# Areas Of Two Similar Triangles

## Areas Of Two Similar Triangles

Theorem 1:  The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides.
Given: Two triangles ABC and DEF such that ∆ABC ~ ∆DEF.
To Prove:  $$\frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=\frac{A{{B}^{2}}}{D{{E}^{2}}}=\frac{B{{C}^{2}}}{E{{F}^{2}}}=\frac{A{{C}^{2}}}{D{{F}^{2}}}$$

Construction: Draw AL ⊥ BC and DM ⊥ EF.
Proof: Since similar triangles are equiangular and their corresponding sides are proportional. Therefore,
∆ABC ~ ∆DEF
⇒  ∠A = ∠D,   ∠B = ∠E,   ∠C = ∠F   and   $$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$$               ….(i)
Thus, in ∆ALB and ∆DME, we have
⇒ ∠ALB = ∠DME            [Each equal to 90º]
and, ∠B = ∠E                 [From (i)]
So, by AA-criterion of similarity, we have
∆ALB ~ ∆DME
$$\Rightarrow \frac{AL}{DM}=\frac{AB}{DE}$$             ….(ii)
From (i) and (ii), we get
$$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{AL}{DM}$$
Now,
$$\Rightarrow \frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=\frac{\frac{1}{2}(BC\times AL)}{\frac{1}{2}(EF\times DM)}$$
$$\Rightarrow \frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=\frac{BC}{EF}\times \frac{AL}{DM}$$
$$\Rightarrow \frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=\frac{BC}{EF}\times \frac{BC}{EF}\text{ }\left[ From\ (iii),\ \frac{BC}{EF}=\frac{AL}{DM} \right]$$
$$\Rightarrow \frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=\frac{B{{C}^{2}}}{E{{F}^{2}}}$$
$$But\frac{BC}{EF}=\frac{AB}{DE}=\frac{AC}{DF}$$
$$\Rightarrow \frac{B{{C}^{2}}}{E{{F}^{2}}}=\frac{A{{B}^{2}}}{D{{E}^{2}}}=\frac{A{{C}^{2}}}{D{{F}^{2}}}$$
$$Hence,\frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=\frac{A{{B}^{2}}}{D{{E}^{2}}}=\frac{B{{C}^{2}}}{E{{F}^{2}}}=\frac{A{{C}^{2}}}{D{{F}^{2}}}$$

Theorem 2:  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 ABC and DEF such that
∆ABC ~ ∆DEF and Area (∆ABC) = Area (∆DEF).
To Prove: We have,
∆ABC ≅ ∆DEF
Proof: ∆ABC ~ ∆DEF
∠A = ∠D,   ∠B = ∠E,   ∠C = ∠F   and   $$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$$
In order to prove that ∆ABC ≅ ∆DEF, it is sufficient to show that AB = DE, BC = EF and AC = DF.
Now, Area (∆ABC) = Area (∆DEF)
$$\Rightarrow \frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=1$$
$$\Rightarrow \frac{A{{B}^{2}}}{D{{E}^{2}}}=\frac{B{{C}^{2}}}{E{{F}^{2}}}=\frac{A{{C}^{2}}}{D{{F}^{2}}}=1\text{ }\left[ \because \ \frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=\frac{A{{B}^{2}}}{D{{E}^{2}}}=\frac{B{{C}^{2}}}{E{{F}^{2}}}=\frac{A{{C}^{2}}}{D{{F}^{2}}} \right]$$
⇒  AB2 = DE2,    BC2 = EF2   and   AC2 = DF2
⇒  AB = DE,   BC = EF   and   AC = DF
Hence, ∆ABC ≅ ∆DEF.

## Areas Of Two Similar Triangles With Examples

Example 1:    The areas of two similar triangles ∆ABC and ∆PQR are 25 cm2 and 49 cm2 respectively. If QR = 9.8 cm, find BC.
Sol.    It is being given that ∆ABC ~ ∆PQR,
ar (∆ABC) = 25 cm2 and ar (∆PQR) = 49 cm2.
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

$$\therefore \text{ }\frac{ar\ (\Delta ABC)}{ar\ (\Delta PQR)}=\frac{B{{C}^{2}}}{Q{{R}^{2}}}$$
$$\Rightarrow \frac{25}{49}=\frac{{{x}^{2}}}{{{(9.8)}^{2}}}$$
$$\Rightarrow {{x}^{2}}=\left( \frac{25}{49}\times 9.8\times 9.8 \right)$$
$$\Rightarrow x=\sqrt{\frac{25}{49}\times 9.8\times 9.8}=\left( \frac{5}{7}\times 9.8 \right)=\left( 5\times 1.4 \right)=7$$
Hence BC = 7 cm.

