Question

The area of the rhombus is 540 cm2; the length of one of its diagonals is 4.5 dm. What is the distance between the point of intersection of the diagonals and the side of the rhombus?

Answer 1

Area of rhombus = $\frac{xy}{2}$

where x and y are diagonals.

Given length of one diagonal is 4.5 dm

So, let x= 4.5dm

1 dm = 10 cm

4.5dm= 45cm

area = $\frac{xy}{2}$

so, $540 = \frac{45y}{2}$

y = 24 cm

The two diagonals are x= 45cm and y = 24cm

Since diagonals bisect each other, we get 22.5cm and 12cm

Using right triangle formula

$\frac{1}{r^{2} } = \frac{1}{22.5^{2}} + \frac{1}{12^{2}}$

$\frac{1}{r^{2} } =\frac{12^{2}+22.5^{2}}{22.5^{2}*12^{2}}$

$r^{2} = \frac{22.5^{2} *12^{2}}{12^{2}+22.5^{2}}$

r = 10.588cm

Distance of center to the side = $\frac{10.588}{2}$  = 5.294 cm