Question

Please help me with this practice problem I’m stuck on

Answer 1

It is given that the width of a rectangle is 2 inches longer than the height.

If the height is h inches,

The width of the rectangle is (2 + h) inches

If the diagonal measurement is 58 inches.

Using Pythagoras theorem, the relationship between the three sides is given as

`h^2+(2+h)^2=58^2`

Expanding the brackets;

`(2+h)^2=(2+h)(2+h)=2(2+h)+h(2+h)=4+2h+2h+h^2=4+4h+h^2`
`\begin{gathered} \Rightarrow h^2+4+4h+h^2=58 \\ \\ \Rightarrow2h^2+4h=58-4 \\ \\ \Rightarrow2h^2+4h=54 \\ \\ \Rightarrow h^2+2h-27=0 \end{gathered}`

Using the quadratic formula,

`\begin{gathered} h=(-2\pm√(2^2-4(1)(-27)))/(2) \\ \\ \Rightarrow h=(-2\pm√(4+108))/(2) \\ \\ \Rightarrow h=(-2\pm4√(7))/(2) \\ \\ \Rightarrow h=-1\pm2√(7) \\ \\ \Rightarrow h\approx4.3\text{ inches} \end{gathered}`

Hence, the height is 4.3 inches