Question

the height of a triangle is 4 in. greater than twice its base. the area of the triangle is no more than 168 in.². which inequality can be used to find the possible lengths, x, of the base of the triangle?

x(x + 2)≥168

x(x + 2)≤168

\\(\frac{1}{2}x(x + 4)≤168\\)

\\(\frac{1}{2}x(x + 4)≥168\\)

Explanation:

Step1: Define base and height

Let the base of the triangle be $x$. The height $h$ is 4 in. greater than twice the base, so $h = 2x+4$.

Step2: Recall area formula

The area formula for a triangle is $A=\frac{1}{2}bh$. Substituting $b = x$ and $h=2x + 4$ into the formula, we get $A=\frac{1}{2}x(2x + 4)=x(x + 2)$.

Step3: Set up the inequality

The area of the triangle is no more than 168 in.$^{2}$, so $x(x + 2)\leq168$.

Answer:

$x(x + 2)\leq168$