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 triangle - area formula

The area formula of a triangle is $A=\frac{1}{2}bh$, where $b$ is the base and $h$ is the height.

Step2: Substitute the given base and height

We have $b = \frac{2x^{2}+5x - 3}{x^{2}-x - 12}$ and $h=\frac{x^{2}-x - 2}{2x^{2}-5x + 2}$. Then $A=\frac{1}{2}\times\frac{2x^{2}+5x - 3}{x^{2}-x - 12}\times\frac{x^{2}-x - 2}{2x^{2}-5x + 2}$.

Step3: Factor the quadratic expressions

Factor $2x^{2}+5x - 3=(2x - 1)(x + 3)$; $x^{2}-x - 12=(x - 4)(x+3)$; $x^{2}-x - 2=(x - 2)(x + 1)$; $2x^{2}-5x + 2=(2x - 1)(x - 2)$.

Step4: Substitute the factored - forms into the area formula

$A=\frac{1}{2}\times\frac{(2x - 1)(x + 3)}{(x - 4)(x + 3)}\times\frac{(x - 2)(x + 1)}{(2x - 1)(x - 2)}$.

Step5: Cancel out the common factors

Cancel out the common factors $(2x - 1)$, $(x + 3)$ and $(x - 2)$ in the numerator and denominator. We get $A=\frac{x + 1}{2(x - 4)}$.

Answer:

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