Question

12. Find the perimeter of the triangle below:
please help show steps thank you :)
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Answer 1

Given:

The lengths of the three sides of the triangle are $\frac{3}{x-4}$, $\frac{16-x}{x^{2}-2 x-8}$ and $\frac{x+5}{x+2}$

We need to determine the perimeter of the triangle.

Perimeter of the triangle:

The perimeter of the triangle can be determined by adding all the three sides.

Thus, we have;

$Perimeter=\frac{3}{x-4}+\frac{16-x}{x^{2}-2 x-8}+\frac{x+5}{x+2}$

Factoring the term $x^2-2x-8$, we get, $(x-4)(x+2)$

Substituting, we get;

$Perimeter=\frac{3}{x-4}+\frac{16-x}{(x-4)(x+2)}+\frac{x+5}{x+2}$

Taking LCM , we have;

$Perimeter=\frac{3(x+2)+16-x+(x+5)(x-4)}{(x-4)(x+2)}$

Simplifying the numerator, we get;

$Perimeter=\frac{3x+6+16-x+x^2-4x+5x-20}{(x-4)(x+2)}$

Adding the like terms in the numerator, we get;

$Perimeter=\frac{x^2+3x+2}{(x-4)(x+2)}$

Factoring the numerator, we get;

$Perimeter=\frac{(x+2)(x+1)}{(x-4)(x+2)}$

Cancelling the common terms, we have;

$Perimeter=\frac{x+1}{x-4}$

Thus, the perimeter of the triangle is $\frac{x+1}{x-4}$