If the sum of the diagonals of a rhombus is L and the perimeter is 4P, find the area of the rhombus?
A. $$\frac{1}{4}\left( {{{\text{L}}^2} - {{\text{P}}^2}} \right)$$
B. $$\frac{1}{4}\left( {{{\text{L}}^2} - 4{{\text{P}}^2}} \right)$$
C. $$\frac{1}{2}\left( {{{\text{L}}^2} - 4{{\text{P}}^2}} \right)$$
D. $$\frac{1}{4}\left( {{{\text{L}}^2} + 3{{\text{P}}^2}} \right)$$
Answer: Option B
Solution (By Examveda Team)
Sum of diagonal of quadrilateral = LPerimeter = 4P
$$\eqalign{ & {\left( {\frac{{{\text{Diagonal}}}}{2}} \right)^2} - {\left( {\frac{{{\text{Semi perimeter}}}}{2}} \right)^2} \cr & = \frac{{{{\text{L}}^2}}}{4} - {\left( {\frac{{2{\text{P}}}}{2}} \right)^2} \cr & = \frac{{{{\text{L}}^2}}}{4} - {{\text{P}}^2} \cr & = \frac{1}{4}\left( {{{\text{L}}^2} - 4{{\text{P}}^2}} \right) \cr} $$
This Question Belongs to Arithmetic Ability >> Mensuration 2D
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