Question

A rectangle has a diagonal with length 8 centimeters, its length is 4 centimeters greater than the width, how do you find the length and width of the rectangle?

Answer 1

The width is `3.29` cm
The length is `7.29` cm

Explanation:

The angles of a rectangle are `90°`

A diagonal bisects the rectangle and forms two right-angled triangles.

The width is the shorter side, let it be `x` cm
The length is `4` cm longer, so it is `x+4` cm
The diagonal is the hypotenuse of the triangle and has a length of `8` cm.

Use Pythagoras' Theorem to write an equation:

`x^2 + (x+4)^2 = 8^2`

`x^2 + x^2 +8x+16 = 64`

`:. 2x^2 +8x -48 =0" "larr div 2`

`x^2 +4x -24 =0" "larr` does not fcatorise

Solve by completing the square.

`x^2 +4x color(blue)(+4) = 24 color(blue)(+4)`

`(x+2)^2 = 28`

`x = +-sqrt28-2`

`x = 3.29cm`

`x = -7.29cm" "larr` reject as a length of a side

The width is `3.29` cm
The length is `3.29+4= 7.29` cm