Area of Similar Triangles: Theorems, Proofs, and Examples

Geometric figures have the same shape, but different sizes are known as similar figures. Two triangles are said to be similar if their corresponding angles are equal and corresponding sides are proportional. The two conditions given in the above definition are independent. If either of the two conditions holds, then the other holds automatically. So any one of the two conditions can be used to define similar triangles. For similar triangles, not only do their angles and sides share a relationship, but also the ratio of their perimeter, areas, and other aspects are in proportion. In this article, we will learn about the relationship between the area of similar triangles.

What are Similar Triangles?

Two triangles are said to be similar if

1. their corresponding angles are equal 
2. corresponding sides are proportional.

If two triangles \(A B C\) and \(D E F\) are similar, then \(\angle A=\angle D, \angle B=\angle E, \angle C=\angle F\) and \(\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}\).

The two conditions given in the above definition are independent. If either of the two conditions holds, then the other holds automatically. So any one of the two conditions can be used to define similar triangles.

If corresponding angles of two triangles are equal, then they are known as equiangular triangles.

Area of Similar Triangles

In two similar triangles, the ratio of their corresponding sides is the same. However, to ensure that the two triangles are similar, we do not necessarily need information about all sides and all angles. For similar triangles, not only do their angles and sides share a relationship, but also the ratio of their perimeter, areas, and other aspects are in proportion. 

In this section, we will discuss some theorems concerning the ratio of areas of similar triangles.

Theorems on Area of Similar Triangles

Theorem 1

Theorem: The ratio of the areas of two similar triangles is equal to the ratio of the square of any two corresponding sides.

Given: Two triangles \(\triangle A B C\) and \(\triangle D E F\) such that \(\triangle A B C^{\sim} \triangle D E F\).

To prove: \(\frac{{{\rm{Area}}\left( {\Delta ABC} \right)}}{{{\rm{Area}}\left( {\Delta DEF} \right)}}=\frac{A B^{2}}{D E^{2}}=\frac{B C^{2}}{E F^{2}}=\frac{A C^{2}}{D F^{2}}\)

Construction: Draw \(A L \perp B C\) and \(D M \perp E F\).

Proof: Similar triangles are equiangular, and their corresponding sides are proportional.

Therefore, \(\triangle A B C^{\sim} \triangle D E F\)
\(\Rightarrow \angle A=\angle D, \angle B=\angle E, \angle C=\angle F\) and \(\frac{A B}{D E}=\frac{B C}{F E}=\frac{A C}{D F} \quad \ldots \ldots(i)\)

Thus, in \(\triangle A L B\) and \(\triangle D M E\), we have
\( \Rightarrow \angle ALB = \angle DME\) (Each equal to \({90^{\rm{o}}}\))
and, \(\angle B=\angle E \) (From \((i)\))
So, by AA criterion of similarity, we have
\(\triangle A L B^{\sim} \triangle D M E\)
\(\Rightarrow \frac{A L}{D M}=\frac{A B}{D E} \quad \ldots \ldots(ii)\)

From \((i)\) and \((ii)\), we get
\(\frac{A B}{D E}=\frac{B C}{F E}=\frac{A C}{D F}=\frac{A L}{D M} \quad \ldots \ldots(iii)\)

Now, \(\frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\triangle D E F)}=\frac{\frac{1}{2}(B C \times A L)}{\frac{1}{2}(E F \times D M)}\)
\(\Rightarrow \frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\triangle D E F)}=\frac{B C}{E F} \times \frac{A L}{D M}\)
\(\Rightarrow \frac{\text { Area }(\triangle A B C)}{\operatorname{Area}(\triangle D E F)}=\frac{B C}{E F} \times \frac{B C}{E F}\) (From \((iii)\))
\( \Rightarrow \frac{{{\rm{Area}}\left( {\Delta ABC} \right)}}{{{\rm{Area}}\left( {\Delta DEF} \right)}} = \frac{{B{C^2}}}{{E{F^2}}}\)

But, \(\frac{B C}{E F}=\frac{A B}{D E}=\frac{A C}{D F}\)
\(\Rightarrow \frac{B C^{2}}{E F^{2}}=\frac{A B^{2}}{D E^{2}}=\frac{A C^{2}}{D F^{2}}\)

Hence, \(\frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\triangle D E F)}=\frac{B C^{2}}{E F^{2}}=\frac{A B^{2}}{D E^{2}}=\frac{A C^{2}}{D F^{2}}\)

Theorem 2

Theorem: The areas of two similar triangles are in the ratio of the squares of corresponding altitudes.

