The area of an isosceles triangle is the area covered by the figure in the two-dimensional plane. The general formula for finding the area of a triangle is half the product of the length of the base and the height of the triangle.

Here, we will look at a detailed explanation of the formula for the area of an isosceles triangle. In addition, we will also look at the derivation of these formulas together with some exercises in which we will apply the formulas to obtain the answer.

## Formula for the area of an isosceles triangle

The area of an isosceles triangle can be calculated if we know the lengths of the base and the height. Multiplying the length of the base by the length of the height and dividing by 2 results in the area of the triangle. Therefore, the area of a triangle is found using the formula:

$latex \text{Area}= \frac{1}{2} \times \text{base} \times \text{height}$ $latex A=\frac{1}{2} \times b \times h$ |

where *b* is the length of the base and *h* is the length of the height.

### How to calculate the area if we only know the sides of an isosceles triangle?

If we know the length of the equal sides and the length of the base of an isosceles triangle, then the height can be calculated using the following formula:

$latex h=\sqrt{{{a}^2}-\frac{{{b}^2}}{4}}$

Therefore, the area can be calculated with the formula:

$latex h=\frac{1}{2}(\sqrt{{{a}^2}-\frac{{{b}^2}}{4}}\times b)$ |

where,

- $latex b$ is the length of the base of the isosceles triangle
- $latex h$ is the length of the height of the triangle
- $latex a$ is the length of the congruent sides of the isosceles triangle

## Area of an isosceles triangle – Examples with answers

The formulas for the area of isosceles triangles detailed above are used to solve the following exercises. Each example has its respective solution, but it is recommended that you try to solve the exercises yourself before looking at the answer.

**EXAMPLE 1**

What is the area of a triangle that has a base of length 6 m and a height of length 7 m?

##### Solution

We can use the following values:

- Height, $latex h=7$ m
- Base, $latex b=6$ m

Therefore, using the formula with these values, we have:

$latex A= \frac{1}{2} \times b \times h$

$latex A= \frac{1}{2} (6)(7)$

$latex A=21$

The area of the triangle is 21 m².

**EXAMPLE 2**

What is the area of a triangle that has a base of length 10 m and a height of length 11 m?

##### Solution

We have the information:

- Height, $latex h=11$ m
- Base, $latex b=10$ m

Substituting these values in the area formula, we have:

$latex A= \frac{1}{2} \times b \times h$

$latex A= \frac{1}{2} (10)(11)$

$latex A=55$

The area of the triangle is 55 m².

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**EXAMPLE 3**

A triangle has a height of 13 m and a base of 15 m. What is its area?

##### Solution

We have the following information:

- Height, $latex h=13$ m
- Base, $latex b=15$ m

Using this information in the formula, we have:

$latex A= \frac{1}{2} \times b \times h$

$latex A= \frac{1}{2} (15)(13)$

$latex A=97.5$

The area of the triangle is 97.5 m².

**EXAMPLE 4**

An isosceles triangle has a base with a length of 8 m and its congruent sides are 10 m long. What is its area?

##### Solution

We have the following lengths:

- Base, $latex b=8$ m
- Congruent sides, $latex a=10$ m

Therefore, we can use the second formula to find the area using the lengths of the sides instead of the height:

$latex h=\frac{1}{2}(\sqrt{{{a}^2}-\frac{{{b}^2}}{4}}\times b)$

$latex h=\frac{1}{2}(\sqrt{{{10}^2}-\frac{{{8}^2}}{4}}\times 8)$

$latex h=\frac{1}{2}(\sqrt{100-\frac{64}{4}}\times 8)$

$latex h=\frac{1}{2}(\sqrt{100-16}\times 8)$

$latex h=\frac{1}{2}(\sqrt{84}\times 8)$

$latex h=\frac{1}{2}(9.17\times 8)$

$latex h=\frac{1}{2}(73.36)$

$latex h=36.68$

The area of the triangle is 36.68 m².

**EXAMPLE 5**

An isosceles triangle has a base with a length of 12 cm and congruent sides that measure 14 cm. What is its area?

##### Solution

We have the following information:

- Base, $latex b=12$ cm
- Congruent sides, $latex a=14$ cm

Therefore, we use these values in the second formula:

$latex h=\frac{1}{2}(\sqrt{{{a}^2}-\frac{{{b}^2}}{4}}\times b)$

$latex h=\frac{1}{2}(\sqrt{{{14}^2}-\frac{{{12}^2}}{4}}\times 12)$

$latex h=\frac{1}{2}(\sqrt{196-\frac{144}{4}}\times 12)$

$latex h=\frac{1}{2}(\sqrt{196-36}\times 12)$

$latex h=\frac{1}{2}(\sqrt{160}\times 12)$

$latex h=\frac{1}{2}(12.65 \times 12)$

$latex h=\frac{1}{2}(151.8)$

$latex h=75.9$

The area of the triangle is 75.9 cm².

## Area of an isosceles triangle – Practice problems

Practice using area formulas to find the area of isosceles triangles. If you need help, you can look at the solved examples above.

## See also

Interested in learning more about isosceles triangles? Take a look at these pages:

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