# The base of a triangle is 6 cm shorter than the altitude to that base. How do you find the length of the base and the altitude if the are of the triangle is 56 cm^2?

Feb 8, 2016

Base $a = 8$
Altitude $h = 14$

#### Explanation:

Let's approach this problem algebraically.

Assume that the triangle's altitude is $h$ and a corresponding base is $a$.
Then from the relationship between their lengths we can derive the following equation:
$a = h - 6$
Since the area is given, the second equation is
$\frac{1}{2} a h = 56$

All we have to do now is to solve this system of two equations with two unknowns.
Use an expression for $a$ from the first equation and substitute it into the second:
$\frac{1}{2} \left(h - 6\right) h = 56$

This is a quadratic equation, which in a more traditional form looks like
${h}^{2} - 6 h - 112 = 0$

Its solutions are
${h}_{1 , 2} = \frac{6 \pm \sqrt{36 + 448}}{2} = \frac{6 \pm 22}{2} = 3 \pm 11$

Negative solution should be discarded since we are dealing with lengths, so the only solution for $h$ is $14$.
Then the base $a$ is $8$.