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