Chapter ISG, Problem 8.6.1.2P

To determine

To find the dimensions of triangle.

Answer to Problem 8.6.1.2P

Base is 10

Height is 20

Explanation of Solution

Given information:

Area of triangle is 64.

And its height is 8 feet more than its base.

Formula used:

Area of triangle $=\frac{1}{2}\times base\times height$

Calculation:

Let base of the triangle is $x$

So,height of the triangle is $x+8$

Substituting the values in the formula

Area of triangle $=\frac{1}{2}\times x\times (x+8)$

According to question,Area of triangle is 64

  $64=\frac{1}{2}\times x\times (x+8)$

On solving,

  $64=\frac{1}{2}(x^2+8x)$

On cross multiplying

  $128=x^2+8x$

So,

  $x^2+8x-128=0$

Split the middle term so that its addition comes out to be $+8x$

And its multiplication $x^2\times 128$ which is equalsto $128x^2$

  $x^2-8x+16x-128=0$

Taking common,

  $x(x-8)+16(x-8)=0$

Therefore,

  $(x+16)(x-8)=0$

Using the Zero Product Property ,

  $(x+16)=0$ or $(x-8)=0$

So,

  $x+16=0$

Or

  $x=-16$

Also,

  $x-8=0$

So,

  $x=8$

So, $x=-16$ and $x=8$ base can never be negative

Therefore, base is 8

So, height is 16