Question

find the fraction of the area of a triangle that is occupied by the largest rectangular that can be drawn in the triangle (with one of its sides along a side of the triangle ). show that this fraction does not depend on the dimensions of given triangle ???????​

Answer 1

½

Explanation:

Draw a picture of the triangle with the rectangle inside it.

Let's say the width and height of the triangle are w and h (these are constants).

Let's say the width and height of the rectangle are x and y (these are variables).

The area of the triangle is ½ wh.

The area of the rectangle is xy.

Using similar triangles, we can say:

(h − y) / h = x / w

x = (w/h) (h − y)

So the rectangle's area in terms of only y is:

A = (w/h) (h − y) y

A = (w/h) (hy − y²)

We want to maximize this, so find dA/dy and set to 0:

dA/dy = (w/h) (h − 2y)

0 = (w/h) (h − 2y)

0 = h − 2y

y = h/2

So the width of the rectangle is:

x = (w/h) (h − y)

x = (w/h) (h − h/2)

x = (w/h) (h/2)

x = w/2

That means the area of the rectangle is:

A = xy

A = ¼ wh

The ratio between the rectangle's area and the triangle's area is:

(¼ wh) / (½ wh)

½

So no matter what the dimensions of the triangle are, the maximum rectangle will always be ½ its area.