a parralelogram has adjacent sides 8cm and 15cm. if the shorter diagnal is 10cm long find the length of the longer diagnal
make a sketch, label you parallelogram ABCD, so that
AB = CD = 8, BC = AD = 15 and AC = 10
Using the cosine law twice, let's find angle C
in triangle ACD,
15^2 = 8^2 + 10^2 - 2(8)(10)cos ACD
160cosACD = -61
angle ACD = 112.411°
in triangle ABC,
8^2 = 10^2 + 15^2 - 2(10)(15)cosACB
300cosACB = 261
cosACB = 261/300
angle ACB = 29.541°
so angle C = 29.541+112.411 = 141.952°
finally,
BD^2 = 8^2 + 15^2 - 2(8)(15)cos141.952
= 477.9996
BD = 21.86 or appr 22
consider the sides as vectors u and v. Then the shorter diagonal is u-v and the longer is u+v
So let u = 15i and v = hi+jk
Then
|u-v|^2 = 10^2
|v|^2 = 64
That is,
(15-h)^2 + k^2 = 100
h^2 + k^2 = 64
Now just solve for h and k, and then find |u+v|. I get about 25.67