A rectangle is inscribed in an equilateral triangle so that one side of the rectangle lies on the base of the triangle. How do I find the maximum area of the rectangle when the triangle has side length of 10?

1 Answer
Dec 24, 2015

A = (25sqrt(3))/2A=2532

Explanation:

First, let's look at a picture.

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Some initial observations:

  • The area AA of the rectangle is A=bhA=bh.

  • By symmetry, the base of the triangle is of length b+2tb+2t, and thus, as it is of length 1010, we have b+2t = 10 => t = 5-b/2b+2t=10t=5b2

  • If we decide bb that also determines hh, and thus we can write hh as a function of bb.

To write hh as a function of bb, we can look at the right triangle with legs tt and hh. As it shares an angle with an equilateral triangle, we know it is a 30^@30-60^@60-90^@90 triangle, and thus t/h = 1/sqrt(3)th=13.
Solving for hh gives us h = sqrt(3)t = sqrt(3)(5-b/2)h=3t=3(5b2) by our initial observation.

Then, we can rewrite our formula for the area as

A = b*sqrt3(5-b/2) = sqrt(3)(-1/2b^2 + 5b)A=b3(5b2)=3(12b2+5b)

If we look at the graph for AA we will see it is a downward facing parabola, and thus will have a maximum at its vertex. Then, we can complete the square to find

A = sqrt(3)(-1/2b^2+5b)A=3(12b2+5b)
= -sqrt(3)/2(b^2-10b)=32(b210b)
= -sqrt(3)/2((b-5)^2-25)=32((b5)225)

And so the vertex, and thus the maximum area, is at b = 5b=5.

Finally, we calculate the area from this to get

A = -sqrt(3)/2((5-5)^2-25) = (25sqrt(3))/2A=32((55)225)=2532