# 10

Legs ( sides other than the hypotenuse) of a right triangle are of lengths 12 and 6 cm. Find the side of the largest square that can be inscribed in the triangle.?

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1. Geometry?

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2. Nice problem.

Let
ABC be the right-triangle, right-angled at C, such that AC=12, BC=6.
Let the "inscribed" square share a common right-angle at C, and meet the hypotenuse AB at D.

Let E be a corner of the square on AC, and F be a corner of the square on BC.

Let x denote the side length of the square.

Then ΔADE is similar to ΔDBF.

By the ratio of the sides, we have
x/(12-x)=(6-x)/x
Cross multiply and solve for x to get
x=4.

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posted by MathMate
3. What happens if the square does not share sides with the triangle?

If the vertices of the triangle are at (0,0),(6,0),(0,12), then if one side of the square is on the hypotenuse, and one corner is at (s,0), another is at (0,2s), so the hypotenuse has length √5 s.

Now, we also know that

s + √(5s^2 + 5/4 s^2) = 6
s = 12/7
and the square has side

x = 12/7 √5 = 3.83

Hmmm. It appears that MathMate has found the maximal square. We'd have to consider the square at various angles of rotation, and find the angle which maximizes its area. I suspect it is indeed such that the square and the triangle share the right angle.

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posted by Steve

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