## Similar Triangles | Areas of Similar Triangles

**Theorem:
**

**The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
**

**Given:**

D*ABC *and D*PQR *such that D*ABC *~ D*PQR*.

**To prove:**

**Construction:***
*Draw

*AD*^

*BC*and

*PS*^

*QR*

**Proof:*** **
*[ Area of triangle base ´ height]

Þ …(i)

In D*ADB* and D*PSQ
*

Ð*B* = Ð*Q* [ D*ABC* ~ D*PQR*]

Ð*ADB* = Ð*PSQ* [Both 90°]

\ D*ADB* ~ D*PSQ* [By *AA* similarity]

Þ …(ii)

[Corresponding sides of similar triangles are proportional]

But [ D*ABC* ~ D*PQR*]

\ [Using (ii)] …(iii)

From (i) and (iii), we have

…(iv)

Since D*ABC* ~ D*PQR
*

\ …(v)

Hence, [From (iv) and (v)]

## Solved Questions based on Areas of Similar Triangles

### Question:

#### The areas of two similar triangles *ABC *and *PQR *are 64 cm^{2} and 36 cm^{2} respectively. If *QR *= 16.5 cm, find *BC*.

### Solution:

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

\ Þ

Þ Þ

Hence *BC* = 22 cm.

### Question:

#### In the given figure, *LM *|| *BC*. *AM *= 3 cm, *MC *= 4 cm. If the ar(D*ALM*) = 27 cm^{2}, calculate the ar(D*ABC*).

### Solution:

**Given:** *LM* || *BC AM *= 3 cm, *MC *= 4 cm and ar(D*ALM*) = 27 cm^{2}

**To Find: ** ar(D*ABC*)

**Proof: **ÐALM = ÐABC and ÐAML = ÐACB [Corresponding Ðs]

\ DALM ~ DABC [By *AA* criterion of similarity]

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Therefore,

Þ

Þ

Þ

Þ ar (D*ABC*) = 147 cm^{2}_{
}

### Question:

*D*, *E*, *F* are the midpoints of the sides *BC*, *CA *and *AB *respectively of D*ABC*. Determine the ratio of the areas of D*DEF *and D*ABC*.

### Solution:

**Given:*** D*, *E* and *F* are the midpoints of the sides *BC,*

*CA* and *AB* respectively of D*ABC*.

**To find:***
*Ratio of the areas of D

*DEF*and D

*ABC*

**Proof:***
*Since

*D*and

*E*are the midpoints of the sides

*BC*and

*CA*respectively of D

*ABC*

Therefore, *DE* || *BA*

Þ

*DE* || *BF* …(i)

Since *F* and *E* are the midpoints of *AB* and *AC* respectively of D*ABC*.

Therefore, *FE* || *BC*

Þ *FE* || *BD* …(ii)

From equations (i) and (ii), we get that *BDEF *is a parallelogram.

\ Ð*B* = Ð*DEF* …(iii)

[Opposite angles of a parallelogram *BDEF*]

Similarly *AFDE *is a parallelogram

\ Ð*A* = Ð*FDE* …(iv)

[Opposite angles of a parallelogram *BDEF*]

In D*ABC *and D*DEF*

Ð*B* = Ð*DEF* [From (iii)]

Ð*A* = Ð*FDE* [From (iv)]

\ D*ABC* ~ D*DEF* [By AA similarity]

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

\ [ By Midpoint Theorem ]

Hence, *ar*(DDEF) : *ar*(DABC) = 1 : 4

### Question:

#### In figure, the line segment *XY *is parallel to side *AC *of D*ABC *and it divides the triangle into two parts of equal areas. Find the ratio

### Solution:

We have *XY* || *AC* [Given]

So, Ð*BXY *= Ð*A* and Ð*BYX *= Ð*C* [Corresponding angles]

Therefore, D*ABC* ~ D*XBY* [By AA similarity]

So, …(i)

[The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides]

ar(*BXY*) = ar(*ACYX*) [Given]

Adding ar(*BXY*) to both sides we have

ar(*ABC*) = 2ar (*XBY*)

So, …(ii)

Therefore, from (i) and (ii),

or,

or,

or,

or =

=

### Question:

#### Prove that ratio of areas of two similar triangles is the same as the ratio of the squares of their corresponding medians.

### Solution:

**Given: ** Two triangles *ABC* and *PQR* such that D*ABC* ~ D*PQR
*

*AL* and *PM* are the medians of D*ABC* and D*PQR* respectively.

