11
Pythagoras Theorem
This unit facilitates you in,
Pythagoras Theorem.
stating Pythagoras theorem
Converse
theorem.
proving Pythagoras theorem logically.
of
Pythagoras
Pythagorean triplets
Proof of Pythagoras theorem
and its converse
Problems and riders based on
Pythagoras theorem.
stating converse of Pythago ras
theorem.
pro ving converse of Pythago ras
theorem logically.
explaining the meaning of Pythagorean
triplets.
reasoning deductively and prove the
riders based on Pythagoras theorem
and its converse.
analysing and solving real life problems
based on pythagoras theorem.
Pythagoras
(569-479 B.C, Greece)
Pythagoras, a pupil of Thales is
perhaps best known for the
theorem on right angled triangle
which bears his name. He helped
to devlop the study of geometry in
a logical way by using undefined
terms, definitions, postualtes and
logical deductions. The influence
of his school in the fields of
mathematics, science, music,
religio n and philosophy is
apparent even today.
There is geometry in the humming of the strings,
there is music in the spacing of spheres.
-Pythagoras
270
4 cm
A
4 cm
cm
Now, count the number of small squares on all the
three sides. They are 9, 16 and 25. If we add 9 and
16, we get 25. Repeat this activity for right angled
triangles with sides, {6cm, 8cm and 10cm}, {5cm,
12cm, 13cm} etc.
5
ABC will be exactly 90°.
Construct squares on all the three sides and divide
them into squares of sides 1cm each.
cm
3 cm
4 cm B
5
C
cm
3 cm
Observe that
5
cm
ABC with AB = 4cm, BC = 3cm, AC = 5cm.
5
Construct a
3 cm
5.
Let us recall some of the properties of a right angled triangle which we have already
learnt.
A
The side opposite to right angle is called hypotenuse.
Hy
po
In a right angled triangle, hypotenuse is the longest side.
te
nu
In an isosceles right angled triangle, each acute angle is 45°.
se
Area of a right angled triangle is half the product of the sides
containing the right angle.
B
C
The perpendicular drawn from the right angled vertex to the
hypotenuse divides the triangle into two similar triangles which are similar to the
given right angled triangle.
Now, let us learn one more very interesting property about right angled triangles.
To understand the property, do the following activity.
4 cm
1.
2.
3.
4.
UNIT-11
3 cm
What do we infer from this? We can infer that,
"In a right angled triangle, the square on the hypotenuse is equal to the sum of
the squares on the other two sides".
This statement which gives the relationship between areas of sides of a right angled
triangle is called Pythagoras theorem, named after the Greek mathematician
Pythagoras, who lived around 500BC. This theorem which is used in many branches
of mathematics has attracted the attention of many mathematicians and today we
have hundreds of varieties of proofs for it.
The statement of Pythagoras theorem can be proved practically by another
way.
5
This was given by Henry Perigal (1830). Hence, it is called Perigal
dissection.
Study the activity given below and do it in groups.
Construct a
ABC, right angled at C on a card board
as shown in the figure.
Draw squares on AB, BC and CA. In these squares, do
the constructions as shown in the figure and steps
mentioned below.
Mark the middle point (P) of the square drawn on the
longest side containing the right angle, i.e. BC. (This
can be obtained by joining the diagonals of the square)
Through P draw DE || AB and FG
DE .
G
A
C
D
EB
F
P
Pythagoras Theorem
271
Mark the quadrilaterals formed as 1,2,3 and 4, and the square on AC as 5. (Observe
the figure)
Cut the quadrilaterals 1,2,3,4, and 5.
Arrange them on the square drawn on the hypotenuse AB. (observe the figure)
What conclusion we can draw about the areas of the squares on sides of the right
angled triangle?
Pythagoras Theorem
In a right angled triangle, the square on the hypotenuse is equal to the sum of
the squares on the other two sides.
A
Data
:
To Prove
:
In
ABC = 90°
AB2 + BC2 = CA2
Construction :
Proof
D
ABC,
Draw BD  AC.
:
B
C
Statement
Compare
Reason
ABC and
ABC
ADB,
ADB
90
(  Data and construction)
BAD is common.
ABC



ADB
AB
AC
=
AD
AB
(  Equiangular triangles)
(  A A similarity criteria)
 AB2 = AC.AD
Compare
ABC and
ABC
.....(1)
BDC,
BDC
90
(  Data and construction)
ACB is common
ABC



