the dimensions of the inscribed retangle

Question

A rectangle is inscribed in a right triangle, if the triangle has sides of length 5, 12, and 13, what are the dimensions of the inscribed rectangle of greatest area?

 

Answer 1

Draw the right triangle with horizontal base 5, vertical leg 12. Draw an inscribed rectangle sharing the right triangle's right angle such that the rectangle's vertex opposite the shared right angle lies on the hypotenuse.

Label the horizontal side of the inscribed rectangle x, and the vertical side y. 

 

This will create a smaller similar rt. triangle having horizontal side 5-x and vertical side y. We write the following proportion based on the similarity:  y/(5-x) = 12/5.

 

The area function equals xy, but if we use the above proportion to solve for y and write the area function with only one variable, x, we get f(x) = x[12 - (12/5)x].

 

Using calculus, x = 2.5  Without using calculus, the maximum of f(x) occurs at the vertex of said upside-down parabola, so x = -b/(2a) = 5/2 (same result).

 

So y = 6.