**Object of this page** To practice applying the conventional area of a triangle formula to find the height, given the triangle's area and a base.

**Example**

##### Example 1

In this triangle, the area is 17.7 square units, and its base is 4.

Let's find the height of this triangle, which is pictured below.

The red measurements are the ones that are relevant to this problem. Remember that each triangle has 3 base/height pairs. So whenever you are talking about the height, we have to make sure we know which of the 3 'bases' (or sides) of the the triangle we are talking about.

We can tell from the picture, that the height is perpendicular to the base whose measure is 4. That is why the side of length 4 is the base and the other sides do not affect this problem.

**Steps to Find Area**Step 1

Substitute known values into the area formula

Area = $$ \frac{1}{2} \cdot base \cdot height$$

17.7 = ½(4)(h)

Find the height by solving for 'h'

$$ 17.7 = 2h \\ h = \frac{17.7}{2} = 8.85 $$

**Practice** Problems

This problem is very similar to example 1

Step 1Substitute known values into the area formula

Area = $$ \frac{1}{2} \cdot base \cdot height$$

658.8 = ½(24)(h)

This problem is very similar to example 1

Find the height by solving for 'h'

$$ 658.8 = 12h \\ \frac{658.8}{12} = h \\ h = 54.9 $$

Substitute known values into the area formula

Area = $$ \frac{1}{2} \cdot base \cdot height$$

11.6 = ½(4)(h)

Solve for the height

$$11.6 = 2h \\ \frac{11.6}{2} =h \\ h = 5.8 $$

This problem is very similar to example 1 . In every triangle, there is 3 base/height pairs , but you can tell from the picture that the base is the side of length 12. Remember that the base and height are perpendicular.

Find the height by solving for 'h'

$$ 35.8 = 6h \\ \frac{35.8}{6} =h \\ h = 5.9 $$

This problem is very similar to example 1 . In every triangle, there is 3 base/height pairs , but you can tell from the picture that the base is the side of length 11.Remember that the base and height are perpendicular.

Solve for the height

$$ 73.5 = 5.5h \\ \frac{73.5}{5.5} = h \\ h = 13.4 $$

This problem is very similar to example 1 . In every triangle, there is 3 base/height pairs , but you can tell from the picture that the base is the side of length 21. Remember that the base and height are perpendicular.

Solve

$$ 141.9 = 10.5 h \\ \frac{141.9}{10.5} = h \\ h = 13.5 $$

### Some Challenge problems

All right, now let's try some more challenging problems involving finding the height of a triangle. You should be comfortable with the properties of equilateral and isosceles triangle types before attempting the following questions.

This problem is very similar to example 1 . In every triangle, there is 3 base/height pairs , but you can tell from the picture that the base is the side of length 21. Remember that the base and height are perpendicular.

Solve for the height.

$$ 15.6 = 3h \\ \frac{15.6}{3} = h \\ h = 5.2 $$