CBSE Tutorials

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:
DABC and DPQR such that DABC ~ DPQR.

To prove:

Construction:
Draw AD ^ BC and PS ^ QR

Proof:    
[ Area of triangle base ´ height]

Þ                    …(i)

In DADB and DPSQ

ÐB = ÐQ            [ DABC ~ DPQR]

ÐADB = ÐPSQ        [Both 90°]

\    DADB ~ DPSQ            [By AA similarity]

Þ                             …(ii)

[Corresponding sides of similar triangles are proportional]

But                [ DABC ~ DPQR]

\                [Using (ii)]        …(iii)

From (i) and (iii), we have

            …(iv)

Since     DABC ~ DPQR

\                    …(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 cm² and 36 cm² 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(DALM) = 27 cm², calculate the ar(DABC).

Solution:

Given:     LM || BC AM = 3 cm, MC = 4 cm and ar(DALM) = 27 cm²

To Find: ar(DABC)

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 (DABC) = 147 cm²

Question:

D, E, F are the midpoints of the sides BC, CA and AB respectively of DABC. Determine the ratio of the areas of DDEF and DABC.

Solution:

Given: D, E and F are the midpoints of the sides BC,
CA and AB respectively of DABC.

To find:
Ratio of the areas of DDEF and DABC

Proof:
Since D and E are the midpoints of the sides BC and CA respectively of DABC

Therefore, DE || BA
Þ
DE || BF            …(i)

Since F and E are the midpoints of AB and AC respectively of DABC.

Therefore, FE || BC
Þ FE || BD            …(ii)

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

\    ÐB = ÐDEF                    …(iii)

[Opposite angles of a parallelogram BDEF]

Similarly AFDE is a parallelogram

\    ÐA = ÐFDE                    …(iv)

[Opposite angles of a parallelogram BDEF]

In DABC and DDEF

ÐB = ÐDEF                    [From (iii)]

ÐA = ÐFDE                    [From (iv)]

\    DABC ~ DDEF                    [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 DABC and it divides the triangle into two parts of equal areas. Find the ratio

Solution:

We have    XY || AC                [Given]

So,        ÐBXY = ÐA and ÐBYX = ÐC        [Corresponding angles]

Therefore,    DABC ~ DXBY                [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 DABC ~ DPQR

AL and PM are the medians of DABC and DPQR respectively.

To Prove:

Proof:          DABC ~ DPQR    [Given]

Þ     [Corresponding sides of similar triangles are proportional]

Þ             [AL and PM are the medians]

Þ                        …(i)

Now, in DABL and DPQM, we have

            [From (i)]

ÐB = ÐQ            [Corr. Ðs of similar triangles]

\    DABL ~ DPQM            [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. DBCE is described on side BC is similar to DACF described on diagonal AC.

To Prove:
ar(DCBE) = ar(DACF)

Proof:
ABCD is a square. Therefore,

and, [ Diagonal = (Side)]

Now,    BCE ~ DACF     [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 DABC,

Since DE || BC                    [Given]

Þ    ÐADE = ÐABC and ÐAED = ÐACB        [Corresponding angles]

In triangles ADE and ABC, we have

ÐA = ÐA                    [Common]

ÐADE = ÐABC                [Proved above]

and,     ÐAED = ÐACB                [Proved above]

\    DADE ~ DABC                    [By AAA similarity]

Þ    

We have,

Þ    

Þ                        [Adding 1 to both sides]

Þ    

Þ    

\                        …(i)

In DDEF and DCFB, we have

Ð1 = Ð3                    [Alternate interior angles]

Ð2 = Ð4                    [Vertically opposite angles]

\     DDFE ~ DCFB                    [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 DAOB and DCOD,

ÐAOB = ÐCOD                [Vertically opposite angles]

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

\    DAOB ~ DCOD                [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(DCOD) = 4 : 1

Question:

In figure, prove that

Solution:

Construction: Draw AX and DY ^ BC

Proof:
In DAOX and DDOY, we have

ÐAXO = ÐDYO                [Both 90°]

ÐAOX = ÐDOY                 [Vert. Opp. angles]

\    DAOX ~ DDOY                [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 DABC and DDEF be the given triangles such that AB = AC and DE = DF, ÐA = ÐD

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 ÐA = ÐD            [Given]

So, by SAS-similarity criterion, we have

DABC ~ DDEF

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

Þ                        [Using (i)]

Þ