text{=}\,\text{47}\] \[\Rightarrow \frac{1}{2}ar\,of\,|{{|}^{gm}}\,ABCD=47\] \[\therefore \]\[\text{ar}\,\text{of}\,|{{|}^{gm}}ABCD=94\,c{{m}^{2}}\]
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