Ratio of areas of two similar triangles is equal to square of their corresponding sides

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Area of Similar Triangles Theorem
Proof of Area of Similar Triangles Theorem
FAQs on Area of Similar Triangles
What Is the Ratio of Area of Similar Triangles?
What is the Relation Between Two Similar Triangles Area and Length of the Sides?
Do Similar Triangles Have Equal Areas?
How Do You Solve For Areas of Two Similar Triangles?
What Is the Areas of Similar Triangles Theorem?
How to Prove Theorem For Areas of Similar Triangles?

We will use the formula to find the area of triangle, \[{\text{Area = }}\dfrac{1}{2} \times {\text{Base}} \times {\text{Height}}\].Now, we will find the area of the triangles \[\Delta {\text{ABC}}\] and \[\Delta {\text{PQR}}\] from the above diagram.\[{\text{Area of }}\Delta {\text{PQR}} = \dfrac{1}{2} \times {\text{QR}} \times {\text{PS ......}}\left( 1 \right)\]\[{\text{Area of }}\Delta {\text{ABC}} = \dfrac{1}{2} \times {\text{BC}} \times {\text{AD ......}}\left( 2 \right)\]Dividing \[\left( 1 \right)\] by \[\left( 2 \right)\], we get\[ \dfrac{{{\text{Area of }}\Delta {\text{PQR}}}}{{{\text{Area of }}\Delta {\text{ABC}}}} = \dfrac{{\dfrac{1}{2} \times {\text{QR}} \times {\text{PS}}}}{{\dfrac{1}{2} \times {\text{BC}} \times {\text{AD}}}} \\ = \dfrac{{{\text{QR}} \times {\text{PS}}}}{{{\text{BC}} \times {\text{AD}}}}{\text{ ......}}\left( 3 \right) \\ \]Since we know that \[\angle {\text{ABC}}\] and \[\angle {\text{PQR}}\] are angles of similar triangles, so \[\angle {\text{ABC}} = \angle {\text{PQR}}\] and both right angled triangles \[\angle {\text{ADB}}\] and \[\angle {\text{PSQ}}\] are equal.Therefore, \[\Delta {\text{PQS}} \sim \Delta {\text{ABD}}\].\[ \Rightarrow \dfrac{{{\text{PS}}}}{{{\text{AD}}}} = \dfrac{{{\text{PQ}}}}{{{\text{AB}}}}{\text{ ......}}\left( 4 \right)\]Substituting this value in equation \[\left( 3 \right)\], we get\[\dfrac{{{\text{Area of }}\Delta {\text{PQR}}}}{{{\text{Area of }}\Delta {\text{ABC}}}} = \dfrac{{{\text{QR}}}}{{{\text{BC}}}} \times \dfrac{{{\text{PQ}}}}{{{\text{AB}}}}{\text{ ......}}\left( 5 \right)\]Since we know that the triangles \[\Delta {\text{PQR}}\] and \[\Delta {\text{ABC}}\] are similar,\[\dfrac{{{\text{PQ}}}}{{{\text{AB}}}} = \dfrac{{{\text{QR}}}}{{{\text{BC}}}} = \dfrac{{{\text{PR}}}}{{{\text{AC}}}}\]Using this value in equation \[\left( 5 \right)\], we get\[ \dfrac{{{\text{Area of }}\Delta {\text{PQR}}}}{{{\text{Area of }}\Delta {\text{ABC}}}} = \dfrac{{{\text{QR}}}}{{{\text{BC}}}} \times \dfrac{{{\text{QR}}}}{{{\text{BC}}}} \\ = {\left( {\dfrac{{{\text{QR}}}}{{{\text{BC}}}}} \right)^2} \\ \]Also from equation \[\left( 5 \right)\], we get\[ \dfrac{{{\text{Area of }}\Delta {\text{PQR}}}}{{{\text{Area of }}\Delta {\text{ABC}}}} = {\left( {\dfrac{{{\text{QR}}}}{{{\text{BC}}}}} \right)^2} \\ = {\left( {\dfrac{{{\text{PQ}}}}{{{\text{AB}}}}} \right)^2} \\ = {\left( {\dfrac{{{\text{RP}}}}{{{\text{CA}}}}} \right)^2} \\ \]

Thus, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Hence, proved.