BC
DC
BDC
AC
BC
BC2 = AC.DC
[  Equiangular Triangles]
[  AA similarity criteria]
.....(2)
By adding (1) and (2) we get
AB2 + BC2 = (AC. AD) + (AC. DC)
AB2 + BC2 = AC (AD + DC)
AB2 + BC2 = AC. AC = AC2
2
2
 AB + BC
2
= AC
[  AD + DC = AC)
QED
272
UNIT-11
Know this!
l
Baudhayana, an ancient Indian mathematician in
his Shulva sutras composed during 800 BC states
that, "The diagonal of a rectangle produces both areas
b
of which its length and breadth produce separately".
d
b
d2 = l2 + b2
l
For this reason, this theorem is also referred to as
the 'Baudhayana Theorem'.
Alternate proof for Pythagoras theorem using area of trapezium
Data
:
To prove
:
Construction
:
In
ABC, ABC  90
A
a2 + c2 = b2
Extend BC to D such that,
E
c
b
CD = AB.
At D, draw a 
a
and mark a point E on it such that,
B a
C
c
D
ED = BC.
Join E, C and E,A.
Proof
:
Statement
Compare
Reason
ABC and
CDE
(  Construction)
AB = CD
ABC =
CDE
90
(  Construction)
BC = DE


In
ABC, BAC +
1)
ABC 
[  CPCT]
2)
BAC =
DCE and
3)
ACB =
CED
DCE + ACB = 90°

(  SAS)
CDE
AC = CE
ACB = 90°
ACE = 90°
(  Data and construction)
[
BAC =
ECD ]
[ adjacent angles of linear pair]
Pythagoras Theorem
273
Area of
ABDE = Area of
ABC + Area of
ACE + Area of
CDE
½(a + c)(a + c) = ½ac + ½b2 + ½ac
(a + c)2 = 2ac + b2
[by cancelling
a2 + c2 + 2ac = 2ac + b2
 a2 + c2 = b2
1
2
throughout]
QED
Know this
To-day we have varieties of proofs for Pythagoras Theorem. It is said that there are
more than 300 proofs, some of them are proved based on trigonometry, co-ordinate
geometry, vectors etc.
Try to collect some of them and discuss in groups.
ILLUSTRATIVE EXAMPLES
Numerical problems based on Pythagoras theorem
Example 1:
B = 90° , AC = 17cm and AB = 8cm, find BC.
In a right angled ABC,
Sol. Given, in



ABC,
B
AC2
BC2
BC2
BC2
90
= AB2 + BC2
 Pythagoras theorem]
2
2
= AC - AB
A
45°
17
= 172 - 82
cm
= 289 - 64 = 225
8cm
BC =
15cm
225
B
Example 2 :
A
?
C
4 cm
 = 45° , AM  BC, AM = 4cm and BC = 7cm. Find the length of?AC.
In ABC, ABC
Sol. In

