# CBSE 10th Mathematics | Similar Triangles | Areas of Similar Triangles

## 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:
$\displaystyle \frac{\text{ar}\,(\Delta ABC)}{\text{ar}\,(\Delta PQR)}=\frac{A{{B}^{2}}}{P{{Q}^{2}}}=\frac{B{{C}^{2}}}{Q{{R}^{2}}}=\frac{C{{A}^{2}}}{R{{P}^{2}}}$

Construction:
Draw AD ^ BC and PS QR

Proof:    $\displaystyle \frac{\text{ar}\,(\Delta ABC)}{\text{ar}\,(\Delta PQR)}=\frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PS}$
[$\displaystyle \because$ Area of triangle $\displaystyle =\frac{1}{2}$ base ´ height]

Þ    $\displaystyle \frac{\text{ar}\,(\Delta ABC)}{\text{ar}\,(\Delta PQR)}=\frac{BC}{QR}\times \frac{AD}{PS}$                …(i)

ÐB = ÐQ            [$\displaystyle \because$ DABC ~ DPQR]

\    DADB ~ DPSQ            [By AA similarity]

Þ     $\displaystyle \frac{AD}{PS}=\frac{AB}{PQ}$                        …(ii)

[Corresponding sides of similar triangles are proportional]

But    $\displaystyle \frac{AB}{PQ}=\frac{BC}{QR}$            [$\displaystyle \because$ DABC ~ DPQR]

\    $\displaystyle \frac{AD}{PS}=\frac{BC}{QR}$            [Using (ii)]        …(iii)

From (i) and (iii), we have

$\displaystyle \frac{\text{ar}\,\,(\Delta ABC)}{\text{ar}\,\,(\Delta PQR)}=\frac{BC}{QR}\times \frac{BC}{QR}=\frac{B{{C}^{2}}}{Q{{R}^{2}}}$            …(iv)

Since     DABC ~ DPQR

\    $\displaystyle \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$                …(v)

Hence, $\displaystyle \frac{\text{ar}\,(\Delta ABC)}{\text{ar}\,(\Delta PQR)}=\frac{A{{B}^{2}}}{P{{Q}^{2}}}=\frac{B{{C}^{2}}}{Q{{R}^{2}}}=\frac{C{{A}^{2}}}{R{{P}^{2}}}$    [From (iv) and (v)]

## Solved Questions based on Areas of Similar Triangles

### Solution:

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

\    $\displaystyle \frac{\text{ar}\,\,(\Delta ABC)}{\text{ar}\,(\Delta PQR)}=\frac{B{{C}^{2}}}{Q{{R}^{2}}}$    Þ    $\displaystyle \frac{\text{64}\,\,\text{c}{{\text{m}}^{\text{2}}}}{\text{36}\,\,\text{c}{{\text{m}}^{\text{2}}}}=\frac{B{{C}^{2}}}{{{(16.5\,\,\text{cm})}^{2}}}$

Þ    $\displaystyle \frac{8\,\,cm}{6\,\,cm}=\frac{BC}{16.5\,\,cm}$    Þ    $\displaystyle BC=\frac{8\times 16.5}{6}\text{cm}=\text{22}\,\,\text{cm}$

Hence    BC = 22 cm.

### 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,

$\displaystyle \frac{\text{ar}\,\,(\Delta ALM)}{\text{ar}\,(\Delta ABC)}=\frac{A{{M}^{2}}}{A{{C}^{2}}}$

Þ    $\displaystyle \frac{\text{ar}\,\,(\Delta ALM)}{\text{ar}\,\,(\Delta ABC)}=\frac{A{{M}^{2}}}{{{(AM+MC)}^{2}}}$

Þ    $\displaystyle \frac{27\,\text{c}{{\text{m}}^{2}}}{\text{ar}\,\text{(}\Delta ABC\text{)}}=\frac{{{(3\,\,\text{cm)}}^{\text{2}}}}{{{(7\,\text{cm)}}^{\text{2}}}}$

Þ    $\displaystyle \text{ar}\,(\Delta ABC)=\left( \frac{27\times 7\times 7}{3\times 3} \right)\,\,\text{c}{{\text{m}}^{2}}$

Þ    ar (DABC) = 147 cm2

### 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.

\    $\displaystyle \frac{\text{ar}\,\,(\Delta DEF)}{\text{ar}\,\,(\Delta ABC)}=\frac{D{{E}^{2}}}{A{{B}^{2}}}=\frac{{{\left( \frac{AB}{2} \right)}^{2}}}{A{{B}^{2}}}=\frac{1}{4}$ [$\displaystyle \because$ By Midpoint Theorem $\displaystyle DE=\frac{1}{2}AB$]

