A) \[75%\]
B) \[120%\]
C) \[8m\times 6m\]
D) \[\frac{5}{2}(9+5\sqrt{3})sq.cm\]
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 \[\Delta BCD\] Here \[=\left( \frac{1}{2}\times BD\times AL \right)+\left( \frac{1}{2}\times BD\times CM \right)\] \[=\frac{1}{2}\times BD\times (AL+CM)\] Area of \[=b\times h\] \[\frac{2\pi rh}{\pi r}=\frac{8}{5}\] Hence, area of \[:h=5:4\]You need to login to perform this action.
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