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Q.
Find the radius of the circle inscribed in a rhombus with diagonals of length 12 and 16 ?
Mensuration
Solution:
Let $d_1=12$ and $d_2=16$ be the length of diagonals and ' $a$ ' be the side of Rhombus and $r=$ radius of inscribed circle.
Using Pythagoras theorem
$a=\sqrt{(\frac{d_1}{2})^2+(\frac{d_2}{2})^2} $
$\Rightarrow a=\sqrt{(\frac{12}{2})^2+(\frac{16}
{2})^2}=\sqrt{6^2+8^2}=\sqrt{100}$
$ \Rightarrow a=10$
$ \text { Area of Rhombus }=\frac{d_1 d_2}{2} \text { and } 2 a r $
$ \text { ia } $
$2 a r \text { is derived using }=\text { area of rhombus }= $
$ 4 \times \text { area of triangle } $
$ \text { Area of Rhombus }=4 \times \frac{1}{2} \times a \times r=2 a r $
$ \text { Using (i), we get } $
$\frac{d_1 d_2}{2}=2 a r $
$ \frac{12 \times 16}{2}=2 \times 10 \times r $
$ r=\frac{16 \times 6}{2 \times 10} \Rightarrow r=4.8 $