AMB,
AMB
90 ,
ABM
45 
BAM
45
M
7cm
AMB is an isosceles right angled triangle
 MA = MB = 4cm
MC = BC – MB = 7 – 4 = 3cm
 MC = 3cm
In
AMC, AC2= AM2 + MC2 [ Pythagoras theorem]
AC2 = 42 + 32
AC2 = 16 + 9
AC2 = 25
AC =
25
5cm
B
C
274
UNIT-11
Example 3 :
In the rectangle WXYZ, XY + YZ = 17cm
XZ + YW = 26 cm,
W
Z
X
Y
and
calculate the length and breadth of the rectangle.
Sol.
XZ + YW = 26 cm
d1 + d2 = 26 cm
[ d1 = d2]
2d = 26cm
d = 13 cm
XZ = YW = 13cm
Let length = XY = x cm
In
 breadth = XW = (17 – x)cm
[ pythagoras theorem]
WXY, WX2 + XY2 = WY2
2
(17 – x) + x
2
(289 – 34x + x ) + x
2
2
= 13
2
= 169
2
(2x – 34x + 120 = 0)  2
x2 – 17x + 60 = 0
x2 – 12x – 5x + 60 = 0
x(x – 12) –5(x – 12) = 0
(x – 12)(x – 5) = 0
x – 12 = 0 or x – 5 = 0
x = 12 or x = 5
Length = 12cm, breadth = 5cm
Let AB = x and BC = y
(2
AC = x + 2 or AC = 2y + 1
or
2)
+x
A
Sol.
(
Example 4 : In the
ABC, the hypotenuse is greater than one of the other side by 2
units and it is twice greater than the another side by 1 unit. Find the measure
of the sides.
x
+
 x + 2 = 2y + 1
1)
y
x + 1 = 2y
1
x
B
=y
2
AB2 + BC2 = AC2
2
x +y
x2
1
x
x2
2
x2
2x
4
= (2y + 1)
= 4y2 + 4y + 1
1
= 4
C
 Pythagoras theorem)
2
2
2
y
x
1
2
2
4
x
1
2
1
Pythagoras Theorem
4x 2
275
x 2 2x
4
1
=
4(x 2
2x
4
1)
4(x 1)
1
2
4x2 + x2 + 2x + 1 = 4(x2 + 2x + 1) + 4(2x + 2) + 4
2
4x 2 + x + 2x + 1 = 4x 2 + 8x + 4 + 8x + 8 + 4
x2 + 2x – 16x + 1 – 16 = 0
x2 – 14x – 15 = 0
(x – 15)(x + 1) = 0
 x =15 or x = – 1
If x = 15, x + 2 = 17 and
y =
x
1
2
16
2
8
 AB = 15 units, BC = 8 units, AC = 17 units.
Example 5 :
An insect 8 m away from the foot of a lamp post which is 6m tall, crawls towards
it. After moving through a distance, its distance from the top of the lamp post is
equal to the distance it has moved. How far is the insect away from the foot of
the lamp post? [Bhaskaracharya's Leelavathi]
A
Sol.
Distance between the insect and
the foot of the lamp post = BD = 8m.
=x
6
The height of the lamp post = AB = 6m.
After moving a distance, let the insect be at C,
x
=
B 8-x C
Let AC = CD = x m.
D
 BC = (8 - x) m.
In
ABC,
B  90
( Pythagoras theorem)
 AC2 = AB2+BC2
2
2
x = 6 + ( 8- x)
x2
2
= 36 + 64 - 16x + x 2
 16x = 100
x =
100
 6.25
16
 BC = 8–x = (8–6.25) = 1.75 m
The insect is 1.75 m. away from the foot of the lamp post.
Note : We can also consider BC = x, then CD = AC = (8 – x)
276
UNIT-11
Riders based on Pythagoras Theorem.
Example 6 : In the given figure, AD BC, Prove that AB2 + CD2 = BD2 + AC2
B
Sol. In
ADC, ADC 90
AC2 = AD2 + CD2 (  Pythagoras theorem) .......(1)
In 
ADB
DBA
D
90
60°

AB = AD + BD (  Pythagoras theorem) .......(2)
Subtracting (1) from (2), we get
C
2
2
2
2
A
2
AB2 – AC2 = A D + BD2 - A D - CD2
AB2 – AC2 = BD2 - CD2
AB2 + CD2 = BD2 + AC2
A
Example 7 :
 = 60° (ii) AD  BC.
In ABC, (i) ACB
E
Derive an expression for AB in terms of AC and BC.

