Chapter 3.1, Problem 37E

To determine

To calculate: .The triangle for the value $x$ .

Answer to Problem 37E

The required value of $x$ is $3$

Explanation of Solution

Given Information:

The area of triangle is $42$ .

Also, the height of the triangle is $x+3$ and base of the triangle is $2x+8$ .

Formula Used:

Area of triangle:

  $Area=\frac{1}{2}\times b\times h$ .

Where $h$ is height and $b$ is base of triangle.

Quadratic formula:

If $a{x}^2+bx+c=0$ then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where $a,b,c$ are constant and $a\ne 0$ .

Calculation:

Consider the area of triangle is $42$ .

To find the value of $x$ ; use area of triangle formula.

  $Area=\frac{1}{2}\times h\times b$

Substitute $42$ for $Area$ , $x+3$ for $h$ and $2x+8$ for $b$ in the above equation.

  $42=\frac{1}{2}\times \left(x+3\right)\times \left(2x+8\right)$

Simplify the above equation.

  $\begin{array}{l}42=\frac{1}{2}\times \left(x+3\right)\times \left(2x+8\right)\\ 84=2x^2+8x+6x+24\\ 2x^2+14x+24-84=0\\ x^2+7x-30=0\end{array}$

Now, use quadratic formula in the above equation.

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $7$ for $b$ and $-30$ for $c$ in the above equation.

  $\begin{array}{c}x=\frac{-7\pm \sqrt{7^2-4\left(1\right)\left(-30\right)}}{2\left(1\right)}\\ =\frac{-7\pm \sqrt{49+120}}{2}\\ =\frac{-7\pm \sqrt{169}}{2}\\ =\frac{-7\pm 13}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-7\pm 13}{2}\\ =\frac{-7+13}{2},\frac{-7-13}{2}\\ =\frac{6}{2},\frac{-20}{2}\\ =3,-10\end{array}$

Since, $x$ is measurement and it is known that measurement cannot be negative.

Thus, the value of $x$ is $3$ .

Hence, the required value of $x$ is $3$ .