posted by Xavier
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
since A = bh, and h/2 = 12,
b = 720/24 = 30
So, two of the sides have length 30, and the others have length 40.
Hmmm. Houston, we have a problem. What have I missed?
well.. i got that too.. but i don't know how to find the longer diagonal with this given sides
OK, assuming that the sides are in fact 30 and 40, and the distance from the center to the side of length 30 is 12, then since the center is half way along the diagonal, the height at the end of the diagonal will be 24.
So, if AB=30 and BC=40, then if E is the base of the altitude from C to AB extended, BE=32 and triangle AEC has hypotenuse √(62^2+24^2) = 2√1105 = 66.48
That is the diagonal AC.
so the longer side is 30 and shorter is 40??
yeah - that was my impression. Can't explain it given the provided data. The height and area indicate that the base of the altitude is the side of length 30, but then the math indicates that the "shorter" side is 40!
how did you get BE=32?
How did you get BE=32, and where did 62 come from?