Area of Isosceles Triangle: The isosceles triangle area is the amount of surface or space enclosed between the sides of the isosceles triangle. The formula of the area of the isosceles triangle is equal to half the product of the base and height of the triangle. Apart from the general formula, there are different formulas used to calculate the area of isosceles triangles.

An isosceles triangle is one of the types of triangles with two equal sides. A triangle in which two sides (legs) are equal and the base angles are equal is known as an isosceles triangle. In this article, we will provide detailed information on the area of an isosceles triangle. Scroll down to learn more!

What is an Isosceles Triangle?

An isosceles triangle is one of the types of triangles with two equal sides. A triangle in which two sides (legs) are equal and the base angles are equal is known as an isosceles triangle. The name isosceles triangle is derived from the Greek words iso means same, and skelos mean legs.

We know that angles opposite to equal sides are equal in measurement. Thus, in an isosceles triangle, the base angles (angles opposite to equal sides) formed by the legs and the base of the triangle are equal in measure.

If base angles are equal to \(45\) degrees and vertex angle (angle other than base angle) is equal to right angle \((90^°)\), then the given isosceles triangle is called a right-angled isosceles triangle. One interesting fact is that an equilateral triangle is also an isosceles triangle (special case).

Properties of Isosceles Triangle

An isosceles triangle is a triangle with two equal sides. There are certain properties of an isosceles triangle, which makes it unique from all the types of triangles like an equilateral triangle, right-angled triangle, and scalene triangle.
1. The equal sides of an isosceles triangle are known as legs.

2. The unequal side, other than the equal sides, is called the base of the isosceles triangle.

3. The equal angles (angles opposite to equal sides) or the angles formed by the equal sides with the base of the triangle are called the base angles.

4. The angle between the equal sides or the angle other than base angles is called vertex angle.

5. In a right-angled isosceles triangle, the vertex angle is equal to \(90^°\) and the base angles equal \(45^°\).

Definition of Area of Isosceles Triangle

A triangle in which two sides (legs) are equal and the base angles are equal is known as an isosceles triangle. The area of an isosceles triangle is the amount of surface or space enclosed between the sides of the isosceles triangle.

Besides the general formula for calculating the area of an isosceles triangle, which is equal to half the product of the base and height of the isosceles triangle, there are different formulas used to find the area of the isosceles triangle.

The general formula for calculating the area of an isosceles triangle, if the height and base values are known, is given by the product of the base and height of the isosceles triangle divided by two.
\({\text{Area}} = \frac{1}{2} \times {\text{base}} \times {\text{height}}\)

Area of Isosceles Triangle Using Sides

The area of an isosceles triangle can be found by calculating the height or altitude of the isosceles triangle if the lengths of legs (equal sides) and base are given.
The formula used to calculate the area of the isosceles triangle by using the lengths of the equal sides and base is given below:
\({\text{Area}} = \frac{1}{2} \times b \times \sqrt {{a^2} – \frac{{{b^2}}}{4}} \)
Here,
\(a -\) length of legs (equal sides of the isosceles triangle)
\(b -\) length of unequal side or base of the isosceles triangle.
Derivation:
Let the isosceles triangle with the length of the equal sides or legs is “\(a\)” units, and the length of the base of the isosceles triangle is “\(b\)” units.

We know that in an isosceles triangle, the altitude drawn divides the base into two equal parts.
In the given isosceles triangle \(ABC\),
\(AB = AC = a\) units and \(BC = b\) units, \(AD = h\), \(BD = DC = \frac{b}{2}\) units.
In the right-angled triangle \(ADB\), by using the Pythagoras theorem,
\(\rm{Hyp}^2 = \rm{side}^2 + \rm{side}^2\)
\( ⇒ AB^2 = AD^2 + BD^2\)
\( \Rightarrow {a^2} = {h^2} + {\left({\frac{b}{2}} \right)^2}\)
\( \Rightarrow {a^2} = {h^2} + \frac{{{b^2}}}{4}\)
\( \Rightarrow {h^2} = {a^2} – \frac{{{b^2}}}{4}\)
\( \Rightarrow h = \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} \)
We know that area of the isosceles triangle is \(\frac{1}{2} × b × h\).
So, the area of the isosceles triangle when the sides of the isosceles triangle are given is
\(\frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} \)

