Question

The base of a triangle is 3cm longer than its corresponding height. if the area is 44cm^2, find the length of its base

Answer 1

To find the length of the base of the triangle, we can use the formula for the area of a triangle:

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

Let the height of the triangle be \( h \) cm. Then, the base of the triangle can be expressed as:

\[
\text{base} = h + 3 \text{ cm} \quad \text{(since the base is 3 cm longer than the height)}
\]

According to the problem, the area of the triangle is 44 cm². We can set up the equation using the area formula:

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

Simplifying this, we multiply both sides by 2 to eliminate the fraction:

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

Expanding the right side:

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

Rearranging the equation gives us a standard quadratic form:

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

Now, we can use the quadratic formula to find \( h \):

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

Here, \( a = 1 \), \( b = 3 \), and \( c = -88 \):

Calculating the discriminant:

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

Now substituting this back into the quadratic formula:

\[
h = \frac{-3 \pm \sqrt{361}}{2 \times 1} = \frac{-3 \pm 19}{2}
\]

Calculating the two possible values for \( h \):

1. \( h = \frac{16}{2} = 8 \)
2. \( h = \frac{-22}{2} = -11 \) (not a valid height since it can't be negative)

So, we have:

\[
h = 8 \text{ cm}
\]

Now, substituting \( h \) back to find the base:

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

Thus, the length of the base is:

\[
\boxed{11 \text{ cm}}
\]