# 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?

Mar 15, 2018

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:

${x}^{2} + {\left(x + 4\right)}^{2} = {8}^{2}$

${x}^{2} + {x}^{2} + 8 x + 16 = 64$

$\therefore 2 {x}^{2} + 8 x - 48 = 0 \text{ } \leftarrow \div 2$

${x}^{2} + 4 x - 24 = 0 \text{ } \leftarrow$ does not fcatorise

Solve by completing the square.

${x}^{2} + 4 x \textcolor{b l u e}{+ 4} = 24 \textcolor{b l u e}{+ 4}$

${\left(x + 2\right)}^{2} = 28$

$x = \pm \sqrt{28} - 2$

$x = 3.29 c m$

$x = - 7.29 c m \text{ } \leftarrow$ reject as a length of a side

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