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What is the area of the largest rectangle that can be inscribed in an isosceles triangle with side lengths $8$, $\sqrt{80}$, and $\sqrt{80}$? Assume that one side of the rectangle lies along the base of the triangle.

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First let's solve the general case for an isosceles triangle of base 2 and height k. If the triangle is drawn so the base is on the x-axis and the apex is on the y-axis, the equation for the line containing the right side is
y = -k(x-1)
Then a rectangle with x-dimension w will have an area that is the product of this width and the height y = -k((w/2)-1).
area = w(-k(w/2 -1)) = (-k/2)w² +kw
The derivative of area with respect to w will be zero where the area is a maximum:
d(area)/dw = 0 = -kw +k
w = 1 . . . . . . add kw and divide by k
Using the formula for area, we find
area = 1(-k(1/2-1)) = k/2
This value is 1/2 the total area of the triangle we started with.

If the side lengths of your triangle are 8, √80, and √80, the height will be given by the Pythagorean theorem as
h = √((√80)² - (1/2·8)²) = √(80-16) = 8
Your triangle's area will be
triangle area = (1/2)(8)(8) = 32

The maximum area of the inscribed rectangle will be half this value,
16 square units
What is the area of the largest rectangle that can be inscribed in an isosceles triangle-example-1
User Stasiaks
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