Question

if side of a rhombus is 61 cm and area is 1320 cm square. then finds its diagonals

Answer 1

Let $d_1,d_2$ be the length of the diagonals of the rhombus. The side of the rhombus is $a=61$.

Using the Law of cosines, the diagonals are

$d_1^2=2a^2(1+\cos \theta)\\</p><p>d_2^2=2a^2(1-\cos \theta)$.

Now the area of the rhombus is

$\frac{1}{2} d_1d_2=1320\\</p><p>\frac{1}{2} \sqrt{2a^2(1+\cos \theta)2a^2(1-\cos \theta)} =1320\\</p><p>\frac{1}{2}2a^2 \sqrt{1-\cos^2 \theta)} =1320\\</p><p>a^2 \sin \theta=1320\\</p><p>61^2 \sin \theta=1320\\</p><p> \sin \theta=\frac{20}{61} \\</p><p>\cos \theta=\sqrt{1-\frac{20}{61} ^2} \\</p><p>\cos \theta=\frac{\sqrt{3321} }{61}$.

The diagonals are

$d_1=\sqrt{2(1+\frac{\sqrt{3321} }{61})} 61 \; cm\\</p><p>d_2=\sqrt{2(1-\frac{\sqrt{3321} }{61})} 61\; cm$.