# CBSE 10th Mathematics | Triangles | Pythagoras Theorem

## Triangles | Pythagoras Theorem

Theorem:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: A right triangle ABC, right angled at B

To prove:
$\displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$

Construction: Draw BD AC

Proof:
In DADB and DABC, we have

ÐADB = ÐABC                    [Each equal to 90°]

ÐA = ÐA                    [Common]

\
DADB ~ DABC                     [By AA similarity]

Þ    $\displaystyle \frac{AD}{AB}=\frac{AB}{AC}$    [Corresponding sides of similar triangles are proportional]

Þ    $\displaystyle A{{B}^{2}}=AD\times AC$                …(i)

In DBCD and DACB, we have

ÐCDB = ÐCBA                [Each equal to 90°]

ÐC = ÐC                    [Common]

By AA similarity criterion

DBCD ~ DACB

\    $\displaystyle \frac{BC}{AC}=\frac{DC}{BC}$

Þ    $\displaystyle B{{C}^{2}}=DC\times AC$                …(ii)

Adding equations (i) and (ii), we get

$\displaystyle A{{B}^{2}}+B{{C}^{2}}=AD\times AC+DC\times AC$

$\displaystyle =AC\,(AD+DC)=AC\times AC=A{{C}^{2}}$

Hence, $\displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}.$

Theorem (Converse of Pythagoras theorem):

In a triangle if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

Given: A triangle ABC such that $\displaystyle A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}.$

To prove:
$\displaystyle \angle B=90{}^\circ$

Construction:    Draw a DPQR right angled at Q such that PQ = AB and QR = BC.

Proof:    In right triangle PQR, we have

$\displaystyle P{{Q}^{2}}+Q{{R}^{2}}=P{{R}^{2}}$                    [By Pythagoras Theorem]

Þ    $\displaystyle A{{B}^{2}}+B{{C}^{2}}=P{{R}^{2}}$                    …(i)

[$\displaystyle \because$
PQ = AB and QR = BC (by construction)]

But    $\displaystyle A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$                [Given]    …(ii)

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

$\displaystyle P{{R}^{2}}=A{{C}^{2}}$

Þ     $\displaystyle PR=AC$                        …(iii)

Now, in DABC and DPQR, we have

AB = PQ and BC = QR            [By construction]

and    $\displaystyle AC=PR$                    [From (iii)]

\    DABC @
DPQR                [By SSS congruency]

\    ÐB = ÐQ = 90°                 [c.p.c.t.]

Hence, ÐB = 90°.

## Solved Questions based on Pythagoras Theorem

### Solution:

If the sum of the squares of the smaller sides of a triangle is equal to the square of the larger side, then the triangle is right angled.

(i)

(7 cm)2 + (24 cm)2 = (49 + 576) cm2 = 625 cm2 and (25 cm)2 = 625 cm2

Hence, the given triangle is a right triangle.

(ii)

(7 cm)2 + (6 cm)2 = (49 + 36) cm2 = 85 cm2 and (8 cm)2 = 64 cm2

Hence, the given triangle is not a right triangle.

### Solution:

Let AB be the ladder, B be the window and BC be the wall

Then BC = 12 m, AC = 5m and ÐACB = 90°

In right triangle ACB, we have

$\displaystyle A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}$    [By Pythagoras Theorem]

Þ    $\displaystyle A{{B}^{2}}={{(5)}^{2}}+{{(12\,)}^{2}}=(25+144)\,\,=169\,$

Þ     $\displaystyle AB=13\,\text{m}$

Hence, the length of the ladder is 13 m.

### Solution:

Let AB be the street and let C and D be the windows at heights of 9 m and 12 m respectively from the ground.

Let E be the foot of the ladder. Then EC and ED are the two positions of the ladder.

