Area of Similar Triangles

Introduction

Area of similar triangles theorem help in establishing the relationship between the areas of two similar triangles. Geometric figures having the same shape and size are known as congruent figures.

Eg: Any two circles with the same radii are congruent.

Any two rectangles with the same length and breadth are congruent.

But, geometric figures having the same shape but different sizes are known as similar figures. The congruent figures are always similar, but two similar figures need not be congruent.

Eg: Any two circles are similar. Any two rectangles are similar.

Similarity of triangles is represented by the symbol "~".

Similar Triangles

• Two triangles are comparable if their angles are equal (corresponding angles) and their sides are in equal proportions or ratios (corresponding sides).

• Similar triangles may have exclusive lengths on the sides of the triangle, but their angles must be equal and the size ratios of the corresponding sides must be equal.

• If two triangles are similar, it means that the angle pairs of the corresponding triangles are all equal. All corresponding sides of the triangle are proportional.

• Use image "~" to represent similarity.

• Therefore, if two triangles are similar, they are represented as △ QPR ~ △ XYZ.

Similar Triangular Region Sets

• Similar Triangular Region Sets help set up the relationship between two comparable triangular regions.

• It states, "The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of corresponding sides."

• Let us look at the diagram below showing two comparable triangles ΔABC and ΔDEF.

Area of similar triangle

From the location of similar triangle theorem,

$$\mathrm{Area\: of\: ΔABC/\: area\: of\: ΔDEF = \frac{(AB)^2}{(DE)^2 }=\frac{(BC)^2}{(EF)^2} =\frac{(AC)^2}{(DF)^2}}$$

Ratio of Areas of similar triangles

Statement − The ratio of the areas of two similar triangles is equal to the square of the ratio of every pair of corresponding sides.

Given − Consider two triangles ΔABC and ΔDEF such that ΔABC∼ΔDEF

Proof − $\mathrm{Area\: of\: ΔABC/\: area\: of\: ΔDEF = \frac{(AB)^2}{(DE)^2 }=\frac{(BC)^2}{(EF)^2} =\frac{(AC)^2}{(DF)^2}}$

Construction: Draw the altitudes AP and DQ on edges BC and EF respectively as shown above.

[ ∵ ΔABC ~ ΔDEF ] and

$$\mathrm{\angle APB =\angle DQE.}$$

[ ∵ AP and DQ are perpendicular to sides BC and EF respectively ⇒ each angles are equal to 90° ]

AA By the similarity of triangles, we say that ΔABP and ΔDEQ are equiangular.

So ΔABP ~ ΔDEQ

So $\mathrm{\frac{AP}{DQ} =\frac{AB}{DE}}$

further

$$\mathrm{\frac{AP}{DQ} = \frac{BC}{EF}----- (1)...[∵ ΔABC\sim ΔDEF \Rightarrow \frac{AB}{DE} = \frac{BC}{EF}]}$$

So

Area(ΔABC)/Area(ΔDEF) $\mathrm{=\frac{[(\frac{1}{2}) × BC × AP]}{[(\frac{1}{2}) × EF × DQ]}}$

$$\mathrm{= (\frac{BC}{EF}) × (\frac{AP}{DQ})= (\frac{BC}{EF}) × (\frac{BC}{EF})....[From (1)]}$$

⇒ Area(ΔABC)/Area(ΔDEF) = $\mathrm{(\frac{BC}{EF})^2}$

Similarly,

Area of ΔABC/Area of ΔDEF =$\mathrm{=\frac{(AB)^2}{(DE)^2}=\frac{(BC)^2}{(EF)^2}=\frac{(AC)^2}{( DF)^2}}$

Solved Examples

1)There are two similar triangles, ΔABC, ΔDEF. Their sides are in ratio 2:5 respectively. If Ar(ΔABC) is 28, what is Ar(ΔDEF)?

Ans. We know,

Area of ΔABC/Area of ΔDEF =$\mathrm{=\frac{(AB)^2}{(DE)^2}=\frac{(BC)^2}{(EF)^2}=\frac{(AC)^2}{( DF)^2}}$

Hence,$\mathrm{\frac{28}{Ar(ΔDEF)} = (\frac{AB}{DE})^2 = (\frac{2}{5})^2}$,

$$\mathrm{\Rightarrow \frac{28}{Ar(ΔDEF)} = \frac{4}{25}}$$

$$\mathrm{\Rightarrow Ar(ΔDEF) = 28×\frac{25}{4}=7×25=175\: sq. units}$$

2)There are two similar triangles, ΔABC, ΔDEF. Their sides are in ratio 2:3 respectively. If Ar(ΔABC) is 12, what is Ar(ΔXYZ)?

Ans. We know,

Area of ΔABC/Area of ΔXYZ =$\mathrm{=\frac{(AB)^2}{(XY)^2}=\frac{(BC)^2}{(YZ)^2}=\frac{(AC)^2}{(XZ)^2}}$

$$\mathrm{Hence, \frac{12}{Ar(ΔXYZ)} = (\frac{AB}{XY})^2 = (\frac{2}{3})^2,}$$

$$\mathrm{\Rightarrow \frac{12}{Ar(ΔDEF)} = \frac{4}{9}}$$

$$\mathrm{\Rightarrow Ar(ΔXYZ) = 12×\frac{9}{4}=3×9=27\: sq. units}$$

3)There are two similar triangles, ΔABC, ΔDEF. Their areas are 20, 50 respectively. If AC is 2, what is DF?

