Area of Similar Triangles: Theorems, Formulas and Examples

Similar figures are geometric figures that have the same shape but different sizes. If two triangles' corresponding angles are equal and their corresponding sides are proportional, they are said to be similar. The two conditions specified in the preceding definition are self-contained. If one of the two conditions is met, the other is met automatically. As a result, either of the two conditions can be used to define comparable triangles. The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the similar triangles' corresponding sides. If two triangles are similar, it means that all of their corresponding angle pairs are equal and all of their corresponding sides are proportional. Let’s discuss the area of similar triangles, theorems along with some important questions.

Key takeaways: Similar Shapes, Similar Triangles, Congruency, Similar Triangle Theorems

Read More: Bayes Theorem Formula

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Similar Triangles

If two triangles are similar, Their respective angles are equal, and Their respective sides are proportional. The two conditions specified in the preceding definition are self-contained. If one of the two conditions is met, the other is met automatically. As a result, either of the two conditions can be used to define comparable triangles. Equiangular triangles are formed when the corresponding angles of two triangles are equal.

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Similar Triangles Theorem 1

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

Given: Two triangles ΔABCΔABC and ΔDEFΔDEF such that ΔABC∼ΔDEFΔABC∼ΔDEF.

To prove: Area(ΔABC)/Area(ΔDEF)=AB²/DE² =BC²/EF² =AC²/DF²

Construction: Draw AL⊥BC AL⊥BC and DM⊥EF

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

Therefore, ΔABC∼ΔDEF

⇒∠A=∠D,∠B=∠E,∠C=∠F and ABDE=BCFE=ACDF……(i)

Thus, in ΔALB and ΔDME, we have

⇒∠ALB=∠DME (Each equal to 90^o)

and, ∠B=∠E (From (i))

So, by AA criterion of similarity, we have

ΔALB∼ΔDME

⇒ALDM=ABDE……(ii)

From (i) and (ii), we get

ABDE=BCFE=ACDF=ALDM……(iii)

Now, Area(ΔABC)/Area(ΔDEF)=12(BC×AL)12(EF×DM)

⇒Area(ΔABC)/Area(ΔDEF)=BCEF×ALDM

⇒ Area (ΔABC)/Area(ΔDEF)=BCEF×BCEF (From (iii))

⇒Area(ΔABC)/Area(ΔDEF)=BC²/EF²

But, BCEF=ABDE=ACDF

⇒BC²/EF²=AB2/DE2=AC²/DF²

Hence, Area(ΔABC)/Area(ΔDEF)=BC²/EF² =AB²/DE² =AC² /DF²

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Similar Triangles Theorem 2

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

Given: Two triangles ΔABC and ΔDEF such that ΔABC∼ΔDEF and AL⊥BC and DM⊥EF

To prove: Area(ΔABC)/ Area (ΔDEF)=AL²DM²

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

Therefore, Area(ΔABC)/Area(ΔDEF)=AB²DE²……(i)

Now, in ΔALB and ΔDME, we have

⇒∠ALB=∠DME[ Each equal to 90⁰]

and, ∠B=∠E[ΔABC∼ΔDEF∴∠A=∠D,∠B=∠E,∠C=∠F]

So, by AA criterion of similarity, we have

ΔALB∼ΔDME

⇒ABDE=ALDM

⇒AB²/DE²=AL²/DM²……(ii)

From (i) and (ii), we get

Area(ΔABC)/Area(ΔDEF)=AL²/DM²

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Similar Triangles Theorem 3

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

Given: Two triangles ΔABC and ΔDEF such that ΔABC∼ΔDEF and AP, DQ are their medians.

To prove: Area(ΔABC) / Area (ΔDEF)=AP²/DQ²

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, Area (ΔABC)/Area(ΔDEF)=AB²/DE²……(i)

Now, ΔABC∼ΔDEF

⇒ABDE=BCEF

⇒ABDE=2BP/2EQ=BPEQ…… (ii)

Thus, in triangles ΔAPB and ΔDQE, we have

ABDE=BPEQ and ∠B=∠E[ΔABC∼ΔDEF]

So, by the SAS criterion of similarity, we have

ΔAPB∼ΔDQE

⇒BPEQ=APDQΔ (iii)

From (ii) and (iii), we get

ABDE=APDQ

⇒AB²/DE²=AP²/DQ²…… (iv)

From (i) and (iv), we get

Area (ΔABC)/Area (ΔDEF) =AP²/DQ²

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Similar Triangles 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 ΔABCΔABC and ΔDEFΔDEF such that ΔABC∼ΔDEFΔABC∼ΔDEF and AXAX and DYDY are bisectors of ∠A∠A and ∠D∠D respectively.

To prove: Area(ΔABC)/Area(ΔDEF)=AX²/DY²

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, Area(ΔABC)/Area(ΔDEF)=AB²/DE²……(i)

Now, ΔABC∼ΔDEF

⇒∠A=∠D

⇒12∠A=12∠D

⇒∠BAX=∠EDY

Thus, in triangles ABX and DEY, we have

∠BAX=∠EDY and ∠B=∠E[ΔABC∼ΔDEF]

So, by AA similarity criterion, we have

ΔABX∼ΔDEY

⇒ABDE=AXDY

⇒AB²/DE²=AX²/DY²……(ii)

From (i) and (ii), we get

Area(ΔABC)/Area(ΔDEF)=AX²/DY²

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Similar Triangles 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 ΔABC and ΔDEF such that ΔABC∼ΔDEF and Area (ΔABC)= Area (ΔDEF)

To prove: ΔABC≅ΔDEF

Proof: We have, ΔABC∼ΔDEF

⇒∠A=∠D,∠B=∠E,∠C=∠F and ABDE=BCFE=ACDF

To prove that ΔABC≅ΔDEF, it is sufficient to show that AB=DE,BC=EF and AC=DF

It is given that Area (ΔABC)= Area (ΔDEF)

⇒Area(ΔABC)Area(ΔDEF)=1

⇒AB²/DE²=BC²/EF²=AC²/DF²=1[ΔArea(ΔABC)/Area(ΔDEF)=AB²/DE²=BC²/EF²=AC²/DF²]

⇒AB²=DE²,BC²=EF²,AC²=DF²

⇒AB=DE, BC=EF and AC=DF

Hence, ΔABC≅ΔDEF

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Things to Remember

• The area of two similar triangles implies that if two triangles are similar to each other, the ratio of their areas will be proportional to the square of their corresponding side ratios. 
• This demonstrates that the ratio of the areas of the two similar triangles is proportional to the squares of their corresponding sides. The symbol “~” represents the similarity of triangles.
• The two comparable triangles have the same shape but may differ in size. The ratio of similar triangles' corresponding sides is the same. Each pair of similar triangles' corresponding angles is equal. 

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