Clearly AC = 9 m, BD = 12 m, EC = ED = 15 m and ÐCAE = ÐDBE = 90°

In right triangle CAE, we have

$\displaystyle C{{E}^{2}}=A{{C}^{2}}+A{{E}^{2}}$ [By Pythagoras Theorem]

Þ    (15)2 = (9)2 + AE2

Þ    AE2 = (15)2 – (9)2 = (225 – 81) = 144

Þ    AE = 12 m                    …(i)

In right triangle DBE, we have

$\displaystyle D{{E}^{2}}=B{{D}^{2}}+E{{B}^{2}}$        [By Pythagoras Theorem]

Þ    (15)2 = (12)2 + EB2

Þ    EB2 = (15)2 – (12)2 = (225 – 144) = 81

Þ    EB = 9 m                    …(ii)

Adding equations (i) and (ii), we get

AE + EB = (12 + 9) m      Þ    AB = 21 m

Hence, the width of the street is 21 m.

### Solution:

Given:
ÐACB = 90° and CD AB

To prove: $\displaystyle \frac{B{{C}^{2}}}{A{{C}^{2}}}=\frac{BD}{AD}.$

Proof:    In DACD and DABC

ÐA = ÐA                 [Common]

ÐADC = ÐACB            [Both 90°]

\    DACD ~ DABC            [By AA similarity]

So,         $\displaystyle \frac{AC}{AB}=\frac{AD}{AC}$

or,        $\displaystyle A{{C}^{2}}=AB.AD$            …(i)

Similarly,     DBCD ~ DBAC

So,         $\displaystyle \frac{BC}{BA}=\frac{BD}{BC}$

or,        $\displaystyle B{{C}^{2}}=BA.BD$                …(ii)

Therefore, from (i) and (ii),

$\displaystyle \frac{B{{C}^{2}}}{A{{C}^{2}}}=\frac{BA.BD}{AB.AD}=\frac{BD}{AD}$

### Solution:

Given: P and Q are the midpoints of the sides CA and CB respectively of DABC, right angled at C

To Prove:

(i)    $\displaystyle 4A{{Q}^{2}}=4A{{C}^{2}}+B{{C}^{2}}$

(ii)    $\displaystyle 4B{{P}^{2}}=4B{{C}^{2}}+A{{C}^{2}}$

(iii)    $\displaystyle 4(A{{Q}^{2}}+B{{P}^{2}})=5A{{B}^{2}}$

Proof:

(i)

From right DACQ, we have

$\displaystyle A{{Q}^{2}}=A{{C}^{2}}+Q{{C}^{2}}$        [By Pythagoras Theorem]

Þ    $\displaystyle A{{Q}^{2}}=A{{C}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}}$        [$\displaystyle \because$ Q is the midpoint of CB]

Þ    $\displaystyle A{{Q}^{2}}=A{{C}^{2}}+\frac{B{{C}^{2}}}{4}$

Þ    $\displaystyle 4A{{Q}^{2}}=4A{{C}^{2}}+B{{C}^{2}}$                …(i)

(ii)

From right DBCP

$\displaystyle B{{P}^{2}}=B{{C}^{2}}+P{{C}^{2}}$        [By Pythagoras Theorem]

Þ    $\displaystyle B{{P}^{2}}=B{{C}^{2}}+{{\left( \frac{AC}{2} \right)}^{2}}$        [$\displaystyle \because$ P is the mid-point of side CA]

Þ    $\displaystyle B{{P}^{2}}=B{{C}^{2}}+\frac{A{{C}^{2}}}{4}$

Þ    $\displaystyle 4B{{P}^{2}}=4B{{C}^{2}}+A{{C}^{2}}$            … (ii)

(iii)

Adding equation (i) and equation (ii), we get

$\displaystyle 4(A{{Q}^{2}}+B{{P}^{2}})=4(A{{C}^{2}}+B{{C}^{2}})+(B{{C}^{2}}+A{{C}^{2}})$

Þ    $\displaystyle 4(A{{Q}^{2}}+B{{P}^{2}})=5\,(A{{C}^{2}}+B{{C}^{2}})$

Þ    $\displaystyle 4(A{{Q}^{2}}+B{{P}^{2}})=5\,A{{B}^{2}}$    [$\displaystyle \because$ In DABC, $\displaystyle A{{C}^{2}}+B{{C}^{2}}=A{{B}^{2}}$]

Hence, $\displaystyle 4(A{{Q}^{2}}+B{{P}^{2}})=5\,\,A{{B}^{2}}.$

### Solution:

Given: O is any point inside a rectangle ABCD

To prove:
OB2 + OD2 = OA2 + OC2

Construction:
Through O, draw PQ || BC so that P lies on AB and Q lies on DC.

