To find the base and height of the triangle, we can start by using the formula for the area of a triangle:
\[ A = \frac{b \times h}{2} \]
We know the area \( A = 38 \) cm². Thus, we can write:
\[ 38 = \frac{b \times h}{2} \]
Multiplying both sides by 2 to eliminate the fraction gives us:
\[ 76 = b \times h \]
Next, we also know the relationship between the base \( b \) and the height \( h \). The problem states that the base is four less than three times the height:
\[ b = 3h - 4 \]
Now we can substitute this expression for \( b \) into the area equation:
\[ 76 = (3h - 4) \times h \]
Expanding this gives:
\[ 76 = 3h^2 - 4h \]
Rearranging the equation gives us a standard quadratic form:
\[ 3h^2 - 4h - 76 = 0 \]
Next, we can use the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Here, \( a = 3 \), \( b = -4 \), and \( c = -76 \).
Calculating the discriminant:
\[ b^2 - 4ac = (-4)^2 - 4 \times 3 \times (-76) = 16 + 912 = 928 \]
Now we can apply the quadratic formula:
\[ h = \frac{-(-4) \pm \sqrt{928}}{2 \times 3} = \frac{4 \pm \sqrt{928}}{6} \]
Calculating \( \sqrt{928} \):
\[ \sqrt{928} \approx 30.5 \]
Now we substitute this back into the equation for \( h \):
\[ h = \frac{4 \pm 30.5}{6} \]
This gives us two potential solutions:
- \( h = \frac{34.5}{6} \approx 5.75 \)
- \( h = \frac{-26.5}{6} \approx -4.42 \) (not a valid solution, since height cannot be negative)
Thus, we take the valid solution:
\[ h \approx 5.75 \]
Now we can find \( b \):
\[ b = 3h - 4 = 3(5.75) - 4 = 17.25 - 4 = 13.25 \]
So the height of the triangle is approximately \( 5.75 \) cm and the base is approximately \( 13.25 \) cm.
To summarize:
- Height \( h \approx 5.75 \) cm
- Base \( b \approx 13.25 \) cm