How can i prove that the area rectangle in a right angled triangle cannot exceed half the area of the triangl?

Answer 1

Position the rectangle on the coordinate axes, leg a along the positive x-axis, leg b along the positive y-axis. Then the equation of the hypotenuse is y = -(b/a)x + b.

Now let's find the rectangle with the largest possible area. Its lower-left vertex will be the origin, and its opposite vertex will be on the line. Its area is then:

A = x * y = x * (-(b/a)x + b)
A = -(b/a)x² + bx
(Take derivative and set to 0)
A' = -(2b/a)x + b = 0
b = 2bx/a
x = a/2

Then y = b/2, and A = xy = ab/4. The area of the triangle is ab/2. So the maximum area of the rectangle is half the triangle's. QED.

Answer 2

Did you draw a diagram? Here's mine: http://i5.tinypic.com/6jkmvep.gif

Suppose side AB has length a and side BC has length b.
Triangle ADE is similar to triangle ABC. So

AD/AB = DE/BC

AD = a-x, DE = y, AB = a, BC = b, so

(a-x)/a = y/b

Solving for y,

y = b(a-x)/a

The area A of the rectangle is given by

A = xy = xb(a-x)/a

Differentiate A with respect to x and set to zero.

Compare area A with the area of triangle ABC