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?

Respuesta :

H=height
X=base
A=area

H=2b+4

A=1/2 bh

A=(1/2)(2b+4)b

(1/2)(2b+4)b<=168

Or

b^2+2b<=168

Answer:

(x² + 2x) < 168

Step-by-step explanation:

Let the height of the triangle is h in. and base is x in.

As per statement of the question, height of the triangle is 4 in. greater than twice twice of its base

The equation will be h = 2x + 4

Second statement is, the area of the triangle is not more than 168 in²

The inequality for this statement will be A < 168

Now we know area of a triangle A = [tex]\frac{1}{2}(\text{Base})({\text{height)}[/tex]

So A = [tex]\frac{1}{2}(x)(h)[/tex]

A = [tex]\frac{1}{2}(x)(2x+4)[/tex]

A = (x).(x + 2)

and A < 148

So inequality to find the length of x will be

x² + 2x < 168