Plain Parallelogram
A parallelogram has sides measuring 7 and 9. Its shorter diagonal has a length of 8.
Find the measure of the longer diagonal.
The answer is 14.
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5 solutions
Isn't this just an extension of Apollonius' theorem...?
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The math is the same, but the Parallelogram Law has a far larger impact on vector spaces, where "median" doesn't have much meaning. Have a look at this wiki about Hilbert Space, and see where that law pops up
Since the point of intersection of the two diagonals bisect these diagonals. Thus, we can use the Apollonius theorem :D
In a parallelogram 2(AB)^2 + 2(BC)^2 = AC^2 + BD^2 So the answer is 14
Let A be the angle opposite the side 7 in 7, 8, 9 triangle. ∴ C o s A = 2 ∗ 8 ∗ 9 8 2 + 9 2 − 7 2 = 3 2 . ( 2 B i g D i a g o n a l ) 2 = 9 2 + ( 2 S m a l l D i a g o n a l ) 2 − 2 ∗ 9 ∗ 4 ∗ 3 2 = 4 9 . B i g D i a g o n a l = 2 ∗ 4 9 = 1 4 .
This involves the Parallelogram Law. One of the easiest ways to prove this law is by the following: Let a , b be the sides of the parallelogram, and p , q be the diagonals. By the Cosine Law, we have:
p 2 = a 2 + b 2 − 2 a b C o s ( θ )
q 2 = a 2 + b 2 − 2 a b C o s ( π − θ )
Adding them together gets us the Parallelogram Law
p 2 + q 2 = 2 ( a 2 + b 2 )
For a = 7 , b = 9 , p = 8 , we find that q = 1 4