Question 1081654: One side of a parallelogram is 10 meters and its diagonals are 16 and 24 meters respectively. Its area is?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! We can get the angle of intersection of the diagonals using Law of Cosines
using 10 as a side
10^2=8^2+12^2-2*8*12 cos C, using the facts the diagonals bisect each other.
100=64+144-192 cos C
-108=-192 cos C
cos C=108/192
arc cos both sides and C=55.77 deg
Vertical angles and supplementary angles makes for 2 triangles of 8-10-12
The other two have a missing side, but the angle is 124.23
therefore, c^2=64+144-2*8*12 cos 124.23, and the cosine is -.5625
c^2=208+108=316
c=17.78 m,
using Heron's formula, area of the smaller triangle has s=15 (half the perimeter)
and A=sqrt (s(s-a)(s-b)(s-c))= sqrt (15*7*3*5)=39.69. Two of these have area 79.37 (rounding at the end)
The area of the larger triangle has s=37.78/2=18.89
sqrt(18.89*(1.11)*10.89*6.89)=39.66, and two of these are 79.32
The total area is 158.69 m^2.
Can check using A=(1/2)pq sin theta, using 55.77 as theta, and p, q are diagonal lengths.
A=(1/2)384*sin 55.77=158.74 m^2.
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