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# Non brutal technique anyone?

There is an equilateral triangle (say $$\Delta ABC$$ ). There is a point $$P$$ inside $$\Delta ABC$$ such that $$AP=3,\,BP=4,\,CP=5$$.

How do we find the area of $$\Delta ABC$$?

This problem was posed by Milind Blaze.

Note by Deeparaj Bhat
5 months ago

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Swing the line segments of length 4 and 5 through 60 degrees to form a congruent triangle on another side of the original triangle:

Because we pivoted the 5-long line segment by 60 degrees, the two segments are two sides of an equilateral triangle. So we fill in the third side, forming not just the equilateral triangle but also one with sides 3,4, and 5. And we do the same with the other pairs of line segments: 3,4 and 3,5.

We now have on the outside of the original triangle three triangles congruent to the three inner triangles formed by the line segments to the interior point, so the overall shape has twice the area of the original, and its area is easy to find because it is composed of equilateral and 3-4-5 right triangles. · 5 months ago

Comment deleted 4 months ago

The 2 are congruent through SAS arent' they? The angles at the vertices are equal, the side of the original triangle are equal and so is the line segment of length 5, no? · 4 months, 4 weeks ago

Never mind, I got it. But, which 2 triangles are you talking about? · 4 months, 4 weeks ago

The inner 45a and outer 45a. Poor notation, i know.... · 4 months, 4 weeks ago

Ah... Yeah. Thanks for that additional justification. · 4 months, 4 weeks ago