# 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]

\    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)

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

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

(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)

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

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

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

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)

$\displaystyle BD+DC=4CD$

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.

Proof:

AB = AC            [Given]

ÐB = ÐC            [Each equal to 60°]

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    $\displaystyle \frac{AC}{AB}=\frac{CF}{DF}$        [$\displaystyle \because$

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.

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