Chapter 9.5, Problem 28PPS

To determine

Find the solution.

Answer to Problem 28PPS

$x=-14.3$ or $x=6.3$

Explanation of Solution

Given Information: Area of Triangle = $45i{n}^2$ , Height $h=x$ in., Base $b=x+8$ in.

Formula Used : Area of Triangle = $\frac{1}{2}\times Base\times Height$ , $a^2+2ab+b^2={(a+b)}^2$ and

  $a^2-b^2=(a+b)(a-b)$

Calculation:

Area of Triangle = $45i{n}^2$ , Height $h=x$ in., Base $b=x+8$ in.

Area of Triangle = $\frac{1}{2}\times Base\times Height$

  $45=\frac{1}{2}\times \left(x+8\right)\times x$

So,

  $90=x^2+8x$

Now, simplify

  $x^2+8x-90=0$

Now, adding & subtracting the square of half the coefficient of w.

  $x^2+8x-90+{\left(\frac{8}{2}\right)}^2-{\left(\frac{8}{2}\right)}^2=0$

Now use formula, $a^2+2ab+b^2={(a+b)}^2$

  ${\left(x+4\right)}^2-90-4^2=0$

So,

  ${\left(x+4\right)}^2-90-16=0$

So,

  ${\left(x+4\right)}^2-106=0$

Now, square root of 106 is $\sqrt{106}$

  ${\left(x+4\right)}^2-{\left(\sqrt{106}\right)}^2=0$

Now use formula, $a^2-b^2=(a+b)(a-b)$

  $\left(x+4+\sqrt{106}\right)\left(x+4-\sqrt{106}\right)=0$

Now,

  $\left(x+14.3\right)\left(x-6.3\right)=0$

Hence,

$x=-14.3$ or $x=6.3$