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

### Solution:

We have    XY || AC                [Given]

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

Therefore,    DABC ~ DXBY                [By AA similarity]

So,        $\displaystyle \frac{\text{ar}\,(ABC)}{\text{ar}\,\,(XBY)}={{\left( \frac{AB}{XB} \right)}^{2}}$            …(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,        $\displaystyle \frac{\text{ar}\,\,(ABC)}{\text{ar}\,\,(XBY)}=\frac{2}{1}$            …(ii)

Therefore, from (i) and (ii),

$\displaystyle {{\left( \frac{AB}{XB} \right)}^{2}}=\frac{2}{1},\,\,i.e.,\,\,\frac{AB}{XB}=\frac{\sqrt{2}}{1}$

or,

$\displaystyle \frac{XB}{AB}=\frac{1}{\sqrt{2}}$

$\displaystyle \frac{AB}{AB}-\frac{AX}{AB}=\frac{1}{\sqrt{2}}$

or,

$\displaystyle 1-\frac{AX}{AB}=\frac{1}{\sqrt{2}}$

or,

$\displaystyle 1-\frac{1}{\sqrt{2}}=\frac{AX}{AB}$

$\displaystyle \frac{\sqrt{2}-1}{\sqrt{2}}=\frac{AX}{AB}$ or $\displaystyle \frac{\sqrt{2}}{\sqrt{2}}\times \frac{\sqrt{2}-1}{\sqrt{2}}$ = $\displaystyle \frac{AX}{AB}$

= $\displaystyle \frac{2-\sqrt{2}}{2}=\frac{AX}{AB}$

### Solution:

Given:     Two triangles ABC and PQR such that DABC ~ DPQR

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

To Prove: $\displaystyle \frac{\text{ar}\,(\Delta ABC)}{\text{ar}\,(\Delta PQR)}=\frac{A{{L}^{2}}}{P{{M}^{2}}}$

Proof:          DABC ~ DPQR    [Given]

Þ $\displaystyle \frac{AB}{PQ}=\frac{BC}{QR}$    [Corresponding sides of similar triangles are proportional]

Þ $\displaystyle \frac{AB}{PQ}=\frac{2BL}{2QM}$             [$\displaystyle \because$AL and PM are the medians]

Þ    $\displaystyle \frac{AB}{PQ}=\frac{BL}{QM}$                    …(i)

Now, in DABL and DPQM, we have

$\displaystyle \frac{AB}{PQ}=\frac{BL}{QM}$            [From (i)]

ÐB = ÐQ            [Corr. Ðs of similar triangles]

\    DABL ~ DPQM            [By SAS similarity]

Þ    $\displaystyle \frac{BL}{QM}=\frac{AL}{PM}$                    …(ii)

[Corresponding sides of similar triangles are proportional]

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

$\displaystyle \frac{AB}{PQ}=\frac{AL}{PM}\Rightarrow \frac{A{{B}^{2}}}{P{{Q}^{2}}}=\frac{A{{L}^{2}}}{P{{M}^{2}}}$             …(iii)

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

Therefore,    $\displaystyle \frac{\text{ar}\,(\Delta ABC)}{\text{ar}\,\text{(}\Delta PQR\text{)}}=\frac{A{{B}^{2}}}{P{{Q}^{2}}}$            …(iv)

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

$\displaystyle \frac{\text{ar}\,\,(\Delta ABC)}{\text{ar}\,\text{(}\Delta PQR\text{)}}=\frac{A{{L}^{2}}}{P{{M}^{2}}}$

### Solution:

Given:
ABCD is a square. DBCE is described on side BC is similar to DACF described on diagonal AC.

To Prove:
ar(DCBE) = $\displaystyle \frac{1}{2}$ ar(DACF)

Proof:
ABCD is a square. Therefore,

$\displaystyle AB=BC=CD=DA$ and, $\displaystyle AC=\sqrt{2}BC$ [$\displaystyle \because$ Diagonal = $\displaystyle \sqrt{2}$ (Side)]

Now,    $\displaystyle \Delta$BCE ~ DACF     [Both are equiangular hence similar by AA criteria]

Þ    $\displaystyle \frac{\text{Area}\,(\Delta BCE)}{\text{Area}\,(\Delta ACF)}=\frac{B{{C}^{2}}}{A{{C}^{2}}}$

