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?

Try this solution (this is not the shortest way!).

Note, the length of the diagonal is 45 cm.=4,5 dm.

the side is 22.5 cm., the EF is 12 cm.

Step-by-step explanation:

area of rhombus = base x height or diagonal 1 x diagonal 2 / 2

one diagonal = 4.5dm = 45cm

the other diagonal = 540 / 45 * 2 = 24cm

the diagonals in a rhombus are perpendicular 2 each other

so half diagonals and one side of the rhombus form a right-angled triangle

side^2 = half-diagonal 1^2 + half-diagonal 2^2

= (45/2)^2 + (24/2)^2

= 650.25

side = sqrt (650.25) = 25.5cm

540 = height x side

height = 540 / 25.5 = 21.1765cm

distance between the point of intersection of the diagonals and the side of the rhombus = height / 2 = 10.59cm

sas (side angle side)

aa (angle angle)

sss(side side side)

sas find two sides with an angle in between

aa is finding two angles and the third will add up to 180

sss find all sides

step-by-step explanation: