The perimeter of an isosceles triangle represents the total length around the triangle. On the other hand, the area represents the two-dimensional space occupied by the figure. We can find the perimeter of an isosceles triangle by adding the lengths of its three sides, and we can find its area by multiplying the product of its base and height by one-half.
In this article, we will learn how to calculate the perimeter and area of an isosceles triangle. We will explore their formulas and use them to solve some practice problems.
GEOMETRY
Relevant for...
Learning about the perimeter and area of an isosceles triangle.
GEOMETRY
Relevant for...
Learning about the perimeter and area of an isosceles triangle.
How to find the perimeter of an isosceles triangle?
We can calculate the perimeter of an isosceles triangle by adding the lengths of its three sides. This means we can use the following formula:
$latex p=a+b+c$
where, $latex a,~b,~c$ are the lengths of the sides of the triangle.
However, since an isosceles triangle has two sides of equal length, we can simplify the formula for the perimeter as follows:
| $latex p=b+2a$ |
where, b is the length of the base and a is the length of the congruent sides.
How to find the area of an isosceles triangle?
We can calculate the area of an isosceles triangle by multiplying one-half times the product of its base and height. That is, we multiply the lengths of its base and its height and divide by 2:
| $latex \text{Area}= \frac{1}{2} \times \text{base} \times \text{height}$ $latex A=\frac{1}{2} \times b \times h$ |
where, b is the length of the base and h is the length of the height.
Calculate the area of an isosceles triangle if we only know its sides
To find the area of an isosceles triangle in terms of its sides only, we need to find an expression for the height of the triangle in terms of its sides and then plug it into the area formula.
The Height of an Isosceles Triangle can be calculated using the following formula:
$latex h=\sqrt{{{a}^2}-\frac{{{b}^2}}{4}}$
Substituting this expression for height into the area formula, we have:
| $latex A=\frac{1}{2}(\sqrt{{{a}^2}-\frac{{{b}^2}}{4}}\times b)$ |
where,
- b is the length of the base of the isosceles triangle
- h is the height of the triangle
- a is the length of the congruent sides of the isosceles triangle
Area and perimeter of an isosceles triangle - Examples with answers
EXAMPLE 1
Find the perimeter of an isosceles triangle that has a base with a length of 11 inches and congruent sides of 8 inches.
Solution
We have the following lengths:
- Base, $latex b=11$ in
- Length, $latex a=8$ in
Using the perimeter formula with these values, we have:
$latex p=b+2a$
$latex p=11+2(8)$
$latex p=11+16$
$latex p=27$
The perimeter of the triangle is equal to 27 in.
EXAMPLE 2
Find the area of an isosceles triangle that has a base with a length of 6 yards and a height of 7 yards.
Solution
We have the following information:
- Height, $latex h=7$ yd
- Base, $latex b=6$ yd
Using the area 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 equal to 21 yd².
EXAMPLE 3
What is the perimeter of an isosceles triangle that has a base of 12 feet and congruent sides of 15 feet?
Solution
We have the following lengths:
- Base, $latex b=12$ ft
- Sides, $latex a=15$ ft
Applying the formula for the perimeter with the given information, we have:
$latex p=b+2a$
$latex p=12+2(15)$
$latex p=12+30$
$latex p=42$
The perimeter of the triangle is equal to 42 ft.
EXAMPLE 4
Find the area of an isosceles triangle that has a base with a length of 10 inches and a height of 11 inches.
Solution
We have the following lengths:
- Height, $latex h=11$ in
- Base, $latex b=10$ in
Substituting these values into the formula for the area, 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 equal to 55 in².
EXAMPLE 5
What is the perimeter of an isosceles triangle that has congruent sides of 22 yards and a base of 15 yards?
Solution
We can observe the following lengths:
- Base, $latex b=15$ yd
- Sides, $latex a=22$ yd
Using the formula for the perimeter with these values, we have:
$latex p=b+2a$
$latex p=15+2(22)$
$latex p=15+44$
$latex p=59$
The perimeter of the triangle is equal to 59 yd.
EXAMPLE 6
What is the area of an isosceles triangle that has a height of 13 feet and a base of 15 feet?
Solution
We have the following lengths:
- Height, $latex h=13$ ft
- Base, $latex b=15$ ft
Using the formula for the area with these values, 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 equal to 97.5 ft².
EXAMPLE 7
What is the length of the base of an isosceles triangle that has a perimeter of 38 inches and congruent sides of 13 inches?
Solution
We have the following:
- Perimeter, $latex p=38$ in
- Sides, $latex a=13$ in
In this case, we know the perimeter of the triangle, and we want to find the length of the base, so we use the formula for the perimeter and solve for b:
$latex p=b+2a$
$latex 38=b+2(13)$
$latex 38=b+26$
$latex b=12$
The length of the base is 12 in.
EXAMPLE 8
What is the area of an isosceles triangle that has a base with a length of 8 feet and congruent sides of 10 feet?
Solution
We have the following lengths:
- Base, $latex b=8$ ft
- Congruent sides, $latex a=10$ ft
Since we only know the lengths of the sides of the triangle and not its height, we can use the second formula for the area, and we have:
$latex h=\frac{1}{2}(\sqrt{{{a}^2}-\frac{{{b}^2}}{4}}\times b)$
$latex h=\frac{1}{2}(\sqrt{{{10}^2}-\frac{{{8}^2}}{4}}\times 8)$
$latex h=\frac{1}{2}(\sqrt{100-\frac{64}{4}}\times 8)$
$latex h=\frac{1}{2}(\sqrt{100-16}\times 8)$
$latex h=\frac{1}{2}(\sqrt{84}\times 8)$
$latex h=\frac{1}{2}(9.17\times 8)$
$latex h=\frac{1}{2}(73.36)$
$latex h=36.68$
The area of the triangle is equal to 36.68 ft².
EXAMPLE 9
If an isosceles triangle has a base with a length of 25 inches and a perimeter of 55 inches, what is the length of one of the triangle's congruent sides?
Solution
We have the following:
- Perimeter, $latex p=55$ in
- Base, $latex b=25$ in
We can use the formula for the perimeter and solve for a:
$latex p=b+2a$
$latex 55=25+2a$
$latex 2a=30$
$latex a=15$
The length of one of the congruent sides of the triangle is equal to 15 in.
EXAMPLE 10
What is the area of an isosceles triangle that has a base with a length of 12 inches and congruent sides with a length of 14 inches?
Solution
We have the following lengths:
- Base, $latex b=12$ in
- Congruent sides, $latex a=14$ in
Using the second formula for the area, we have:
$latex h=\frac{1}{2}(\sqrt{{{a}^2}-\frac{{{b}^2}}{4}}\times b)$
$latex h=\frac{1}{2}(\sqrt{{{14}^2}-\frac{{{12}^2}}{4}}\times 12)$
$latex h=\frac{1}{2}(\sqrt{196-\frac{144}{4}}\times 12)$
$latex h=\frac{1}{2}(\sqrt{196-36}\times 12)$
$latex h=\frac{1}{2}(\sqrt{160}\times 12)$
$latex h=\frac{1}{2}(12.65 \times 12)$
$latex h=\frac{1}{2}(151.8)$
$latex h=75.9$
The area of the triangle is equal to 75.9 in².
Area and perimeter of an isosceles triangle - Practice problems
See also
Interested in learning more about isosceles triangles? Take a look at these pages:
