Question

The area of a triangle is 17.5 square inches. The length of its base is 3 inches longer than 2 times its height.
Which equation can you use to find the height of the triangle, h?
What is the height of the triangle?
inches

Answer 1

To find the height of the triangle, we can use the formula for the area of a triangle, which is:

\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

According to the problem, the area of the triangle is 17.5 square inches, and the base is 3 inches longer than 2 times the height. We can express the base \( b \) in terms of the height \( h \):

\[
b = 2h + 3
\]

Now, substituting the expression for the base into the area formula, we have:

\[
17.5 = \frac{1}{2} \times (2h + 3) \times h
\]

Next, we can eliminate the fraction by multiplying both sides by 2:

\[
35 = (2h + 3) \times h
\]

Expanding the right-hand side gives us:

\[
35 = 2h^2 + 3h
\]

Rearranging this equation leads to:

\[
2h^2 + 3h - 35 = 0
\]

Now, we can use the quadratic formula to solve for \( h \):

\[
h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 2 \), \( b = 3 \), and \( c = -35 \).

Calculating the discriminant:

\[
b^2 - 4ac = 3^2 - 4 \times 2 \times (-35) = 9 + 280 = 289
\]

So the solution simplifies to:

\[
h = \frac{-3 \pm \sqrt{289}}{2 \times 2} = \frac{-3 \pm 17}{4}
\]

Calculating the two possible values for \( h \):

1. \( h = \frac{14}{4} = 3.5 \)
2. \( h = \frac{-20}{4} = -5 \) (not physically possible since height cannot be negative)

Thus, the height of the triangle is

\[
\boxed{3.5} \text{ inches}
\]