Prove that the Area of ParallelogramACC'A'=b^2

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- Oct 31st 2010, 12:00 PMMATNTRNGArea Question...
Prove that the Area of Parallelogram

*ACC'A'*=*b*^2

- Oct 31st 2010, 02:04 PMArchie Meade
If we take it that the triangle with sides a, b, c is right-angled such that $\displaystyle a^2+b^2=c^2$

and the inner shape with sides c is a square, then if we label the angle and side as shown in the attachment..

and the area of the required shape equals the rectangle area,

we only need calculate $\displaystyle kc$.

$\displaystyle Cos\theta=\frac{k}{b}\Rightarrow\ k=bCos\theta$

Using the larger triangle... $\displaystyle Cos\theta=\frac{b}{c}$

Hence, it's straightforward to calculate $\displaystyle kc$ in terms of $\displaystyle b$.