Thread: [SOLVED] Square Problem

Please help me with this problem.

Originally Posted by learnmath

Please help me with this problem.

You know that:

BC² + CD² = BD²
1² + 1² = BD²
2 = BD²
BD = √2

Since BE = BC, and BC =1, then BE = 1, per transitivity.

Have you arrived this far?

Yes, I can understand this part. Please help me with the other steps to arrive at the final answer.

Hi

Consider triangle BPE. PR is one height of this triangle. The area of BPE is

The same for triangle BPC :

The area of triangle BCE is the sum of the areas above :

Since BE = BC, PQ + PR)\times BC" alt="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



Since ABCDE is a square, angle CBE is 45 degrees. Therefore



Hence PQ + PR)\times BC = \frac12\:\frac{\sqrt{2}}{2}\times BC^2" alt="A(BCE) = \frac12\PQ + PR)\times BC = \frac12\:\frac{\sqrt{2}}{2}\times BC^2" />

And finally

Thank you!