Question

The base of a triangle is 3cm longer than the height. If the area is 44cm². Find the length

Answer 1

To find the dimensions of the triangle, let's define the height as \( h \) cm. According to the problem, the base \( b \) is 3 cm longer than the height, so we can express the base as:

\[
b = h + 3
\]

The area \( A \) of a triangle is given by the formula:

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

Substituting the expressions for area, height, and base into the formula, we have:

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

Now, we can simplify this equation. First, multiply both sides by 2 to eliminate the fraction:

\[
88 = (h + 3) \times h
\]

Expanding the right side gives:

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

Rearranging the equation to one side sets it to zero:

\[
h^2 + 3h - 88 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 3 \), and \( c = -88 \):

1. Calculate the discriminant:

\[
b^2 - 4ac = 3^2 - 4 \times 1 \times (-88) = 9 + 352 = 361
\]

2. Now, plug the values into the quadratic formula:

\[
h = \frac{-3 \pm \sqrt{361}}{2 \times 1}
\]

Calculating \( \sqrt{361} = 19 \):

\[
h = \frac{-3 \pm 19}{2}
\]

This gives us two potential solutions:

\[
h = \frac{16}{2} = 8 \quad \text{and} \quad h = \frac{-22}{2} = -11
\]

Since height cannot be negative, we discard \( h = -11 \) and take \( h = 8 \) cm.

Now, we can find the base:

\[
b = h + 3 = 8 + 3 = 11 \text{ cm}
\]

Thus, the height and base of the triangle are:

- Height = 8 cm
- Base = 11 cm

The length we would report, depending on the context, can refer to either the height or the base. Would you like any further assistance on this problem?