Ans. We know,

Area of ΔABC/Area of ΔDEF =$\mathrm{=\frac{(AB)^2}{(DE)^2}=\frac{(BC)^2}{(EF)^2}=\frac{(AC)^2}{( DF)^2}}$

$$\mathrm{Hence,\frac{20}{50}=\frac{2}{DF}; DF=5}$$

So, the length of DF is 5 units.

4)There are two similar triangles, ΔABC, ΔXYZ. Their areas are 30, 40 respectively. If AC is 3, what is XZ?

Ans. We know,

Area of ΔABC/Area of ΔXYZ =$\mathrm{=\frac{(AB)^2}{(XY)^2}=\frac{(BC)^2}{(YZ)^2}=\frac{(AC)^2}{(XZ)^2}}$

$$\mathrm{Hence, \frac{30}{40}=\frac{3}{XZ}; XZ=4}$$

So, the length of XZ is 4 units.

5)There are two similar triangles, ΔABC, ΔDEF. Their sides are in ratio 1:2 respectively. If Ar(ΔPQR) is 10, what is Ar(ΔXYZ)?

Ans. We know,

$$\mathrm{Area\: of\: ΔPQR/Area\: of\: ΔXYZ = \frac{(PQ)^2}{(XY)^2}=\frac{(QR)^2}{(YZ)^2}=\frac{(PR)^2}{(XZ)^2}}$$

$$\mathrm{Hence, \: \frac{10}{Ar(ΔXYZ)} = (\frac{PQ}{XY})^2 = (\frac{1}{2})^2,}$$

$$\mathrm{\Rightarrow \frac{10}{Ar(ΔDEF)} =\frac{1}{4}}$$

$$\mathrm{\Rightarrow Ar(ΔXYZ) = 10×\frac{4}{1}=10×4=40\: sq. units}$$

6)There are two similar triangles, ΔABC, ΔDEF. Their areas are 100, 600 respectively. If AC is 10, what is DF?

Ans. We know,

Area of ΔABC/Area of ΔDEF =$\mathrm{=\frac{(AB)^2}{(DE)^2}=\frac{(BC)^2}{(EF)^2}=\frac{(AC)^2}{( DF)^2}}$

Hence, $\mathrm{\frac{100}{600}=\frac{10}{DF}}$; DF=60

So, the length of DF is 60 units.

7)There are two similar triangles, ΔPQR, ΔXYZ. Their areas are 80, 90 respectively. If PR is 8, what is XZ?

Ans. We know,

$$\mathrm{Area\: of\: ΔPQR/Area\: of\: ΔXYZ = \frac{(PQ)^2}{(XY)^2}=\frac{(QR)^2}{(YZ)^2}=\frac{(PR)^2}{(XZ)^2}}$$

Hence, $\mathrm{\frac{80}{90}=\frac{8}{XZ}}$; XZ=9

So, the length of XZ is 9 units.

8)There are two similar triangles, ΔPQR, ΔDEF. Their areas are 30, 50 respectively. If PR is 3, what is DF?

Ans. We know,

$$\mathrm{Area\: of\: ΔPQR/Area\: of\: ΔDEF = \frac{(PQ)^2}{(DE)^2}=\frac{(QR)^2}{(EF)^2}=\frac{(PR)^2}{(DF)^2}}$$

Hence, $\mathrm{\frac{30}{50}=\frac{3}{DF}}$; DF=5

So, the length of DF is 5 units.

Conclusion

In this tutorial, we learned about similar triangles and the theorem relating the ratio of area of two similar triangles to the ratio of their corresponding sides. The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of corresponding sides. For similar triangles ΔABC and ΔDEF, location of ΔABC/ area of ΔDEF $\mathrm{=\frac{(AB)^2}{(DE)^2}=\frac{(BC)^2}{(EF)^2}=\frac{(AC)^2}{( DF)^2}}$ related angles are all equal, and all corresponding sides are proportional for similar triangles.

FAQs

1. Briefly explain what similar triangles are in your own words.

The geometric figures having the same shape, but different sizes are known as similar figures. The congruent figures are always similar, but two similar figures need not be congruent. E.g., any two circles are similar, but if their radii are different, then they are not congruent.

2. Briefly explain what the symbol is for a similar triangle.

Let’s suppose there are two triangles XYZ and GHI. Then we can represent them by ∆XYZ ~ ∆GHI.

3. Can we assume that the similar and the congruent triangles are all but the same?

The congruent figures are always similar, but two similar figures need not be congruent. E.g., any two circles are similar, but if their radii are different, then they are not congruent.

4. In the case of similar triangles, what are the three theorems that can be applied to prove that they are similar?

In the case of similar triangles, the three theorems that can be applied to prove that they are similar are −

• Angle-angle (AA)

• Side-angle-side (SAS)

• Side-side-side (SSS)

5. How can we understand the proportionality in which the sides of similar triangles are related?

The proportionality in which the sides of similar triangles are related if they have the sides P,Q,R and p,q,r respectively are −

P : p = Q : q = R : r.