posted by math on .
What is the area of the largest rectangle that can be placed in a 5-12-13 right triangle (as shown)?
Let ABC be the vertices of the triangle, right-angled at B, AB=5, BC=12 (vertical side), AC=13.
Draw a rectangle BDEF, where D is on AB, E is on AC and F is on BC.
x=DE= height of rectangle
Width of rectangle = DB = 5-(5x/12)
Area of rectangle,
A'(x) = 5-10x/12
For A(x) to be maximum,
A'(x) = 0 = 5-10x/12
x=6, 5-5(6)/12 = 2.5
The maximum area is 6*2.5=15
Note that the aspect ratio of the rectangle is the same as that of the right sides of the triangle.