Example 2:    In two similar triangles ABC and PQR, if their corresponding altitudes AD and PS are in the ratio 4 : 9, find the ratio of the areas of ∆ABC and ∆PQR.
Sol.    Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.
$$\therefore \text{ }\frac{Area\ (\Delta ABC)}{Area\ (\Delta PQR)}=\frac{A{{D}^{2}}}{P{{S}^{2}}}$$
$$\Rightarrow \frac{Area\,(\Delta ABC)}{Area\ (\Delta PQR)}={{\left( \frac{4}{9} \right)}^{2}}=\frac{16}{81}$$          [∵  AD : PS = 4 : 9]
Hence, Area (∆ABC) : Area (∆PQR) = 16 : 81

Example 3:    If ∆ABC is similar to ∆DEF such that ∆DEF = 64 cm2, DE = 5.1 cm and area of ∆ABC = 9 cm2. Determine the area of AB.
Sol.    Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
$$\therefore \text{ }\frac{Area\ (\Delta ABC)}{Area\,\,(\Delta DEF)}=\frac{A{{B}^{2}}}{D{{E}^{2}}}$$
$$\Rightarrow \frac{9}{64}=\frac{A{{B}^{2}}}{{{(5.1)}^{2}}}$$
$$\Rightarrow AB=\sqrt{3.65}$$
⇒  AB = 1.912 cm

Example 4:    If ∆ABC ~ ∆DEF such that area of ∆ABC is 16cm2 and the area of ∆DEF is 25cm2 and
BC = 2.3 cm. Find the length of EF.
Sol.    We have,
$$\frac{\text{Area}\ \text{(}\Delta \text{ABC})}{Area\ (\Delta DEF)}=\frac{B{{C}^{2}}}{E{{F}^{2}}}$$
$$\Rightarrow \frac{16}{25}=\frac{{{(2.3)}^{2}}}{E{{F}^{2}}}$$
$$\Rightarrow EF=\sqrt{8.265}~~=\text{ }2.875\text{ }cm$$

Example 5:    In a trapezium ABCD, O is the point of intersection of AC and BD, AB || CD and AB = 2 × CD. If the area of ∆AOB = 84 cm2. Find the area of ∆COD.
Sol.    In ∆AOB and ∆COD, we have

∠OAB = ∠OCD         (alt. int. ∠s)
∠OBA = ∠ODC         (alt. int. ∠s)
∴ ∆AOB ~ ∆COD      [By AA-similarity]
$$\Rightarrow \frac{ar\ (\Delta AOB)}{ar\ (\Delta COD)}=\frac{A{{B}^{2}}}{C{{D}^{2}}}=\frac{{{(2CD)}^{2}}}{C{{D}^{2}}}$$        [∵  AB = 2 × CD]
$$\Rightarrow \frac{4\times C{{D}^{2}}}{C{{D}^{2}}}=4$$
⇒  ar (∆COD) = 1/4 × ar (∆AOB)
$$\Rightarrow \left( \frac{1}{4}\times 84 \right)c{{m}^{2}}=21c{{m}^{2}}$$
Hence, the area of ∆COD is 21 cm2.

Example 6:    Prove that the area of the triangle BCE described on one side BC of a square ABCD as base is one half the area of the similar triangle ACF described on the diagonal AC as base.
Sol.    ABCD is a square. ∆BCE is described on side BC is similar to ∆ACF described on diagonal AC.
Since ABCD is a square. Therefore,
AB = BC = CD = DA and, AC = √2 BC           [∵ Diagonal = √2 (Side)]

Now, ∆BCE ~ ∆ACF
$$\Rightarrow \frac{Area\ (\Delta BCE)}{Area\ (\Delta ACF)}=\frac{B{{C}^{2}}}{A{{C}^{2}}}$$
$$\Rightarrow \frac{Area\ (\Delta BCE)}{Area\ (\Delta ACF)}=\frac{B{{C}^{2}}}{{{(\sqrt{2}BC)}^{2}}}=\frac{1}{2}$$
⇒  Area (∆BCE) = $$\frac { 1 }{ 2 }$$  Area (∆ACF)

Example 7:    D, E, F are the mid-point of the sides BC, CA and AB respectively of a ABC. Determine the ratio of the areas of ∆DEF and ∆ABC.
Sol.    Since D and E are the mid-points of the sides BC and AB respectively of ∆ABC. Therefore,
DE || BA
DE || FA                ….(i)