Area of Isosceles Triangle Using Heron’s Formula

Heron’s formula is used to calculate the area of a triangle when the length of all three sides is given. The area of the isosceles triangle using Heron’s formula is given below:
\(\frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} \)
Derivation:
We know that area of the triangle with sides \(a, b, c\), and semi perimeter \(s\) is given by \(\sqrt {s \left({s – a} \right)\left({s – b} \right)\left({s – c} \right)} \)

Consider an isosceles triangle with sides “\(a\)” and base “\(b\)”,
Then semi perimeter \((s) = \frac{{a + a + b}}{2} = a + \frac{b}{2}\)
The area of the isosceles triangle by Heron’s formula is given by
\( \Rightarrow {\text{Area}} = \sqrt {\left({a + \frac{b}{2}} \right)\left({a + \frac{b}{2} – a} \right)\left({a + \frac{b}{2} – a} \right)\left({a + \frac{b}{2} – b} \right)} \)
\( \Rightarrow {\text{Area}} = \sqrt {\left({a + \frac{b}{2}} \right){{\left({\frac{b}{2}} \right)}^2}\left({a – \frac{b}{2}} \right)} \)
\( \Rightarrow {\text{Area}} = \sqrt {{{\left({\frac{b}{2}} \right)}^2}\left({{a^2} – \frac{{{b^2}}}{4}} \right)} \)
\({\text{Area}} = \frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} \)

Area of Isosceles Triangle Using Trigonometric Ratios

The formula for area of isosceles triangle when the length of two sides and the angle between them is given or when two angles and the length of side included between them are given can be found by using trigonometric ratios in the following manner:

When two sides and angle between them is given:
\({\text{Area}} = \frac{1}{2} \times b \times a \times \sin \alpha \)
Here,
\(a -\) length of the equal side of the triangle
\(b -\) length of the base (unequal side)
\(α -\) angle between the legs and base (base angle)
When two angles and side included is given:
\({\text{Area}} = {a^2} \times \sin \alpha \times \sin \frac{\beta }{2}\)
Here,
\(a -\) length of the equal side of the triangle
\(α -\) base angle
\(β -\) vertex angle

Area of an Isosceles Right Triangle

A triangle in which one angle is a right angle \((90^°)\) and with two equal sides other than the hypotenuse is called an isosceles right triangle.
The area of the isosceles right triangle is
\({\text{Area}} = \frac{1}{2} \times {a^2}\)
Where \(a -\) the length of equal sides.
Derivation:
In the isosceles right triangle, the base and height of the triangle are “\(a\)” units.

We know that area of the triangle is \(\frac{1}{2} × \rm{base} × \rm{height}\).
\({\text{Area}} = \frac{1}{2} \times a \times a = \frac{{{a^2}}}{2}\,{\text{sq}}{\text{.units}}\)

Area of Isosceles Triangle Formulas

The various formulas to be used to find the area of isosceles triangle are given below:

Parameters are given for the isosceles triangle	Formula to be used to calculate the area of the isosceles triangle
When three sides are given	\(\frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}}  \right)} \)
When the lengths of base and height are given	\(\frac{1}{2} \times  {\text{base}} \times  {\text{height}}\)
When two sides and included angle between them is given (SAS)	\(\frac{1}{2} \times b \times a \times \sin \,\alpha \)
When two angles and included side between them is given (ASA)	\( {a^2} \times \sin \,\alpha  \times \sin \,\frac{\beta }{2}\)
Isosceles right triangle	\(\frac{1}{2} \times {a^2}\)

Solved Examples on Area of Isosceles Triangle

Below we have provided some solved examples on Area of Isosceles Triangle:

Q.1. Find the area of triangle with the length of the sides \(3\,\rm{cm}\), \(3\,\rm{cm}\), \(4\,\rm{cm}\).
Ans: Given that the length of the sides of the triangle are \(3\,\rm{cm}\), \(3\,\rm{cm}\), \(4\,\rm{cm}\).

So, \(a = 3\,\rm{cm}\) and \(b = 4\,\rm{cm}\)
We know that area of the isosceles triangle is \(\frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} \).
\(= \frac{1}{2} \times 4 \times \sqrt {\left({{3^2} – \frac{{{4^2}}}{4}} \right)} \)
\(= 2 × \sqrt {9 – 4}\)
\( = 2\sqrt 5 \,{\text{c}}{{\text{m}}^2}\)
Hence, the area of the given triangle is \( 2\sqrt 5 \,{\text{c}}{{\text{m}}^2}\)

Q.2. Find the height \((h)\) of the triangle with the length of the sides \(10\,\rm{cm}\), \(10\,\rm{cm}\), \(12\,\rm{cm}\).
Ans: Given that the length of the sides of the triangle are \(10\,\rm{cm}\), \(10\,\rm{cm}\), \(12\,\rm{cm}\).