Given: Two triangles \(\triangle A B C\) and \(\triangle D E F\) such that \(\triangle A B C^{\sim} \triangle D E F\) and \(A L \perp B C\) and \(D M \perp E F\)

To prove: \(\frac{\operatorname{Area}(\triangle A B C)}{\text { Area }(\triangle D E F)}=\frac{A L^{2}}{D M^{2}}\)

Proof: The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Therefore, \(\frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\triangle D E F)}=\frac{A B^{2}}{D E^{2}} \ldots \ldots(i)\)

Now, in \(\triangle A L B\) and \(\triangle D M E\), we have
\(\Rightarrow \angle A L B=\angle D M E \quad\left[\right.\) Each equal to \(\left.90^{\circ}\right]\)

and, \(\angle B=\angle E \quad\left[\because \triangle A B C^{\sim} \triangle D E F \therefore \angle A=\angle D, \angle B=\angle E, \angle C=\angle F\right]\)

So, by AA criterion of similarity, we have
\(\triangle A L B \sim \triangle D M E\)
\(\Rightarrow \frac{A B}{D E}=\frac{A L}{D M}\)
\(\Rightarrow \frac{A B^{2}}{D E^{2}}=\frac{A L^{2}}{D M^{2}} \quad \ldots \ldots (ii)\)

From \((i)\) and \((ii)\), we get
\(\frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\Delta D E F)}=\frac{A L^{2}}{D M^{2}}\)

Theorem 3

Theorem: The areas of two similar triangles are in the ratio of the squares of the corresponding medians.

Given: Two triangles \(\triangle A B C\) and \(\triangle D E F\) such that \(\triangle A B C^{\sim} \triangle D E F\) and \(A P, D Q\) are their medians.

To prove: \(\frac{\operatorname{Area}(\triangle A B C)}{\text { Area }(\triangle D E F)}=\frac{A P^{2}}{D Q^{2}}\)

Proof: Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Therefore, \(\frac{{{\rm{Area}}\left( {\Delta ABC} \right)}}{{{\rm{Area}}\left( {\Delta DEF} \right)}}=\frac{A B^{2}}{D E^{2}} \ldots \ldots(i)\)

Now, \(\triangle A B C^{\sim} \triangle D E F\)
\(\Rightarrow \frac{A B}{D E}=\frac{B C}{E F}\)
\(\Rightarrow \frac{A B}{D E}=\frac{2 B P}{2 E Q}=\frac{B P}{E Q} \quad \ldots \ldots(ii)\)

Thus, in triangles \(\triangle A P B\) and \(\triangle D Q E\), we have
\(\frac{A B}{D E}=\frac{B P}{E Q}\) and \(\angle B=\angle E \quad\left[\because \triangle A B C^{\sim} \triangle D E F\right]\)

So, by SAS criterion of similarity, we have
\(\triangle A P B^{\sim} \triangle D Q E\)
\(\Rightarrow \frac{B P}{E Q}=\frac{A P}{D Q} \quad \cdots \cdots(iii)\)

From \((ii)\) and \((iii)\), we get
\(\frac{A B}{D E}=\frac{A P}{D Q}\)
\(\Rightarrow \frac{A B^{2}}{D E^{2}}=\frac{A P^{2}}{D Q^{2}} \quad \ldots \ldots(iv)\)

From \((i)\) and \((iv)\), we get
\(\frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\triangle D E F)}=\frac{A P^{2}}{D Q^{2}}\)

Theorem 4

Theorem: The areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments.