Proof:    PQ || BC            [By construction]

Therefore,    PQ AB and PQ DC (ÐB = 90° and ÐC = 90°)

So,         ÐBPQ = 90° and ÐCQP = 90°

Therefore, BPQC and APQD are both rectangles.

Now, from DOPB,

$\displaystyle O{{B}^{2}}=B{{P}^{2}}+O{{P}^{2}}$                …(i)

Similarly, from DOQD,

$\displaystyle O{{D}^{2}}=O{{Q}^{2}}+D{{Q}^{2}}$                …(ii)

From DOQC, we have

$\displaystyle O{{C}^{2}}=O{{Q}^{2}}+C{{Q}^{2}}$                …(iii)

and from DOAP, we have

$\displaystyle O{{A}^{2}}=A{{P}^{2}}+O{{P}^{2}}$                …(iv)

Adding (i) and (ii),

$\displaystyle O{{B}^{2}}+O{{D}^{2}}=B{{P}^{2}}+O{{P}^{2}}+O{{Q}^{2}}+D{{Q}^{2}}$

$\displaystyle =C{{Q}^{2}}+O{{P}^{2}}+O{{Q}^{2}}+A{{P}^{2}}$

(As BP = CQ and DQ = AP)

$\displaystyle =\left( C{{Q}^{2}}+O{{Q}^{2}} \right)+\left( O{{P}^{2}}+A{{P}^{2}} \right)$

$\displaystyle =O{{C}^{2}}+O{{A}^{2}}$            [From (iii) and (iv)]

### Solution:

(a)

Taking c as the base and p as the altitude, we have

area of DABC
$\displaystyle =\frac{1}{2}pc$                …(i)

Taking b as the base and a as the altitude, we have

area DABC = $\displaystyle \frac{1}{2}ab$                 …(ii)

\    $\displaystyle \frac{1}{2}pc=\frac{1}{2}ab$        [From (i) and (ii)]

Þ     pc = ab    Hence Proved.

(b)

$\displaystyle \because$
ABC is a right triangle-angled at C.

\    $\displaystyle {{c}^{2}}={{a}^{2}}+{{b}^{2}}$        …(iii)

[By Pythagoras Theorem]

pc = ab     [proved above]

\    $\displaystyle p=\frac{ab}{c}$

or    $\displaystyle \frac{1}{p}=\frac{c}{ab}$

Þ    $\displaystyle \frac{1}{{{p}^{2}}}=\frac{{{c}^{2}}}{{{a}^{2}}{{b}^{2}}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}$        [From equation (iii)]

$\displaystyle =\frac{{{a}^{2}}}{{{a}^{2}}{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}=\frac{1}{{{b}^{2}}}+\frac{1}{{{a}^{2}}}.$

Þ    $\displaystyle \frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}.$

### Solution:

Given:
DABC in which ÐACB > 90°.

To prove: $\displaystyle A{{B}^{2}}=B{{C}^{2}}+A{{C}^{2}}+2CA\times CD.$

Construction:    Draw BD AC (produced).

