Areas of Similar Triangles
Two triangles are said to be similar when one can be obtained from the other by uniformly scaling. If two triangles are similar it means that: All corresponding angle pairs are equal and all corresponding sides are proportional. However, in order to be sure that the two triangles are similar, we do not necessarily need to have information about all sides and all angles. For similar triangles, not only their angles and sides share a relationship, but also the ratio of their perimeter, altitudes, angle bisectors, areas, and other aspects are in proportion. Let us study and understand the relation between the areas of similar triangles in the following sections.
1.  Areas of Similar Triangles Theorem 
2.  Proof of Areas of Similar Triangles Theorem 
3.  FAQs on Areas of Similar Triangles 
Areas of Similar Triangles Theorem
Area of similar triangle theorem helps in establishing the relationship between the areas of two similar triangles. It states that "The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides". Consider the following figure, which shows two similar triangles, ΔABC and ΔDEF.
According to the theorem for areas of similar triangles,
Area of ΔABC/Area of ΔDEF = (AB)^{2}/(DE)^{2} =(BC)^{2}/(EF)^{2} = (AC)^{2}/(DF)^{2}
We will understand the proof of this theorem in the next section.
Proof of Areas of Similar Triangles Theorem
Statement: The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides.
Given: Consider two triangles, ΔABC and ΔDEF, such that ΔABC∼ΔDEF
To prove: Area of ΔABC/Area of ΔDEF = (AB)^{2}/(DE)^{2} = (BC)^{2}/(EF)^{2} = (AC)^{2}/(DF)^{2 }
Construction: Draw the altitudes AP and DQ to the sides BC and EF respectively, as shown below:
Proof: Since, ∠B = ∠E, .[ ∵ ΔABC∼ΔDEF ] and,
∠APB = ∠DQE.....[ ∵ AP and DQ are perpendicular on sides BC and EF respectively. ⇒ Both equal to 90º ]
By, AA property of similarity, we can note that ΔABP and ΔDEQ are equiangular.
Hence, ΔABP∼ΔDEQ
Thus, AP/DQ = AB/DE
This further implies that,
AP/DQ = BC/EF → (1)....[ ∵ ΔABC∼ΔDEF ⇒ AB/DE = BC/EF]
Thus,
Ar(ΔABC)/Ar(ΔDEF) = [(1/2) × BC × AP]/[(1/2) × EF × DQ]
=(BC/EF) × (AP/DQ)
= (BC/EF) × (BC/EF)....[from (1)]
⇒Ar(ΔABC)/Ar(ΔDEF) = (BC/EF)^{2}
Similarly, we can show that,
Area of ΔABC/Area of ΔDEF =(AB)^{2}/(DE)^{2} = (BC)^{2}/(EF)^{2} = (AC)^{2}/(DF)^{2}
Challenging Question:
It is given that ΔABC∼ΔXYZ. The area of ΔABC is 45 sq units and the area of ΔXYZ is 80 sq units. YZ=12 units. Find BC? Hint: Use Theorem for Areas of Similar Triangles.
Solved Examples on Areas of Similar Triangles

Example 1: Consider two similar triangles, ΔABC and ΔDEF, as shown below:
AP and DQ are medians in the two triangles. Show that
ArΔ(ABC)/AP^{2 }= ArΔ(DEF)/DQ^{2}
Solution: Since ΔABC∼ΔDEF,
AB/DE = BC/EF
⇒AB/DE = (1/2)BC/(1/2)EF
⇒AB/DE = BP/EQ →(1)
Also,
∠B = ∠E →(2)...[ ∵ ΔABC∼ΔDEF]
From (1) and (2) and by SAS similarity criterion, We can note that,
ΔABP∼ΔDEQ
⇒AB/DE = AP/DQ →(3)
Now, by theorem for areas of similar triangles,
ArΔ(ABC)/ArΔ(DEF) = AB^{2}/DE^{2} = AP^{2}/DQ^{2}....[from (3)]
⇒ArΔ(ABC)/AP^{2} = ArΔ(DEF)/DQ^{2} 
Example 2: Consider the following figure:
It is given that XY∥BC and divides the triangle into two parts of equal areas. Find the ratio AX:XB.
Solution: Since XY∥BC, ∠X = ∠B and ∠Y = ∠C...[Corresponding angles]
⇒ΔAXY must be similar to ΔABC...[By AA similarity criterion in triangles]Now, by theorem for areas of similar triangles,
Ar(ΔABC)/Ar(ΔAXY) = AB^{2}/AX^{2} → (1)
Also, XY divides the triangle into two parts of equal areas. Thus,
Ar(ΔABC)/Ar(ΔAXY) = 2 → (2)
From (1) and (2), we have,
AB^{2}/AX^{2 }= 2
⇒AB/AX = √2
⇒(AB/AX) − 1 = √2 − 1
(AB  AX) / (AX) = √2 − 1
⇒XB/AX = √2 − 1
⇒AX/XB = 1/(√2 − 1)
FAQs on Areas of Similar Triangles
What Is the Area of Two Similar Triangles?
The areas of two similar triangles share a relationship with the ratio of the corresponding sides of the similar triangles. According to the areas of similar triangles theorem, we can state that "the ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides".
What Is the Ratio of Areas of Similar Triangles?
The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles. For example, for any two similar triangles ΔABC and ΔDEF,
Area of ΔABC/Area of ΔDEF = (AB)^{2}/(DE)^{2} = (BC)^{2}/(EF)^{2} = (AC)^{2}(DF)^{2}.
Do Similar Triangles Have Equal Areas?
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 be necessarily equal. But note that congruent triangles always have equal areas.
How Do You Solve For Areas of Two Similar Triangles?
Areas of similar triangles can be solved by relating their ratio with the ratio of the pair of corresponding sides. For any two similar triangles, the ratio of the areas is equal to the square of the ratio of corresponding sides.
What Is the Areas of Similar Triangles Theorem?
The areas of similar triangles theorem states that "the ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides"
How to Prove Theorem For Areas of Similar Triangles?
The theorem for the areas of similar triangles can be proved by constructing altitudes for both triangles and comparing the area thus obtained with the ratio of corresponding sides of both the similar triangles. To understand the proof in detail, refer to section Proof of Areas of Similar Triangles Theorem of this page.