Given: Two triangles \(\triangle A B C\) and \(\triangle D E F\) such that \(\triangle A B C \sim \triangle D E F\) and \(A X\) and \(D Y\) are bisectors of \(\angle A\) and \(\angle D\) respectively.

To prove: \(\frac{{{\rm{Area}}\left( {\Delta ABC} \right)}}{{{\rm{Area}}\left( {\Delta DEF} \right)}}=\frac{A X^{2}}{D Y^{2}}\)

Proof: Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Therefore, \(\frac{{{\rm{Area}}\left( {\Delta ABC} \right)}}{{{\rm{Area}}\left( {\Delta DEF} \right)}}=\frac{A B^{2}}{D E^{2}} \ldots \ldots(i)\)

Now, \(\triangle A B C^{\sim} \triangle D E F\)
\(\Rightarrow \angle A=\angle D\)
\(\Rightarrow \frac{1}{2} \angle A=\frac{1}{2} \angle D\)
\(\Rightarrow \angle B A X=\angle E D Y\)

Thus, in triangles \(A B X\) and \(D E Y\), we have
\(\angle B A X=\angle E D Y\) and \(\angle B=\angle E\left[\because \triangle A B C^{\sim} \triangle D E F\right]\)

So, by AA similarity criterion, we have
\(\triangle A B X^{\sim} \triangle D E Y\)
\(\Rightarrow \frac{A B}{D E}=\frac{A X}{D Y}\)
\(\Rightarrow \frac{A B^{2}}{D E^{2}}=\frac{A X^{2}}{D Y^{2}} \quad \ldots \ldots(ii)\)

From \((i)\) and \((ii)\), we get
\(\frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\Delta D E F)}=\frac{A X^{2}}{D Y^{2}}\)

Theorem 5

Theorem: If the areas of two similar triangles are equal, then the triangles are congruent, i.e., equal and similar triangles are congruent.

Given: Two triangles \(\triangle A B C\) and \(\triangle D E F\) such that \(\triangle A B C^{\sim} \triangle D E F\) and Area \((\triangle A B C)=\) Area \((\triangle D E F)\)

To prove: \(\triangle A B C \cong \triangle D E F\)

Proof: We have, \(\triangle A B C^{\sim} \triangle D E F\)
\(\Rightarrow \angle A=\angle D, \angle B=\angle E, \angle C=\angle F\) and \(\frac{A B}{D E}=\frac{B C}{F E}=\frac{A C}{D F}\)

To prove that \(\triangle A B C \cong \triangle D E F\), it is sufficient to show that \(A B=D E, B C=E F\) and \(A C=D F\)

It is given that Area \((\triangle A B C)=\) Area \((\triangle D E F)\)
\(\Rightarrow \frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\Delta D E F)}=1\)
\(\Rightarrow \frac{A B^{2}}{D E^{2}}=\frac{B C^{2}}{E F^{2}}=\frac{A C^{2}}{D F^{2}}=1 \quad\left[\because \frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\triangle D E F)}=\frac{A B^{2}}{D E^{2}}=\frac{B C^{2}}{E F^{2}}=\frac{A C^{2}}{D F^{2}}\right]\)
\(\Rightarrow A B^{2}=D E^{2}, B C^{2}=E F^{2}, A C^{2}=D F^{2}\)
\(\Rightarrow A B=D E, B C=E F\) and \(A C=D F\)

Hence, \(\triangle A B C \cong \triangle D E F\)

Solved Examples – Area of Similar Triangles

Q.1. In the given figure, the line segment \(X Y\) is parallel to the side \(A C\) of \(\triangle A B C\), and it divides the triangle into two parts of equal areas. Find the ratio \(\frac{A X}{A B}\).

