**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: **

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 ….(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

….(ii)

From (i) and (ii), we get

Now,

**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

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)

⇒ AB^{2} = DE^{2}, BC^{2} = EF^{2 } and AC^{2} = DF^{2}

⇒ AB = DE, BC = EF and AC = DF

Hence, ∆ABC ≅ ∆DEF.

**Read More:**

- Angle Sum Property of a Triangle
- Median and Altitude of a Triangle
- The Angle of An Isosceles Triangle
- Area of A Triangle
- To Prove Triangles Are Congruent
- Criteria For Similarity of Triangles
- Construction of an Equilateral Triangle
- Classification of Triangles

**Areas Of Two Similar Triangles With Examples**

**Example 1: **The areas of two similar triangles ∆ABC and ∆PQR are 25 cm^{2} and 49 cm^{2} respectively. If QR = 9.8 cm, find BC.

**Sol. **It is being given that ∆ABC ~ ∆PQR,

ar (∆ABC) = 25 cm^{2} and ar (∆PQR) = 49 cm^{2}.

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

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.

[∵ AD : PS = 4 : 9]

Hence, Area (∆ABC) : Area (∆PQR) = 16 : 81

**Example 3: **If ∆ABC is similar to ∆DEF such that ∆DEF = 64 cm^{2}, DE = 5.1 cm and area of ∆ABC = 9 cm^{2}. 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.

⇒ AB = 1.912 cm

**Example 4: **If ∆ABC ~ ∆DEF such that area of ∆ABC is 16cm^{2} and the area of ∆DEF is 25cm^{2} and

BC = 2.3 cm. Find the length of EF.

**Sol.** We have,

**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 cm^{2}. 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]

[∵ AB = 2 × CD]

⇒ ar (∆COD) = 1/4 × ar (∆AOB)

Hence, the area of ∆COD is 21 cm^{2}.

**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

⇒ Area (∆BCE) = 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

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 (∆ADE) + Area (∆ADE)

= 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

⇒ AB = √2 AD AB = √2 (AB – BD)

⇒ (√2 – 1) AB = √2 BD

**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 …….(i)

Draw AL ⊥ BC and DM ⊥ EF.

Now, AB = AC, DE = DF

Thus, in triangles ABC and DEF, we have

and ∠A = ∠D [Given]

So, by SAS-similarity criterion, we have

∆ABC ~ ∆DEF

[Using (i)]

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.

Let ar (∆ADE) = 9x sq units

Then, ar (∆ABC) = 25x sq units

ar (trap. BCED) = ar (∆ABC) – ar (∆ADE)

= (25x – 9x) = (16x) sq units