Proof:     In right-triangle DBDA, we get

$\displaystyle A{{B}^{2}}=B{{D}^{2}}+A{{D}^{2}}$…(i) [By Pythagoras theorem]

$\displaystyle =B{{D}^{2}}+{{(AC+CD)}^{2}}$            [$\displaystyle \because$AD = AC + CD]

$\displaystyle =B{{D}^{2}}+A{{C}^{2}}+C{{D}^{2}}+2AC.CD$

$\displaystyle =(B{{D}^{2}}+C{{D}^{2}})+A{{C}^{2}}+2AC.CD$

[$\displaystyle \because$ In right-angled DBDC, BD2 + CD2 = BC2] [By Pythagoras theorem]

$\displaystyle =B{{C}^{2}}+A{{C}^{2}}+2AC.CD$

Hence, $\displaystyle A{{B}^{2}}=B{{C}^{2}}+A{{C}^{2}}+2CA\times CD.$

### Solution:

Given:
DABC in which ÐB < 90° and AD is perpendicular to BC

To prove: $\displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2BC.BD.$

Proof:     In right-angled DABD, we have

$\displaystyle A{{D}^{2}}+B{{D}^{2}}=A{{B}^{2}}$ [By Pythagoras theorem]…(i)

In right-angled DADC, we have

$\displaystyle A{{C}^{2}}=A{{D}^{2}}+D{{C}^{2}}$

$\displaystyle =A{{D}^{2}}+{{(BC-BD)}^{2}}$            [$\displaystyle \because$DC = BC – BD]

$\displaystyle =A{{D}^{2}}+B{{C}^{2}}+B{{D}^{2}}-2BC.BD$

$\displaystyle =(A{{D}^{2}}+B{{D}^{2}})+B{{C}^{2}}-2BC.BD$    [$\displaystyle \because$In D
ABD, AB= AD2 + BD2]

$\displaystyle =A{{B}^{2}}+B{{C}^{2}}-2BC.BD.$

Þ    $\displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2BC.BD.$

### Solution:

Given:     In DABC, AD BC and BD = 3 CD.

To prove:    $\displaystyle 2A{{B}^{2}}=2A{{C}^{2}}+B{{C}^{2}}$

Proof:         $\displaystyle BD=3CD\,$            (given)

Adding CD on both sides

$\displaystyle BD+DC=4CD$

BC = 4CD

$\displaystyle \therefore CD=\frac{BC}{4}$            …(i)

In DABD, we have,

$\displaystyle A{{B}^{2}}=B{{D}^{2}}+A{{D}^{2}}$ [Pythagoras Theorem]

$\displaystyle =A{{D}^{2}}+{{(BC-DC)}^{2}}$

Þ    $\displaystyle A{{B}^{2}}=A{{D}^{2}}+B{{C}^{2}}+D{{C}^{2}}-2BC.CD$

$\displaystyle =B{{C}^{2}}+(A{{D}^{2}}+D{{C}^{2}})-2BC.CD$

$\displaystyle =B{{C}^{2}}+A{{C}^{2}}-2BC.CD$

[$\displaystyle \because$ In DADC, $\displaystyle A{{C}^{2}}=A{{D}^{2}}+C{{D}^{2}}$]

$\displaystyle =B{{C}^{2}}+A{{C}^{2}}-\frac{2BC\times BC}{4}$        [From (i)]

$\displaystyle =B{{C}^{2}}+A{{C}^{2}}-\frac{B{{C}^{2}}}{2}$

Þ    $\displaystyle A{{B}^{2}}=A{{C}^{2}}+\frac{B{{C}^{2}}}{2}$

or    $\displaystyle 2A{{B}^{2}}=2A{{C}^{2}}+B{{C}^{2}}$

### Solution:

Given: In DABC is an equilateral triangle and D is a point on BC such that $\displaystyle BD=\frac{1}{3}BC.$

To prove:    $\displaystyle 9A{{D}^{2}}=7A{{B}^{2}}$

Construction:    Draw $\displaystyle AE\bot BC$ and join A to D.

Proof:    In DAEB and DAEC, we have

$\displaystyle AE=AE$                [Common]

ÐAEB = ÐAEC            [Each = 90°]

$\displaystyle AB=AC$                [ABC is an equilateral triangle]

\    DAEB DAEC                [R.H.S. congruency]

\    BE = EC                [c.p.c.t.]

