Question

A triangle has base 2x+1 and height 6x-3. What value of x would give an area of 240m2?

Answer 1

Answer:

The values of x which would give an area of 240m² would be:

$x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}$

Step-by-step explanation:

Given

The base of triangle b = 2x+1

The height of triangle h = 6x-3

The Area of the triangle A = 240 m²

The Area of the triangle has the formula

A = 1/2 × b × h

substituting b = 2x+1, h = 6x-3 and A = 240

$240\:=\:\frac{1}{2}\:\left(2x+1\right)\:\times \left(6x-3\right)$

$480=\left(2x+1\right)\left(6x-3\right)$

$480=12x^2-3$

Subtract 480 from both sides

$12x^2-3-480=480-480$

$12x^2-483=0$

$3\left(4x^2-161\right)=0$

$3\left(2x+\sqrt{161}\right)\left(2x-\sqrt{161}\right)=0$

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

$2x+\sqrt{161}=0\quad \mathrm{or}\quad \:2x-\sqrt{161}=0$

solving

$2x+\sqrt{161}=0$

$2x=-\sqrt{161}$

Divide both sides by 2

$\frac{2x}{2}=\frac{-\sqrt{161}}{2}$

$x=-\frac{\sqrt{161}}{2}$

also solving

$2x-\sqrt{161}=0$

$2x=\sqrt{161}$

Divide both sides by 2

$\frac{2x}{2}=\frac{\sqrt{161}}{2}$

$x=\frac{\sqrt{161}}{2}$

Therefore, the values of x which would give an area of 240m² would be:

$x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}$