Since D and F are mid-points of the sides BC and AB respectively of ∆ABC. Therefore,
DF || CA   ⇒   DF || AE
From (i), and (ii), we conclude that AFDE is a parallelogram.
Similarly, BDEF is a parallelogram.
Now, in ∆DEF and ∆ABC, we have
∠FDE = ∠A     [Opposite angles of parallelogram AFDE]
and, ∠DEF = ∠B       [Opposite angles of parallelogram BDEF]
So, by AA-similarity criterion, we have
∆DEF ~ ∆ABC
$$\Rightarrow \frac{Area\ (\Delta DEF)}{Area\ (\Delta ABC)}=\frac{D{{E}^{2}}}{A{{B}^{2}}}=\frac{{{(1/2AB)}^{2}}}{A{{B}^{2}}}=\frac{1}{4}\text{ }\left[ \because \ DE\ =\frac{1}{2}AB \right]$$
Hence, Area (DDEF) : Area (DABC) = 1 : 4.

Example 8:    D and E are points on the sides AB and AC respectively of a ∆ABC such that DE || BC and divides ∆ABC into two parts, equal in area. Find .
Sol.    We have,
Area (∆ADE) = Area (trapezium BCED)
= Area (trapezium BCED) + Area (∆ADE)
⇒ 2 Area (∆ADE) = Area (∆ABC)

In ∆ADE and ∆ABC, we have
∠ADE = ∠B                [∵  DE || BC ∴  ∠ADE = ∠B (Corresponding angles)]
and, ∠A = ∠A             [Common]
∴  ∆ADE ~ ∆ABC
$$\Rightarrow \frac{Area\ (\Delta ADE)}{Area\ (\Delta ABC)}=\frac{A{{D}^{2}}}{A{{B}^{2}}}$$
$$\Rightarrow \frac{Area\ (\Delta ADE)}{2\,Area\,(\Delta ADE)}=\frac{A{{D}^{2}}}{A{{B}^{2}}}$$
$$\Rightarrow \frac{1}{2}={{\left( \frac{AD}{AB} \right)}^{2}}\Rightarrow \frac{AD}{AB}=\frac{1}{\sqrt{2}}$$
⇒ AB = √2 AD  AB = √2 (AB – BD)
⇒ (√2 – 1) AB = √2 BD
$$\Rightarrow \frac{BD}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{2-\sqrt{2}}{2}$$

Example 9:    Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. Find the ratio of their corresponding heights.
Sol.    Let ∆ABC and ∆DEF be the given triangles such that AB = AC and DE = DF, ∠A = ∠D.
and  $$\frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=\frac{16}{25}$$    …….(i)
Draw AL ⊥ BC and DM ⊥ EF.
Now, AB = AC, DE = DF
$$\Rightarrow \frac{AB}{AC}=1\text{ and }\frac{DE}{DF}=1$$

$$\Rightarrow \frac{AB}{AC}=\frac{DE}{DF}\text{ }\Rightarrow \text{ }\frac{AB}{DE}=\frac{AC}{DF}$$
Thus, in triangles ABC and DEF, we have
$$\frac{AB}{DE}=\frac{AC}{DF}\text{ and }$$       and ∠A = ∠D    [Given]
So, by SAS-similarity criterion, we have
∆ABC ~ ∆DEF
$$\Rightarrow \frac{Area\ (\Delta ABC)}{Area\ (\Delta DEF)}=\frac{A{{L}^{2}}}{D{{M}^{2}}}$$
$$\Rightarrow \frac{16}{25}=\frac{A{{L}^{2}}}{D{{M}^{2}}}$$           [Using (i)]
$$\frac{AL}{DM}=\frac{4}{5}$$
AL : DM = 4 : 5

Example 10:    In the given figure, DE || BC and DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED.

Sol.    ∆ADE ~ ∆ABC.
$$\therefore \frac{ar(\Delta ADE)}{ar(\Delta ABC)}=\frac{D{{E}^{2}}}{B{{C}^{2}}}={{\left( \frac{DE}{BC} \right)}^{2}}={{\left( \frac{3}{5} \right)}^{2}}=\frac{9}{25}$$
Let ar (∆ADE) = 9x sq units
Then, ar (∆ABC) = 25x sq units
ar (trap. BCED) = ar (∆ABC) – ar (∆ADE)
= (25x – 9x) = (16x) sq units
$$\therefore \frac{ar(\Delta ADE)}{ar(trap.BCED)}=\frac{9x}{16x}=\frac{9}{16}$$