Sol. Const: Draw DE such that EDC
D
60
 DEC is an equilateral  [ Data and
construction]
 DE = DC = EC
B
DEA is an isosceles   AED 120 and
ADE 30
DAE
 DE = EA
DC = EA = EC [ Axiom - 1]
 AC = 2DC
Now in
2
2
2
AB = (AC – DC ) + (BC – DC)
2
2
[ pythagoras theorem]
AB2 = AD2 + BD2
ADB,
2
2
[ pythagoras theorem]
2
AB = 4DC – DC + BC – 2BC . DC + DC
2
[ AC = 2DC]
AB2 = 4DC2 + BC2 – 2BC.DC
AB2 = 4DC2 – 2BC.DC + BC2
AB2 = 2DC (2DC – BC) + BC2
AB2 = AC (AC – BC) + BC2
2
2
AB = AC – AC.BC + BC
[ 2DC = AC)
2
or AB2 = AC2 + BC2 – AC.BC
By this we have expressed AB interms of AC and BC only.
Example 8 :
Derive the formula for height and area of an equilateral triangle.
Sol. In the equilateral ABC, let AB = BC = CA = 'a' units.
Draw AD  BC Let AD = 'h' units
C
a
2
2
Pythagoras Theorem
277
[ RHS theorem]
 BD = DC
 BD = DC =
A
a
2
a
Now there are two right angled triangles,
ADB and
ADC.
B
[ AD  BC]
[ Pythagoras theorem]
In
ADB,
ADB = 90°
 AB2= AD2 + BD2
2
a
 a=h +
2
2
2
a2
4
a2 = h2 +
a2
1
a2
= h2
4
4a 2
a2
4
3a 2
4
= h2
= h2
Take square root on either sides.
3a 2
4
=
h2
3 .a
=h
2
 h=
a 3
2
 Height of an equilateral triangle of side a units, h =
Let us now find the area of equialteral ABC.
Area of ABC =
A=
1
2
A=
a2 3
4
a
1
2
base
height
a 3
2
 Area of an equilateral triangle, A =
a2 3
4
a 3
2
a
a
D
C
2
278
UNIT-11
R
x
2
ABC
y
B
90
+
ABC,
x
y
: In
3x 2
4
P
2
Data
3
4
A
x
Example 9 :
Equilateral triangles are drawn on the sides of a right
angled triangle. Show that the area of the triangle
on the hypotenuse is equal to sum of the area of
triangles on the other two sides.
Sol.
y
C
3
y
4
les
Equilateral  PAB, QBC and RAC are drawn
on sides AB, BC and CA respectively.
Q
To prove : Area (PAB) + Area (QBC) = Area (RAC)
Proof
Consider
[ Equiangular les are similar]
: PAB  QBC  RAC
Area ( PAB)
Area ( RAC)
Area( QBC)
AB2
=
Area ( RAC)
AC2
BC2
AC2
[ Theorem]
Area( PAB) Area( QBC)
AB2 BC2
=
Area( RAC)
AC2
Area ( PAB) Area( QBC)
Area( RAC)
[ pythagoras theorem
AC2 = AB2 + BC2]
AC 2
AC2
Area ( PAB) Area( QBC)
=1
Area( RAC)
 Area (PBA) + Area (QBC) = Area (RAC)
Alternate Proof :
APB + BQC =
3 2
x
4
3
x2
4
 APB + BQC = RAC
RAC =
Example 10 :
3 2
y
4
y2
2
3 2
x
4
y2
3 2
x
4
y2
 area of equilateral le=
3
side
4
2
In ABD, C is a point on BD such that BC : CD = 1 : 2 and ABC is an equilateral
triangle Prove that AD2 = 7AC2
A
Sol. Data:
In ABD, BC : CD = 1 : 2
In ABC, AB = BC = CA
a
a 3 a
To Prove:
AD2 = 7AC2
2
Construction:
Draw AE  BC
2a
Proof:
In ABC,
D
B a E a C
2
a
BE = EC =
and
2
AE =
a 3
2
2
Pythagoras Theorem
279
In ADE, AED
[ Construction]
90
2
2
AD2 =
a 3
2
AD = AE + ED

[ Pythagoras theorem]
2
2
2a
2
2
AD2 =
3a 2
4
5a
2
AD2 =
3a 2
4
25a 2
4
3a 2
25a 2
4
AD2 =
a
2
28a 2
4
AD2 = 7a2
AD2 = 7AC 2
AD2 =

