Question

Set up an equation and solve the problem The area of a triangle sheet of paper is 124 square inches. One side of the triangle is 7 inches more than three times the length of the altitude to that side find the length of that side and the altitude to the sideSide. InAltitude. In

Answer 1

Answer:

• Altitude = 8 inches

,

• Side = 31 inches

Explanation:

Let the altitude to the given side = x inches

One side of the triangle is 7 inches more than 3 times the length of the altitude.

• Length of the side = (3x+7) inches

,

• The area if the triangle = 124 square inches

If the altitude is to the given side, then the given side is the base of the triangle.

$\begin{gathered} \text{Area}=\frac{1}{2}\times\text{Base}\times Height \\ 124=\frac{1}{2}(3x+7)(x) \end{gathered}$

We then solve for x:

$\begin{gathered} 124\times2=3x^2+7x \\ 248=3x^2+7x \\ 3x^2+7x-248=0 \end{gathered}$

Since the equation is quadratic, we solve it using the quadratic formula:

$\begin{gathered} x=\dfrac{-b\pm\sqrt[]{b^2-4ac}}{2a}\; where\; a=3,b=7,c=-248 \\ =\dfrac{-7\pm\sqrt[]{7^2-4(3)(-248)}}{2\times3}=\dfrac{-7\pm\sqrt[]{49-(-2976)}}{6} \\ =\dfrac{-7\pm\sqrt[]{49+2976}}{6}=\dfrac{-7\pm\sqrt[]{3025}}{6} \\ =\dfrac{-7\pm55}{6} \end{gathered}$

Therefore, the values of x are:

$\begin{gathered} x=\dfrac{-7-55}{6}\text{ or x}=\dfrac{-7+55}{6} \\ x=-\dfrac{62}{6}\text{ or x }=\dfrac{48}{6} \\ \text{ x }=8\text{ inches (x cannot be negative)} \end{gathered}$

Thus, we have that:

• Length of the altitude = 8 inches

,

• Length of the side = (3x+7)=3(8)+7=31 inches

CHECK

$\text{Area}=\frac{1}{2}\times\text{Base}\times Height=\frac{1}{2}\times8\times31=124\; in^2$