Question: Perimeter of a rhombus is 2p unit and sum of length of diagonals is m units, then area of the rhombus is
Options:
A) $ \frac{1}{4}m^{2}psqunits $
B) $ \frac{1}{4}mp^{2}squnits $
C) $ \frac{1}{4}(m^{2}-p^{2})squnits $
D) $ \frac{1}{4}(p^{2}-m^{2})squnits $
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Answer:
Correct Answer: C
Solution:
- In a rhombus
$ d_1^{2}+d_2^{2}=4a^{2} $
Here, $ d _1 $ and $ d _2 $ are diagonals,
and a = Length of edge $ =\frac{2p}{4}=\frac{p}{2} $
$ \therefore $ $ d_1^{2}+d_2^{2}=4\times {{( \frac{p}{2} )}^{2}} $
$ \Rightarrow $ $ d_1^{2}+d_2^{2}=p^{2} $
On adding $ 2d _1d _2 $ both sides, we get
$ d_1^{2}+d_2^{2}+2d _1d _2=p^{2}+2d _1d _2 $
$ \Rightarrow $ $ {{(d _1+d _2)}^{2}}=p^{2}+2d _1d _2 $
$ \Rightarrow $ $ m^{2}=p^{2}+2d _1d _2 $
$ [d _1+d _2=m,given] $
$ \Rightarrow $ $ 2d _1d _2=m^{2}-p^{2} $
On dividing the whole expression by 4, we get
$ \frac{1}{2}d _1d _2=\frac{1}{4}(m^{2}-p^{2}) $
and $ \frac{1}{2}d _1d _2= $ Area of rhombus
So, the required area of rhombus $ =\frac{1}{4}(m^{2}-p^{2}) $ sq units
BETA