Þ    $\displaystyle \frac{\text{Area}\,(\Delta BCE)}{\text{Area}\,\,(\Delta ACF)}=\frac{B{{C}^{2}}}{{{(\sqrt{2}BC)}^{2}}}=\frac{1}{2}$

Þ    $\displaystyle \text{Area}\,(\Delta BCE)=\frac{1}{2}\text{Area}\,(\Delta ACF)$

### Solution:

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

To find: $\displaystyle \frac{\text{Area}\,(\Delta DEF)}{\text{Area}\,\,(\Delta CFB)}.$

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]

and,     ÐAED = ÐACB                [Proved above]

\    DADE ~ DABC                    [By AAA similarity]

Þ    $\displaystyle \frac{AD}{AB}=\frac{DE}{BC}$

We have,

$\displaystyle \frac{AD}{DB}=\frac{5}{4}$

Þ    $\displaystyle \frac{DB}{AD}=\frac{4}{5}$

Þ    $\displaystyle \frac{DB}{AD}+1=\frac{4}{5}+1$                    [Adding 1 to both sides]

Þ    $\displaystyle \frac{DB+AD}{AD}=\frac{9}{5}$

Þ    $\displaystyle \frac{AB}{AD}=\frac{9}{5}$

\    $\displaystyle \frac{DE}{BC}=\frac{5}{9}$                    …(i)

In DDEF and DCFB, we have

Ð1 = Ð3                    [Alternate interior angles]

Ð2 = Ð4                    [Vertically opposite angles]

\     DDFE ~ DCFB                    [By AA similarity]

Þ    $\displaystyle \frac{\text{Area}\,\,(\Delta \,DFE)}{\text{Area}\,(\Delta CFB)}=\frac{D{{E}^{2}}}{B{{C}^{2}}}$

Þ    $\displaystyle \frac{\text{Area}\,(\Delta DFE)}{\text{Area}\,(\Delta CFB)}={{\left( \frac{5}{9} \right)}^{2}}=\frac{25}{81}$            [From (i)]

### 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

\    $\displaystyle \frac{\text{ar}\,(\Delta AOB)}{\text{ar}\,(\Delta COD)}=\frac{A{{B}^{2}}}{C{{D}^{2}}}=\frac{{{(2\,CD)}^{2}}}{C{{D}^{2}}}$            [$\displaystyle \because$ AB = 2 CD]

$\displaystyle =\frac{4\,C{{D}^{2}}}{C{{D}^{2}}}=\frac{4}{1}$

Hence, ar(AOB) : ar(DCOD) = 4 : 1

### 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    $\displaystyle \frac{AX}{DY}=\frac{AO}{DO}$                    …(i)

Now    $\displaystyle \frac{\text{Area}\,(\Delta ABC)}{\text{Area}\,\text{(}\Delta DBC\text{)}}=\frac{\frac{1}{2}BC\times AX}{\frac{1}{2}BC\times DY}=\frac{AX}{DY}$

Thus,    $\displaystyle \frac{Area\,(\Delta ABC)}{Area\,(\Delta DBC)}=\frac{AO}{DO}$                [Using (i)]

### Solution:

Given:
Let DABC and DDEF be the given triangles such that AB = AC and DE = DF, ÐA = ÐD

and,    $\displaystyle \frac{\text{Area}\,(\Delta ABC)}{\text{Area}\,(\Delta DEF)}=\frac{16}{25}$        …(i)

To Find : $\displaystyle \frac{AL}{DM}$

Construction: Draw AL BC and DM EF

Proof: Now,    AB = AC, DE = DF

Þ    $\displaystyle \frac{AB}{AC}=1$ and $\displaystyle \frac{DE}{DF}=1$

Þ    $\displaystyle \frac{AB}{AC}=\frac{DE}{DF}$

Þ    $\displaystyle \frac{AB}{DE}=\frac{AC}{DF}$

Thus, in triangles ABC and DEF, we have

$\displaystyle \frac{AB}{DE}=\frac{AC}{DF}$ and ÐA = ÐD            [Given]

So, by SAS-similarity criterion, we have

DABC ~ DDEF

Þ    $\displaystyle \frac{Area\,(\Delta ABC)}{Area\,(\Delta DEF)}=\frac{A{{L}^{2}}}{D{{M}^{2}}}$ [Ratio of areas of two similar triangles is equal to ratio of squares of their corresponding altitudes]

Þ    $\displaystyle \frac{16}{25}=\frac{A{{L}^{2}}}{D{{M}^{2}}}$                    [Using (i)]

Þ     $\displaystyle \frac{4}{5}=\frac{AL}{DM}$

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