# 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?

Jul 23, 2017

Its area is $6 {\text{cm}}^{2}$

#### Explanation:

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 $\left(0 , 0\right)$, another vertex at $\left(8 , 0\right)$ and the final vertex at $\left(0 , 3\right)$

We can represent points on the hypotenuse parametrically as:

$\left(8 t , 3 \left(1 - t\right)\right)$

where $t \in \left[0 , 1\right]$

Then the area of the rectangle with vertices:

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

$f \left(t\right) = 8 t \cdot 3 \left(1 - t\right) = 24 t \left(1 - t\right) = 24 \left(t - {t}^{2}\right) = 24 \left(\frac{1}{4} - {\left(t - \frac{1}{2}\right)}^{2}\right)$

This takes its maximum value when $t = \frac{1}{2}$ and $f \left(t\right) = 24 \cdot \frac{1}{4} = 6$