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

i think it is everything has to equal 180 so, if i correct 4=c. 6=d. 8=a. 7=b. and the last question would be c. because they are parallel lines and they are congruent.

step-by-step explanation:

the reason everything has to equal to 180 is because that is how it works. i'm guessing that is its rule. 4 and 8 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 6 and 7 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 60+60 is 180. so, now you have your answer