Please help me with this problem.

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- May 7th 2009, 02:29 AMlearnmath[SOLVED] Square Problem
Please help me with this problem.

- May 7th 2009, 06:10 AMSodapop
- May 7th 2009, 08:50 PMlearnmath
Yes, I can understand this part. Please help me with the other steps to arrive at the final answer.

- May 8th 2009, 01:32 AMrunning-gag
Hi

Consider triangle BPE. PR is one height of this triangle. The area of BPE is $\displaystyle A(BPE) = \frac12\:PR \times BE$

The same for triangle BPC : $\displaystyle A(BPC) = \frac12\:PQ\times BC$

The area of triangle BCE is the sum of the areas above : $\displaystyle A(BCE) = A(BPE) + A(BPC) = \frac12\:PR\times BE + \frac12\:PQ\times BC$

Since BE = BC, $\displaystyle A(BCE) = \frac12\:(PQ + PR)\times BC$

On the other hand the area of triangle BCE can be calculated using the projection of E on [BC], say H

$\displaystyle A(BCE) = \frac12\:EH\times BC$

Since ABCDE is a square, angle CBE is 45 degrees. Therefore $\displaystyle EH = BE\times \sin(45) = BC \:\frac{\sqrt{2}}{2}$

$\displaystyle A(BCE) = \frac12\:\frac{\sqrt{2}}{2}\times BC^2$

Hence $\displaystyle A(BCE) = \frac12\:(PQ + PR)\times BC = \frac12\:\frac{\sqrt{2}}{2}\times BC^2$

And finally $\displaystyle PQ + PR = \frac{\sqrt{2}}{2}\times BC$ - May 10th 2009, 07:27 PMlearnmath
Thank you!