Now,    $\displaystyle BD=\frac{1}{3}BC$ and $\displaystyle BE=EC=\frac{1}{2}BC.$

In DABE

$\displaystyle A{{B}^{2}}=A{{E}^{2}}+B{{E}^{2}}$                [By Pythagoras theorem]

$\displaystyle A{{B}^{2}}$    $\displaystyle =A{{E}^{2}}+{{\left( BD+DE \right)}^{2}}$

$\displaystyle A{{B}^{2}}$    =$\displaystyle A{{E}^{2}}+B{{D}^{2}}+D{{E}^{2}}+2BD.DE$

$\displaystyle A{{B}^{2}}$    = $\displaystyle A{{D}^{2}}+B{{D}^{2}}+2\,BD.DE$        [$\displaystyle \because$In DADE
$\displaystyle A{{E}^{2}}+O{{E}^{2}}=A{{D}^{2}}$]

$\displaystyle A{{B}^{2}}$    $\displaystyle =A{{D}^{2}}+{{\left[ \frac{1}{3}BC \right]}^{2}}+2\times \frac{1}{3}BC\left[ BE-BD \right]$         [$\displaystyle \because BD=\frac{1}{3}BC$]

$\displaystyle A{{B}^{2}}$    $\displaystyle =A{{D}^{2}}+\frac{1}{9}A{{B}^{2}}+\frac{2}{3}AB\left[ \frac{1}{2}AB-\frac{1}{3}AB \right]$        [$\displaystyle \because AB=BC$ and $\displaystyle BE=\frac{1}{2}BC$]

$\displaystyle A{{B}^{2}}$    $\displaystyle =A{{D}^{2}}+\frac{1}{9}A{{B}^{2}}+\frac{2}{3}AB\left[ \frac{AB}{6} \right]$

$\displaystyle A{{B}^{2}}=A{{D}^{2}}+\frac{A{{B}^{2}}}{9}+\frac{A{{B}^{2}}}{9}$

$\displaystyle 9A{{B}^{2}}=9A{{D}^{2}}+2\,A{{B}^{2}}$

$\displaystyle 7A{{B}^{2}}=9A{{D}^{2}}$

### Solution:

Given:
ABC be an equilateral triangle and let AD BC.

To prove: 3AB2 = 4AD2

Proof:
In DADB and DADC, we have

AB = AC            [Given]

ÐB = ÐC            [Each equal to 60°]

and,    ÐADB = ÐADC [Each equal to 90°]

\    DADB DADC [By AAS congruency]

Þ     BD = DC            [c.p.c.t.]

Þ     $\displaystyle BD=DC=\frac{1}{2}BC$

Since DADB is a right triangle right-angled at D.

\    $\displaystyle A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}$

Þ    $\displaystyle A{{B}^{2}}=A{{D}^{2}}+{{\left( \frac{1}{2}BC \right)}^{2}}$

Þ    $\displaystyle A{{B}^{2}}=A{{D}^{2}}+\frac{B{{C}^{2}}}{4}$        [$\displaystyle \because$
BC = AB]

Þ    $\displaystyle 4A{{B}^{2}}=4A{{D}^{2}}+A{{B}^{2}}$        [$\displaystyle \because$
BC = AB]

Þ    $\displaystyle 4A{{B}^{2}}-A{{B}^{2}}=4A{{D}^{2}}$

Þ    $\displaystyle 3A{{B}^{2}}=4A{{D}^{2}}$

### Solution:

Given:
ABC is a right triangle right-angled at C and $\displaystyle AC=\sqrt{3}\,\,BC.$

To prove:
ÐABC = 60°

Construction: Let D be the mid-point of AB. Join CD.

Proof:
ABC is a right triangle right-angled at C.

