CBSE 10th Mathematics | Triangles | Pythagoras Theorem

Triangles | Pythagoras 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

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


\    \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


The sides of some triangles are given below. Determine which of them are right triangles.

(i)    7 cm, 24 cm, 25 cm

(ii)    7 cm, 8 cm, 6 cm


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.


(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.


(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.


A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its top reaches a window 12 m above the ground. Determine the length of the ladder.


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.


A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.


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.


In figure, ÐACB = 90° and CD
AB. Prove that \displaystyle \frac{B{{C}^{2}}}{A{{C}^{2}}}=\frac{BD}{AD}.


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}


P and Q are the midpoints of the sides CA and CB respectively of DABC, right angled at C. Prove that

(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}}


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}}



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)


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)


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}}.


O is any point inside a rectangle ABCD in figure. Prove that OB2 + OD2 = OA2 + OC2.


Given: O is any point inside a rectangle ABCD

To prove:
OB2 + OD2 = OA2 + OC2

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)]


ABC is a triangle right-angled at C and p is the length of the perpendicular from C to AB. Show that

(a) pc = ab

(b) \displaystyle \frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}; where a = BC, b = AC and c = AB.



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.


\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}}}.


In DABC, ÐC > 90° and side AC is produced to D such that segment BD is perpendicular to segment AD. Prove that \displaystyle A{{B}^{2}}=A{{C}^{2}}+2CA\times CD.


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.


In DABC, ÐB < 90° and AD is drawn perpendicular to BC. Prove that \displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2BC.BD.


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.


The perpendicular from A on the side BC of a DABC intersects BC at D such that DB = 3 CD. Prove that \displaystyle 2A{{B}^{2}}=2A{{C}^{2}}+B{{C}^{2}}.


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}}


In an equilateral triangle ABC, D is a point on BC such that \displaystyle BD=\frac{1}{3}BC. Prove that \displaystyle 9A{{D}^{2}}=7A{{B}^{2}}.


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.


\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}}


Prove that three times the square of any side of an equilateral-triangle is equal to four times the square of the altitude.


ABC be an equilateral triangle and let AD BC.

To prove: 3AB2 = 4AD2    

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}}


ABC is a right triangle right-angled at C and \displaystyle AC=\sqrt{3}\,\,BC. Prove that ÐABC = 60°.


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.

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°


The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.


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}


ABCD is a quadrilateral in which AB = AD. The bisector of ÐBAC and ÐCAD intersect the sides BC and CD at the points E and F respectively. Prove that EF || BD.


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.


Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out in figure? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?



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.


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}}

\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


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

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}}}

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]

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


The areas of two similar triangles ABC and PQR are 64 cm2 and 36 cm2 respectively. If QR = 16.5 cm, find BC.


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.


In the given figure, LM || BC. AM = 3 cm, MC = 4 cm.     If the ar(DALM) = 27 cm2, calculate the ar(DABC).


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.


\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


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.


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

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

Therefore, DE || BA
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]


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


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 \displaystyle \frac{AX}{AB}.


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}


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

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


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


\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}


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


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}}}


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.


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)

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)


In figure, DE || BC and AD : DB = 5 : 4. Find \displaystyle \frac{\text{Area}\,(\Delta DEF)}{\text{Area}\,\,(\Delta CFB)}.


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)]


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.


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


In figure, prove that \displaystyle \frac{\text{Area}\,(\Delta ABC)}{\text{Area}\,(\Delta DBC)}=\frac{AO}{DO}.


Construction: Draw AX and DY BC

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)]


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


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


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}