# Find the area of the largest rectangle that can be inscribed in a right triangle

Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3 cm and 4 cm if two sides of the rectangle lie along the legs.

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Melinda McCombs
Step 1

Step 2
Let x and y be the side lengths of the rectangle, then the area is
$$\displaystyle{A}={x}{y}$$
and by similar triangles (ABC and FBE, note: BF is $$\displaystyle{3}-{y}$$)
$$\displaystyle{\frac{{{x}}}{{{4}}}}={\frac{{{3}-{y}}}{{{3}}}}$$
$$\displaystyle{\frac{{{3}{x}}}{{{4}}}}={3}-{y}$$
$$\displaystyle{\frac{{{3}{x}}}{{{4}}}}-{3}=-{y}$$
$$\displaystyle{y}={3}-{\frac{{{3}}}{{{4}}}}{x}$$
Step 3
Substitute for y in the area equation, find $$\displaystyle{A}'$$
$$\displaystyle{A}={x}{\left({3}-{\frac{{{3}}}{{{4}}}}{x}\right)}={3}{x}-{\frac{{{3}}}{{{4}}}}{x}^{{{2}}}$$
$$\displaystyle{A}'={3}-{\frac{{{3}}}{{{2}}}}{x}$$
Find where it is 0
$$\displaystyle{0}={3}-{\frac{{{3}}}{{{2}}}}{x}$$
$$\displaystyle{\frac{{{3}}}{{{2}}}}{x}={3}$$
$$\displaystyle{x}={2}$$
Step 4
Find te corresponding y (with the similar triangle equation)
$$\displaystyle{y}={3}-{\frac{{{3}}}{{{4}}}}{\left({2}\right)}={3}-{\frac{{{3}}}{{{2}}}}={\frac{{{6}-{3}}}{{{2}}}}={\frac{{{3}}}{{{2}}}}$$
So the maximum rectangle area is
$$\displaystyle{A}={2}\cdot{\frac{{{3}}}{{{2}}}}={3}{c}{m}^{{{2}}}$$
###### Not exactly what youâ€™re looking for?
Becky Harrison
Step 1
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:
$$\displaystyle{\left({8}{t},\ {3}{\left({1}-{t}\right)}\right)}$$
where $$\displaystyle{t}\in{\left[{0},{1}\right]}$$
Then the area of the rectangle with vertices:
$$\displaystyle{\left({0},{0}\right)},{\left({8}{t},{0}\right)},{\left({8}{t},{3}{\left({1}-{t}\right)}\right)},{\left({0},{3}{\left({1}-{t}\right)}\right)}$$ is:
$$\displaystyle{f{{\left({t}\right)}}}={8}{t}\cdot{3}{\left({1}-{t}\right)}={24}{t}{\left({1}-{t}\right)}={24}{\left({t}-{t}^{{{2}}}\right)}={24}{\left({\frac{{{1}}}{{{4}}}}-{\left({t}-{\frac{{{1}}}{{{2}}}}\right)}^{{{2}}}\right)}$$
This takes its maximum value when $$\displaystyle{t}={\frac{{{1}}}{{{2}}}}$$ and $$\displaystyle{f{{\left({t}\right)}}}={24}\cdot{\frac{{{1}}}{{{4}}}}={6}$$