How do you find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3 cm and 8 cm if two sides of the rectangle lie along the legs?

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Answer 1 · 2017-07-23T00:13:29

Its area is $6"cm"^2$

The largest possible rectangle must have a vertex that touches the hypotenuse of the triangle at one point.

Let us put the right angle of the triangle at the point $(0, 0)$ , another vertex at $(8, 0)$ and the final vertex at $(0, 3)$

We can represent points on the hypotenuse parametrically as:

$(8t, 3(1-t))$

where $t in [0, 1]$

Then the area of the rectangle with vertices:

$(0, 0)$ , $(8t, 0)$ , $(8t, 3(1-t))$ , $(0, 3(1-t))$ is:

$f(t) = 8t*3(1-t) = 24t(1-t) = 24(t-t^2) = 24(1/4-(t-1/2)^2)$

This takes its maximum value when $t=1/2$ and $f(t) = 24*1/4 = 6$