Question

The distance between the center of symmetry of a parallelogram and its longer side is equal to 12 cm. The area of the parallelogram is 720 cm^2 and its perimeter is 140 cm. Determine the length of the longer diagonal of the parallelogram.

Answer 1

diagram pls :)

Answer 2

@Orion1213 Ive just saw your reply Ill solve for the area and perimeter first :)

Answer 3

@where did you get 24?

Answer 4

already got it then how can I solve that triangle ?

Answer 5

x=sq W-576

Answer 6

so L =30 ?

Answer 7

and W= 40

Answer 8

how will I find the longer diagonal then?

Answer 9

how can I find the longest diagonal?

Answer 10

wait Im trying to understand this :)

Answer 11

where is x?

Answer 12

in the first diagram

Answer 13

how did you get your riht triangle

Answer 14

may I just ask why did the length became smaller than the width?

Answer 15

hmmm.... OK let's try this one...
EG = 2xEF = 24cm
CD = AB = EG = 24cm, also the base of the parallelogram
using the Area A = 720 = base CD x height BH, we can solve BH = 720/24 = 30cm

let's find the longest side AC or BD using the given parameter P = 140 where....
P = 140 = 2xAC + 2xCD = 2xAC + 2x24 = 2AC + 48
AC = (140 - 48)/2 = 46cm
BD = AC = 46cm

we can see from our diagram that side BD, height BH and line DH form a right triangle... length of line DH = sqrt (BD^2 - BH^2) = sqrt (46^2 - 30^2) = sqrt(2116 - 900)
DH = 34.87cm

base CD + line DH and height BH form the legs of another right triangle whose hypotenuse is the longest diagonal BC (can be solve via Pythagorean Theorem)...
BC = sqrt((24+34.87)^2 + (30)^2)
BC = 66.07cm // answer.... :)

i think this one is more acceptable than the previous one. I deleted previous entries to avoid confusions. :)