Question

Please help this one has stumped two tutors and myself

Answer 1

Let w be the width of the rectangle, let d be the diagonal of the rectangle, angle and let h be the height of the rectangle.

Thus, the width is 2 inches longer than the height and the diagonal is 58 inches;

From the right angle triangle, we would apply the Pythagorean Theorem given as;

`\begin{gathered} h^2=o^2+a^2 \\ \\ Where\text{ }h=hypotenuse; \\ o=opposite,a=adjacent \end{gathered}`

Thus, we have;

`\begin{gathered} 58^2=h^2+(h+2)^2 \\ \\ 3364=h^2+h^2+4h+4 \\ \\ 0=2h^2+4h-3360 \\ \\ Divide\text{ }through\text{ }by\text{ }2; \\ \\ h^2+2h-1680=0 \end{gathered}`

Now, we would solve the quadratic equation by factorization; we have;

`\begin{gathered} h^2+42h-40h-1680=0 \\ \\ h(h+42)-40(h+42)=0 \\ \\ h+42=0,h-40=0 \\ \\ h=-42,h=40 \end{gathered}`

Since the height of a rectangle cannot be a negative value, the height of the rectangle is 40 inches.

ANSWER: 40 inches