Ans: We have \(X Y || A C\)
So, \(\angle B X Y=\angle A\) and \(\angle B Y X=\angle C\) (Corresponding angles)
Therefore, \(\triangle A B C^{\sim} \triangle X B Y\) (AA similarity criterion)
So, \(\frac{{{\rm{Area}}\left( {\Delta ABC} \right)}}{{{\rm{Area}}\left( {\Delta DEF} \right)}}=\left(\frac{A B}{X B}\right)^{2}\) (Theorem \(1\)) …….(i)

Also, Area \((\triangle A B C)=2 \times\) Area \((\triangle X B Y)\) (Given)
So, \(\frac{{{\rm{Area}}\left( {\Delta ABC} \right)}}{{{\rm{Area}}\left( {\Delta XBY} \right)}}=\frac{2}{1} \ldots \ldots(ii)\)
Therefore, from \((i)\) and \((ii)\)
\(\left(\frac{A B}{X B}\right)^{2}=\frac{2}{1} \Rightarrow \frac{A B}{X B}=\frac{\sqrt{2}}{1}\)
\(\Rightarrow \frac{X B}{A B}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow 1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\Rightarrow \frac{A B-X B}{A B}=\frac{\sqrt{2}-1}{\sqrt{2}}\)
\(\Rightarrow \frac{A X}{A B}=\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{2-\sqrt{2}}{2}\)
Therefore, \(\frac{A X}{A B}=\frac{2-\sqrt{2}}{2}\)

Q.2. Prove that the area of an equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
Ans:
Given: A square \(A B C D\). Equilateral triangles \(\triangle B C E\) and \(\triangle A C F\) have been described on side \(B C\) and diagonal \(A C\) respectively.

To prove: Area(\(\triangle B C E)=\frac{1}{2}\) Area(\(\triangle A C F\))
Proof: Since \(\triangle B C E\) and \(\triangle A C F\) are equilateral. Therefore, they are equiangular (each angle being equal to \(60^{\circ}\)) and hence
\(\triangle B C E^{\sim} \triangle A C F\)
\(\Rightarrow \frac{\operatorname{Area}(\triangle B C E)}{\operatorname{Area}(\triangle A C F)}=\frac{B C^{2}}{A C^{2}}\)
We know that, diagonal of a square \(=\sqrt{2} \times\) side \(\Rightarrow A C=\sqrt{2} B C\)
\(\Rightarrow \frac{\operatorname{Area}(\Delta B C E)}{\operatorname{Area}(\triangle A C F)}=\frac{B C^{2}}{\sqrt{2} B C^{2}}\)
\(\Rightarrow \frac{\operatorname{Area}(\Delta B C E)}{\operatorname{Area}(\triangle A C F)}=\frac{1}{2}\)

Q.3. \(D, E, F\) are the mid-points of the sides \(B C, C A\), and \(A B\), respectively, of a \(\triangle A B C\). Determine the ratio of the areas of \(\triangle D E F\) and \(\triangle A B C\).
Ans: Since \(D\) and \(E\) are the mid-points of the sides \(B C, C A\), and \(A B\), respectively, of a \(\triangle A B C\).
Therefore, \(D E||B A \Rightarrow D E||F A \ldots \ldots(\mathrm{i})\)

Since \(D\) and \(F\) are mid-points of the sides \(B C\) and \(A B\) respectively of \(\triangle A B C\). Therefore,
\(D F||C A \Rightarrow D F|| A E\)
From \((i)\) and \((ii)\), we conclude that \(A F D E\) is a parallelogram.
Similarly, \(B D E F\) is a parallelogram.
In \(\triangle D E F\) and \(\triangle A B C\), we have
\(\angle F D E=\angle A\) (Opposite angles of parallelogram \(A F D E\))
and, \(\angle D E F=\angle B\) (Opposite angles of parallelogram \(B D E F\))
So, by the AA similarity criterion, we have
\(\triangle D E F^{\sim} \triangle A B C\)
\(\Rightarrow \frac{{{\rm{Area}}\,{\rm{of}}\,\Delta DEF}}{{{\rm{Area}}\,{\rm{of}}\,\Delta ABC}}=\frac{D E^{2}}{A B^{2}}=\frac{\left(\frac{1}{2} A B\right)^{2}}{A B^{2}}=\frac{1}{4} \quad\left[\because D E=\frac{1}{2} A B\right]\)
Hence, the Area of \(\triangle D E F:\) Area of \(\triangle A B C=1: 4\)

