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?

Area of rhombus =

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 =

so,

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

r = 10.588cm

Distance of center to the side =  = 5.294 cm

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (