Question

5.the base of a triangle is $\frac{2x^{2}+5x - 3}{x^{2}-x - 12}$ and the height is $\frac{x^{2}-x - 2}{2x^{2}-5x + 2}$

Explanation:

Step1: Recall the area formula for a triangle

The area formula of a triangle is $A=\frac{1}{2}bh$, where $b$ is the base and $h$ is the height. Here, $b = \frac{2x^{2}+5x - 3}{x^{2}-x - 12}$ and $h=\frac{x^{2}-x - 2}{2x^{2}-5x + 2}$.

Step2: Factor the quadratic - expressions

Factor $2x^{2}+5x - 3=(2x - 1)(x + 3)$; $x^{2}-x - 12=(x-4)(x + 3)$; $x^{2}-x - 2=(x - 2)(x+1)$; $2x^{2}-5x + 2=(2x - 1)(x - 2)$.
So, $b=\frac{(2x - 1)(x + 3)}{(x-4)(x + 3)}$ and $h=\frac{(x - 2)(x + 1)}{(2x - 1)(x - 2)}$.

Step3: Simplify the base and height expressions

Cancel out the common factors:
$b=\frac{2x - 1}{x-4}$ (after canceling out $x + 3$) and $h=\frac{x + 1}{2x - 1}$ (after canceling out $x - 2$ and $2x - 1$).

Step4: Calculate the area

$A=\frac{1}{2}bh=\frac{1}{2}\times\frac{2x - 1}{x-4}\times\frac{x + 1}{2x - 1}$.
Cancel out the common factor $2x - 1$:
$A=\frac{x + 1}{2(x-4)}$.

Answer:

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