Q.4. If \(\triangle A B C^{\sim} \triangle D E F\) such that \(A B=1.2 \mathrm{~cm}\) and \(D E=1.4 \mathrm{~cm}\). Find the ratio of areas of \(\triangle A B C\) and \(\triangle D E F\).
Ans: We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
Therefore, \(\frac{\text {Area of } \triangle A B C}{\text { Area of } \triangle D E F}=\frac{A B^{2}}{D E^{2}}\)
\(\Rightarrow \frac{\text { Area of } \triangle A B C}{\text { Area of } \triangle D E F}=\frac{(1.2)^{2}}{(1.4)^{2}}=\left(\frac{12}{14}\right)^{2}=\frac{36}{49}\)
Therefore, \(\frac{\text { Area of } \triangle A B C}{\text { Area of } \triangle D E F}=\frac{36}{49}\)

Q.5. In two similar triangles \(A B C\) and \(P Q R\), if their corresponding altitudes \(A D\) and \(P S\) are in the ratio \(4: 9\), find the ratio of the areas of \(\triangle A B C\) and \(\triangle P Q R\).
Ans: Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.
Therefore, \(\frac{\text { Area of } \triangle A B C}{\text { Area of } \triangle P Q R}=\frac{A D^{2}}{P S^{2}}\)
\(\Rightarrow \frac{\text { Area of } \triangle A B C}{\text { Area of } \triangle P Q R}=\left(\frac{4}{9}\right)^{2}=\frac{16}{81}\)
Hence, the Area of \(\triangle A B C:\) Area of \(\triangle P Q R=16: 81\)

Q.6. If \(\triangle A B C\) is similar to \(\triangle D E F\) such that \(B C=3 \mathrm{~cm}, E F=4 \mathrm{~cm}\) and area of \(\triangle A B C=54 \mathrm{~cm}^{2}\). Determine the area of \(\triangle D E F\).
Ans: Since the ratios of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
Therefore, \(\frac{\text { Area of } \triangle A B C}{\text { Area of } \triangle D E F}=\frac{A B^{2}}{E F^{2}}\)
\(\Rightarrow \frac{54}{\text { Area of } \triangle D E F}=\frac{3^{2}}{4^{2}}\)
\(\Rightarrow\) Area of \(\triangle D E F=\frac{54 \times 16}{9}=96 \mathrm{~cm}^{2}\)
Therefore, the area of \(\triangle D E F=96 \mathrm{~cm}^{2}\)

Summary

In the above article, we have studied the meaning of similarity of triangles and the concept of area of similar triangles. Also, we have proved some theorems on the area of similar triangles and solved some example problems on the area of similar triangles.

Frequently Asked Questions (FAQ) – Area of Similar Triangles

Q.1. Do similar triangles have equal areas?
Ans: No, similar triangles will have the ratio of their areas equal to the square of the ratio of their pair of corresponding sides. So, the areas of two triangles cannot certainly be equal. But congruent triangles always have equal areas.

Q.2. How do you find the area of similar figures?
Ans: If two figures are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Q.3. How do you find missing sides of similar triangles?
Ans: We can write ratios to compare the lengths of sides. First, identify the corresponding sides of two similar triangles, then place the first side in the numerator and the corresponding side in the denominator.

Q.4. How to find the ratio of the area of similar triangles?
Ans: If two triangles are similar, then the ratio of two similar triangles is equal to the ratio of the square of any two corresponding sides.
Example: If \(\triangle A B C^{\sim} \triangle D E F\). Then,
\(\frac{\operatorname{Area}(\triangle A B C)}{\operatorname{Area}(\triangle D E F)}=\frac{A B^{2}}{D E^{2}}=\frac{B C^{2}}{E F^{2}}=\frac{A C^{2}}{D F^{2}}\)

Q.5. How to find the perimeter of similar triangles?
Ans: If sides of two similar triangles have a scale factor of \(a: b\), then the ratio of their perimeters is \(a: b\).

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