So, \(a = 10\,\rm{cm}\) and \(b = 12\,\rm{cm}\)
We know that height of the isosceles triangle is \(\sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} .\)
\( = \sqrt {{{10}^2} – \frac{{{{12}^2}}}{4}} \)
\( = \sqrt {{100} – \frac{{144}}{4}} \)
\( = \sqrt {100 – 36}\)
\( = \sqrt {64} = 8\,\rm{cm}\)
Hence, the height \(\left( h \right)\) of the given triangle is \(8\,{\rm{cm}}.\)

Q.3. Find the length of the base of an isosceles right triangle, whose area is \(243\,\rm{cm}^2\) and the length of the altitude is \(9\,\rm{cm}\).
Ans: Given the length of the altitude \((h) = 9\,\rm{cm}\).
The area of the isosceles triangle is \(243\,\rm{cm}^2\).
Let the length of the base is \(b\,\rm{cm}\).
We know that area of the isosceles triangle is \(\frac{1}{2} × b × h\).
\( \Rightarrow 243 = \frac{1}{2} \times 9 \times b\)
\( \Rightarrow b = 243 \times \frac{2}{9}\)
\( \Rightarrow b = 54\,\rm{cm}\)
Hence, the length of the base of the given triangle is \(54\,\rm{cm}\).

Q.4. Find the area of the isosceles triangle in which the length of equal sides is \(3\,\rm{cm}\), an unequal side of length \(4\,\rm{cm}\) and angle between the sides is \(60^°\).
Ans: Let \(a = 3\,\rm{cm}\), \(b = 4\,\rm{cm}\) and \(α = 60^°\)
We know that the area of the isosceles triangle is \(\frac{1}{2} × b × a × \rm{sin}\,⁡α\).
\( = \frac{1}{2} \times 4 \times 3 \times \sin \,{60^ \circ }\)
\( = \frac{1}{2} \times 4 \times 3 \times \frac{{\sqrt 3 }}{2} = 3\sqrt 3 \,{\text{c}}{{\text{m}}^2}\)
Hence, the area of the given isosceles triangle is \(3\sqrt 3 \,{\text{c}}{{\text{m}}^2}\)

Q.5. Find the area of the right isosceles triangle, in which the length of the legs is \(4\,\rm{cm}\).
Ans: Given the length of the legs \(a = 4\,\rm{cm}\).
We know that the area of the given triangle is \(\frac{a^2}{2}\).
\( = \frac{{{4^2}}}{2} = \frac{{16}}{2} = 8\,{\text{c}}{{\text{m}}^2}\)
Hence, the area of the given right isosceles triangle is \(8\,{{\text{cm}}^2}\).

Summary

In this article, we have studied the definition of the isosceles triangle and the unique properties of the isosceles triangle. We have discussed the formula for area of isosceles triangle, which tells the amount of surface or space enclosed between the sides of the isosceles triangle.

The general formula for calculating the area of isosceles triangle is \({\text{Area}} = \frac{1}{2} \times {\text{base}} \times {\text{height}}\). The area of the isosceles triangle using Heron’s formula is \(\frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} \). The area of the isosceles right triangle is \({\text{Area}} = \frac{1}{2} \times {a^2}\).

Learn All the Concepts on Area of Triangles

FAQs on Area of Isosceles Triangle

Below we have provided some frequently asked questions related to the Area of the Isosceles Triangle:

Q.1. What is the formula of the area of an isosceles triangle?
Ans: The general formula used to calculate the area of an isosceles triangle with base “\(b\)” and legs “\(a\)” is given by \(\frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} \).

Q.2. How do you find the height of an isosceles triangle?
Ans: The height of the isosceles triangle can be found by using the formula \(\sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} \), where \(a -\) length of equal sides and \(b -\) length of the base.

Q.3. What is the area and perimeter of the isosceles triangle?
Ans: The perimeter and area of the isosceles triangle with legs “\(a\)” and base “\(b\)” is given by \(2a + b\) and \(\frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}}\right)} \), respectively.

Q.4. How do you find the area of the isosceles triangle by using height and base?
Ans: The area of the isosceles triangle is \(\frac {1}{2} × \rm{base} × \rm{height}\).

Q.5. What does the area of the isosceles triangle mean?
Ans: The area of an isosceles triangle is the amount of surface or space enclosed between the sides of the isosceles triangle.

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