**To Prove:**

**Proof: ** D*ABC* ~ D*PQR* [Given]

Þ [Corresponding sides of similar triangles are proportional]

Þ [*AL* and *PM* are the medians]

Þ …(i)

Now, in D*ABL* and D*PQM*, we have

[From (i)]

Ð*B* = Ð*Q* [Corr. Ðs of similar triangles]

\ D*ABL* ~ D*PQM* [By *SAS* similarity]

Þ …(ii)

[Corresponding sides of similar triangles are proportional]

From equation (i) and equation (ii), we have

…(iii)

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Therefore, …(iv)

From equation (iii) and equation (iv), we have

### Question:

#### Prove that the area of the equilateral triangle *BCE* described on one side *BC* of a square *ABCD* as base is half the area of the equilateral triangle *ACF* described on the diagonal *AC* as base.

### Solution:

**Given:**

*ABCD* is a square. D*BCE* is described on side *BC* is similar to D*ACF* described on diagonal *AC*.

**To Prove:***
*ar(D

*CBE*) = ar(D

*ACF*)

**Proof:**

*ABCD* is a square. Therefore,

and, [ Diagonal = (Side)]

Now, *BCE* ~ D*ACF* [Both are equiangular hence similar by AA criteria]

Þ

Þ

Þ

### Question:

#### In figure, *DE* || *BC* and *AD* : *DB* = 5 : 4. Find

### Solution:

**Given:**

*DE* || *BC* and *AD* : *DB* = 5 : 4

**To find:**

**Proof:** In D*ABC*,

Since *DE* || *BC *[Given]

Þ Ð*ADE* = Ð*ABC* and Ð*AED* = Ð*ACB* [Corresponding angles]

In triangles *ADE* and *ABC*, we have

Ð*A* = Ð*A* [Common]

Ð*ADE* = Ð*ABC* [Proved above]

and, Ð*AED* = Ð*ACB* [Proved above]

\ D*ADE* ~ D*ABC* [By *AAA* similarity]

Þ

We have,

Þ

Þ [Adding 1 to both sides]

Þ

Þ

\ …(i)

In D*DEF* and D*CFB*, we have

Ð1 = Ð3 [Alternate interior angles]

Ð2 = Ð4 [Vertically opposite angles]

\ D*DFE* ~ D*CFB* [By AA similarity]

Þ

Þ [From (i)]

### Question:

#### In the given figure, *ABCD* is a trapezium in which *AB* || *DC* and *AB* = 2 *CD*. Find the ratio of the areas of triangles *AOB *and *COD*.

### Solution:

**Given:*** ABCD* is a trapezium in which *AB* || *DC* and *AB* = 2 *CD *

**To Find:** Ratio of the areas of triangles *AOB *and *COD*

**Proof:** In D*AOB *and D*COD*,

Ð*AOB = ÐCOD *[Vertically opposite angles]

ÐOAB = ÐOCD [Alternate angles as *AB* || *DC*]

\ D*AOB *~ D*COD *[By *AA* similarity]

Since the ratio of the areas of two similar triangles is equal to the ratio of squares of any two corresponding sides

\ [ *AB* = 2 *CD*]

Hence, ar(*AOB*) : ar(D*COD*) = 4 : 1

### Question:

#### In figure, prove that

### Solution:

**Construction:** Draw *AX* and *DY *^ *BC*

**Proof:***
*In D

*AOX*and D

*DOY*, we have

Ð*AXO* = Ð*DYO* [Both 90°]

Ð*AOX* = Ð*DOY* [Vert. Opp. angles]

\ D*AOX* ~ D*DOY* [By AA similarity]

Þ …(i)

Now

Thus, [Using (i)]

### Question:

#### Two isosceles triangles have equal vertical angles and their area are in the ratio 16 : 25. Find the ratio of their corresponding heights.

### Solution:

**Given:***
*Let D

*ABC*and D

*DEF*be the given triangles such that

*AB*=

*AC*and

*DE*=

*DF*, Ð

*A*= Ð

*D*

and, …(i)

**To Find :**

**Construction:** Draw *AL *^ *BC* and *DM *^ *EF*

**Proof:** Now, *AB* = *AC*, *DE* = *DF*

Þ and

Þ

Þ

Thus, in triangles *ABC* and *DEF*, we have

and Ð*A* = Ð*D* [Given]

So, by *SAS*-similarity criterion, we have

D*ABC* ~ D*DEF*

Þ [Ratio of areas of two similar triangles is equal to ratio of squares of their corresponding altitudes]

Þ [Using (i)]

Þ