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:

1. First, recall the area - formula of a triangle:

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

1. Factor the quadratic expressions:

• Factor \(2x^{2}+5x - 3\):
• \(2x^{2}+5x - 3=2x^{2}+6x - x - 3 = 2x(x + 3)-(x + 3)=(2x - 1)(x + 3)\).
• Factor \(x^{2}-x - 12\):
• \(x^{2}-x - 12=x^{2}-4x+3x - 12=x(x - 4)+3(x - 4)=(x - 4)(x + 3)\).
• Factor \(x^{2}-x - 2\):
• \(x^{2}-x - 2=x^{2}-2x+x - 2=x(x - 2)+(x - 2)=(x - 2)(x+1)\).
• Factor \(2x^{2}-5x + 2\):
• \(2x^{2}-5x + 2=2x^{2}-4x - x + 2=2x(x - 2)-(x - 2)=(2x - 1)(x - 2)\).

1. Substitute the factored - forms into the area formula:

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

1. Cancel out the common factors:

• Cancel out \((2x - 1)\), \((x + 3)\), and \((x - 2)\) in the numerator and denominator.
• \(A=\frac{1}{2}\times\frac{1}{x - 4}\times\frac{x + 1}{1}=\frac{x + 1}{2(x - 4)}\).

Step1: Recall area formula

\(A=\frac{1}{2}bh\)

Step2: Factor quadratic expressions

\(2x^{2}+5x - 3=(2x - 1)(x + 3)\), \(x^{2}-x - 12=(x - 4)(x + 3)\), \(x^{2}-x - 2=(x - 2)(x + 1)\), \(2x^{2}-5x + 2=(2x - 1)(x - 2)\)

Step3: Substitute factored - forms

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

Step4: Cancel common factors

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

Answer:

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