Note:

In this question, students should write the sides of the triangles appropriately. Since the general area of any triangle is \[{\text{Area = }}\dfrac{1}{2} \times {\text{Base}} \times {\text{Height}}\], so we need to construct the perpendicular triangles for height. Students should know that when two triangles are similar then the ratio of their corresponding sides are same with the ratio of their corresponding altitudes and sides. The measurement of their corresponding angles is also the same.

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Consider ∆ABC ~ ∆DEF

and height of ∆ABC be h1 and ∆DEF be h2

`"AB"/"DE" = "BC"/"EF" = "AC"/"DF"` ....corresponding sides of similar triangles (i)

∠ABC = ∠DEY … corresponding angles of similar triangles(ii)

Consider ∆ABX and ∆DEY

∠AXB = ∠DYE = 90°

From equation (ii)

∠ABC = ∠DEY

∴by AA test for similarity ∆ABX ~ ∆DEY

`"AB"/"DE" = "BX"/"EY" = "AX"/"DY"`....corresponding sides of similar triangles

But from figure AX = h1 and DY = h2

`"AB"/"DE" = "BX"/"EY" = "h"_1/"h"_2`...(iii)

A(∆ABC) = (1/2)×BC×h1

A(∆DEF) = (1/2)×EF×h2

`(A(∆ABC))/(A(∆DEF)) = (AB)/(DE) xx (AB)/(DE) = (AB^2)/(DE^2)` ...(iv)

Squaring equation (i) and using it in (iv)

`(A(∆ABC))/(A(∆DEF)) = (AB^2)/(DE^2) = (BC^2)/(EF^2) = (AC^2)/(DF^2)`

Hence proved

Therefore, the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Two triangles are said to be similar when one can be obtained from the other by uniformly scaling. 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. 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 do 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 area of similar triangles in the following sections.

Area of Similar Triangles Theorem

Area of similar triangles theorem help 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 area 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 Area 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 angles are equal to 90º ]

By, AA property of similarity of triangles, 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,

Area(ΔABC)/Area(ΔDEF) = [(1/2) × BC × AP]/[(1/2) × EF × DQ] = (BC/EF) × (AP/DQ) = (BC/EF) × (BC/EF) ....[from (1)]

⇒ Area(ΔABC)/Area(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 Area of Similar Triangles.

Important Notes on Area of Similar Triangles

The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides.
For similar triangles ΔABC and ΔDEF, Area of ΔABC/Area of ΔDEF = (AB)2/(DE)2 = (BC)2/(EF)2 = (AC)2/(DF)2
All corresponding angle pairs are equal and all corresponding sides are proportional for similar triangles.

Related Topics on Area of Similar Triangles

Similar Triangles
Similar Triangles Formulas
What is Similarity?

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)/AP2 = ArΔ(DEF)/DQ2 using areas of similar triangles theorem.

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) = AB2/DE2 = AP2/DQ2 ....[from (3)]
⇒ArΔ(ABC)/AP2 = ArΔ(DEF)/DQ2
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 using the area of similar triangles theorem.

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 area of similar triangles,

Ar(ΔABC)/Ar(ΔAXY) = AB2/AX2 → (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,

AB2/AX2 = 2

⇒AB/AX = √2

⇒(AB/AX) − 1 = √2 − 1

(AB - AX) / (AX) = √2 − 1

⇒XB/AX = √2 − 1

⇒AX/XB = 1/(√2 − 1)

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FAQs on Area of Similar Triangles

The area of two similar triangles shares a relationship with the ratio of the corresponding sides of the similar triangles. According to the area 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 Area 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.

What is the Relation Between Two Similar Triangles Area and Length of the Sides?

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

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 state that "the ratio of the area 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.