\    $\displaystyle A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}$     [By Pythagoras theorem]

Þ    $\displaystyle A{{B}^{2}}={{(\sqrt{3}BC)}^{2}}+B{{C}^{2}}$ [$\displaystyle \because$
$\displaystyle AC=\sqrt{3}BC$ (Given)]

$\displaystyle A{{B}^{2}}=4B{{C}^{2}}$

Þ     $\displaystyle AB=2BC$

But,     $\displaystyle BD=\frac{1}{2}AB$ i.e., AB = 2BD

\     BD = BC

But mid-point of the hypotenuse of a right triangle is equidistant from the vertices.

\    CD = AD = BD

Þ    CD = BC            [$\displaystyle \because$ BD = BC]

Thus, in DBCD, we have

BD = CD = BC

Þ    DBCD is equilateral    Þ    ÐABC = 60°

### Solution:

Given: A DABC in which AD is the internal bisector of ÐA and meets BC in D.

To Prove:    $\displaystyle \frac{BD}{DC}=\frac{AB}{AC}$

Construction: Draw CE || DA to meet BA produced at E.

Proof:        Ð2 = Ð3            …(i)

[Alternate angles]

and,    Ð1 = Ð4                …(ii)

[Corresponding angles]

But,    Ð1 = Ð2 [$\displaystyle \because$
AD is the bisector of ÐA]

From (i) and (ii), we get

Ð3 = Ð4

Þ    AE = AC                …(iii)

[Sides opposite to equal angles are equal]

Now, in DBCE, we have

$\displaystyle DA||CE$        [By construction]

Þ    $\displaystyle \frac{BD}{DC}=\frac{BA}{AE}$        [Using Basic Proportionality Theorem]

Þ    $\displaystyle \frac{BD}{DC}=\frac{AB}{AC}$        [(From (iii)]

Hence, $\displaystyle \frac{BD}{DC}=\frac{AB}{AC}$

### Solution:

Given: A quadrilateral ABCD in which AB = AD and the bisectors of ÐBAC and ÐCAD meet the sides BC and CD at E and F respectively.

To Prove:
EF || BD

Construction: Join AC, BD and EF.

Proof:     In DCAB, AE is the bisector of ÐBAC.

\    $\displaystyle \frac{AC}{AB}=\frac{CE}{BE}$                    …(i)

In DACD, AF is the bisector of ÐCAD.

\    $\displaystyle \frac{AC}{AD}=\frac{CF}{DF}$

Þ    $\displaystyle \frac{AC}{AB}=\frac{CF}{DF}$        [$\displaystyle \because$
AD = AB]        …(ii)

From (i) and (ii), we get

$\displaystyle \frac{CE}{BE}=\frac{CF}{DF}$

Þ    $\displaystyle \frac{CE}{EB}=\frac{CF}{FD}$

Therefore, by the converse of BPT Theorem, we have

EF || BD.

### Solution:

(i)

In right angled DABC, AB = 1.8 cm, BC = 2.4 cm

\    $\displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$

$\displaystyle A{{C}^{2}}={{(1.8)}^{2}}+{{(2.4)}^{2}}=3.24+5.76=9$    Þ    $\displaystyle AC=\sqrt{9}=3\,\text{m}$

Hence the original length of the string AC is 3 m.

(ii)

When Nazima pulls in the string at the rate of 5 cm/sec, then the length of the string decreases = 5 ´ 12 cm = 60 cm = 0.60 m in 12 seconds.

\ Remaining length of the string (AD) after 12 seconds = (3 – 0.60) = 2.40 m

Now in right angled DABD,

$\displaystyle A{{D}^{2}}=D{{B}^{2}}+A{{B}^{2}}$

Þ
$\displaystyle D{{B}^{2}}=A{{D}^{2}}-A{{B}^{2}}={{(2.40)}^{2}}-{{(1.80)}^{2}}=2.52$ m Þ
$\displaystyle DB=\sqrt{2.52}\text{m}$ = 1.587 m

Þ Horizontal distance (DE) of the fly from Nazima

= (1.587 + 1.2) m = 2.787 m = 2.79 m.

# 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)

In DADB and DPSQ

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

ÐADB = ÐPSQ        [Both 90°]

\    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]

ÐADE = ÐABC                [Proved above]

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