Question

The base of a triangle is four less than three times the height. The area of the triangle is 38 square centimeters. A=(b×h)/2
(3 points)

Answer 1

To solve for the base and height of the triangle, we can use the formula for the area of a triangle:

\[
A = \frac{b \times h}{2}
\]

Given:
- The area \( A = 38 \) square centimeters.
- The base \( b \) is four less than three times the height \( h \).

We can express the base \( b \) in terms of the height \( h \):

\[
b = 3h - 4
\]

Now we can substitute this expression for \( b \) into the area formula:

\[
38 = \frac{(3h - 4) \times h}{2}
\]

Next, we'll multiply both sides by 2 to eliminate the fraction:

\[
76 = (3h - 4) \times h
\]

Expanding the right side:

\[
76 = 3h^2 - 4h
\]

Now, rearranging the equation to set it to zero:

\[
3h^2 - 4h - 76 = 0
\]

Next, we will use the quadratic formula to solve for \( h \). The quadratic formula is:

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

Here, \( a = 3 \), \( b = -4 \), and \( c = -76 \). Plugging in these values:

\[
h = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-76)}}{2 \cdot 3}
\]
\[
h = \frac{4 \pm \sqrt{16 + 912}}{6}
\]
\[
h = \frac{4 \pm \sqrt{928}}{6}
\]
\[
h = \frac{4 \pm 4\sqrt{58}}{6}
\]
\[
h = \frac{2 \pm 2\sqrt{58}}{3}
\]

Calculating the positive value since height cannot be negative:

\[
h = \frac{2 + 2\sqrt{58}}{3}
\]

Now, we can calculate \( b \) using \( b = 3h - 4 \):

\[
b = 3\left(\frac{2 + 2\sqrt{58}}{3}\right) - 4
\]
\[
b = (2 + 2\sqrt{58}) - 4
\]
\[
b = 2\sqrt{58} - 2
\]

Thus, the height \( h = \frac{2 + 2\sqrt{58}}{3} \) and the base \( b = 2\sqrt{58} - 2 \).

To approximate the numerical values:

\[
\sqrt{58} \approx 7.616
\]
Thus:

\[
h \approx \frac{2 + 15.232}{3} \approx \frac{17.232}{3} \approx 5.744 \text{ cm}
\]
\[
b \approx 2(7.616) - 2 \approx 15.232 - 2 \approx 13.232 \text{ cm}
\]

Finally, the height and base of the triangle are approximately:

\[
h \approx 5.74 \text{ cm}, \quad b \approx 13.23 \text{ cm}
\]