Question
A triangle with an area of 136 square centimeters has a height that is 2 more than 4 times its base. Find the base and height, in centimeters, of the triangle.
Answers
GPT-4o mini
Let the base of the triangle be denoted as \( b \) cm. According to the problem, the height \( h \) of the triangle can be expressed in terms of the base as follows:
\[
h = 4b + 2
\]
The area \( A \) of a triangle is given by the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
We are given that the area of the triangle is 136 square centimeters. Substituting for \( A \), \( b \), and \( h \):
\[
136 = \frac{1}{2} \times b \times (4b + 2)
\]
To eliminate the fraction, we multiply both sides by 2:
\[
272 = b \times (4b + 2)
\]
Next, we expand the right-hand side:
\[
272 = 4b^2 + 2b
\]
Rearranging this equation gives us a standard form quadratic equation:
\[
4b^2 + 2b - 272 = 0
\]
To simplify, we can divide every term by 2:
\[
2b^2 + b - 136 = 0
\]
Next, we will use the quadratic formula to solve for \( b \). The quadratic formula is given by:
\[
b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
\]
Where \( A = 2 \), \( B = 1 \), and \( C = -136 \). First, we compute the discriminant:
\[
B^2 - 4AC = 1^2 - 4 \cdot 2 \cdot (-136) = 1 + 1088 = 1089
\]
Now we take the square root of 1089:
\[
\sqrt{1089} = 33
\]
Now we can substitute back into the quadratic formula:
\[
b = \frac{-1 \pm 33}{2 \cdot 2} = \frac{-1 \pm 33}{4}
\]
Calculating the two potential values for \( b \):
1. \( b = \frac{-1 + 33}{4} = \frac{32}{4} = 8 \)
2. \( b = \frac{-1 - 33}{4} = \frac{-34}{4} = -8.5 \) (not a valid solution since the base cannot be negative)
Thus, the base of the triangle is:
\[
b = 8 \text{ cm}
\]
Now that we have \( b \), we can find the height \( h \):
\[
h = 4b + 2 = 4(8) + 2 = 32 + 2 = 34 \text{ cm}
\]
In conclusion, the dimensions of the triangle are:
\[
\text{Base} = 8 \text{ cm}, \quad \text{Height} = 34 \text{ cm}
\]
Thus, the final answer is:
\[
\text{Base} = 8 \text{ cm}, \quad \text{Height} = 34 \text{ cm}
\]
\[
h = 4b + 2
\]
The area \( A \) of a triangle is given by the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
We are given that the area of the triangle is 136 square centimeters. Substituting for \( A \), \( b \), and \( h \):
\[
136 = \frac{1}{2} \times b \times (4b + 2)
\]
To eliminate the fraction, we multiply both sides by 2:
\[
272 = b \times (4b + 2)
\]
Next, we expand the right-hand side:
\[
272 = 4b^2 + 2b
\]
Rearranging this equation gives us a standard form quadratic equation:
\[
4b^2 + 2b - 272 = 0
\]
To simplify, we can divide every term by 2:
\[
2b^2 + b - 136 = 0
\]
Next, we will use the quadratic formula to solve for \( b \). The quadratic formula is given by:
\[
b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
\]
Where \( A = 2 \), \( B = 1 \), and \( C = -136 \). First, we compute the discriminant:
\[
B^2 - 4AC = 1^2 - 4 \cdot 2 \cdot (-136) = 1 + 1088 = 1089
\]
Now we take the square root of 1089:
\[
\sqrt{1089} = 33
\]
Now we can substitute back into the quadratic formula:
\[
b = \frac{-1 \pm 33}{2 \cdot 2} = \frac{-1 \pm 33}{4}
\]
Calculating the two potential values for \( b \):
1. \( b = \frac{-1 + 33}{4} = \frac{32}{4} = 8 \)
2. \( b = \frac{-1 - 33}{4} = \frac{-34}{4} = -8.5 \) (not a valid solution since the base cannot be negative)
Thus, the base of the triangle is:
\[
b = 8 \text{ cm}
\]
Now that we have \( b \), we can find the height \( h \):
\[
h = 4b + 2 = 4(8) + 2 = 32 + 2 = 34 \text{ cm}
\]
In conclusion, the dimensions of the triangle are:
\[
\text{Base} = 8 \text{ cm}, \quad \text{Height} = 34 \text{ cm}
\]
Thus, the final answer is:
\[
\text{Base} = 8 \text{ cm}, \quad \text{Height} = 34 \text{ cm}
\]