1. ## Geometrical Problem

Figure here shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long.

a. Express the y coordinate of P in terms of x. I get 1 - x ... Correct?
b. Express the area of the rectangle in terms of x. Not sure how to go about solving this one.

2. a) Correct.

b) Area of rectangle is one side multiplied by the other.

One of the sides is 2x as it's x one way from the origin and x the other way.

You've already worked out the length of the other side. You did it in part a).

3. Originally Posted by VitaX

Figure here shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long.

a. Express the y coordinate of P in terms of x. I get 1 - x ... Correct? <<<<< Yes
b. Express the area of the rectangle in terms of x. Not sure how to go about solving this one.
The area of a rectangle is calculated by: $A = length \cdot width$

With your problem the length = 2x
and the width is w = 1 - x

Plug in these terms into the formula above.

4. Originally Posted by earboth
The area of a rectangle is calculated by: $A = length \cdot width$

With your problem the length = 2x
and the width is w = 1 - x

Plug in these terms into the formula above.
would the area be better expressed as 2x(1-x) units squared or 2x-2x^2 units squared.

5. That really makes no difference, but if you want it to look neater then write $2x(1-x)$