EXERCISE 11.1
4. In
LAW, LAW 90 , LNA 90 and LW = 26cm,
LN = 6cm and AN = 8 cm. Calculate the length of WA.
cm
26 N 8c
m
W
5. A door of width 6 meter has an arch above it having a
height of 2 meter. Find the radius of the arch.
6c
m
Numerical problems based on Pythagoras theorem.
1. The sides of a right angled triangle containing the right angle are 5cm and 12cm,
find its hypotenuse.
2. Find the length of the diagonal of a square of side 12cm.
3. The length of the diagonal of a rectangular playground is 125m and the length of
one side is 75m. Find the length of the other side.
L
A
h = 2m
6m
6. The sides of a right angled triangle are in an arithmetic progression. Show that the
sides are in the ratio. 3 : 4 : 5.
7. A peacock on a pillar of 9 feet height on seeing a snake coming towards its hole
situated just below the pillar from a distance of 27 feet away from the pillar will fly to
catch it. If both posess the same speed, how far from the pillar they are going to
meet?
280
UNIT-11
Riders based on Pythagoras theorem.
M
8. In MGN, MP  GN. If MG = a units, MN = b units,
GP = c units and PN = d units.
Prove that (a + b)(a – b) = (c + d)(c – d).
9. In
LAB,
ALB
2
Prove that
LA
LB2
b
a
Gc P
90 and LM  AB
L
AM
MB
A
10. In
ABC,  ABC = 900, BD  AC. If AB = 'c' units,
BC = 'a' units, BD = 'p' units , CA = 'b' units.
1
Prove that 2
a
N
d
1
c2
M
A
D
1
.
p2
c
Pythagorean triplets
B
b
p
B
a
C
You have learnt that the measures of three sides of a right angled triangle are
related in a special way.
If the three numbers, which are the measures of three sides of a right angled
triangle are natural numbers then they are called Pythagorean triplets.
Some of them are listed in the table given below. study them.
A
B
3, 4, 5
6, 8, 10
5, 12, 13
15, 36, 39
7, 24, 25
14, 48, 50
11, 60, 61
44, 240, 244
Compare each of the Pythagorean triplets in column A with the corresponding triplet
in column B.
We can conclude that; if ( x , y, z) is a py thagore an triplet, then
(kx, ky, kz) is also a Pythagorean triplet where k  N.
Know this!
Pythagorean triplets can be found using the following general form.
For natural number: 2n, (n2 – 1), (n2 + 1) Here 'n' may be even or odd.
For odd natural numbers: n,
1 2
(n
2
1),
1 2
(n
2
1), Here n is odd
where n N.
From the above general forms any number of pythagorean triplets can be generated by
giving values to 'n'.
Pythagoras Theorem
281
We have learnt that a square number can be equal to sum of two square numbers.
Does this type of relationship apply to cubes or other powers?
Know this!
Fermat's last theorem [Pierre de Fermat (1601-1665), French
mathematician]
"It is impossible to write
- a cube as the sum of two cubes.
- a fourth power as the sum of fourth powers"
or
In general, "it is impossible to write any power beyond the second as the
sum of two similar powers."
Th er e ar e n o a, b, c,  N for which an + bn = cn where n  N and n > 2
Andrew Wiles (Born in cambridge, England - 1963) saw this problem, when
he was ten years old.
Thirty years later and with seven years of intense work Andrew wiles,
proved Fermat's last theorem.
Converse of Pythagoras Theorem
Let us now learn the converse of pythagoras theorem. For this, consider the two
examples.
Example 1
•
•
•
•
•
5 cm
cm
P
4
cm
•
cm
3
4
B
Example 2
A
C
Let ABC be a triang le with
BC = 5cm, AB = 4cm and AC = 3cm.
Imagine squares on all the three sides
and find their areas.
Here, the longest side is BC and its
length is 5cm. The area of the square
on the longest side BC is 25 sq. cm
The sum of the squares on the other
two sides AB and AC will be (16 + 9) sq
cm which is also 25 sq. cm.
•
•
•
•
Now measure BAC which is opposite
to the longest side BC. you will find
 = 90°.
that BAC
•
So, ABC is a right angled triangle,
right angled at the vertex A.
•
4.
3c
m
Q
5 cm
R
Let PQR be another triangle with QR =
5cm, PQ = 4cm and PR = 4.3 cm.
Imagine squares on all the three sides
and find their areas.
Here, the longest side is QR and its
length is 5 cm. The area of the square
on the longest side QR is 25 sq. cm.
The sum of the squares on the other
two sides PQ and PR will be (16 + 18.49)
sq cm = 34.49 sq. cm which is not
equal to the area of the square on the
longest side QR.
QPR which is
Now measure
opposite to the longest side QR, you will

find that QPR=74°
but not 90°.
So, PQR is not a right angled triangle.
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UNIT-11
So, what is the basic condition for a triangle to be a right angled triangle? Compare
the areas on the sides of two triangles in the above examples and try to figure out the
necessary condtion.
The condition is: "If the square on the longest side of a triangle is equal to the
sum of the squares on the other two sides then those two sides contain a right angle."
This is converse of Pythagoras theorem. It was first mentioned and proved by Euclid.
Now let us prove the converse of Pythagoras theorem logically.
Converse of pythagoras theorem
"If the square on the longest side of a triangle is equal to the sum of the squares
on the other two sides, then those two sides contain a right angle."
A
2
Data :
C
B
D
2
2
In ABC, AC = AB + BC
ABC = 90°
To prove :
Construction : Draw a perpendicular on AB at B. Select a point D
on it such that, DB = BC. Join 'A' and 'D'.
Statement
Proof :
In
Reason
ABD,
(  Construction)
ABD = 90°
 AD2 = AB2 + BD2(  Pythagoras theorem)
But in ABC,
AC2 = AB2 + BC2(  Data)
 AD2 = AC2
 AD = AC
Compare
ABD and ABC
AD = AC
(  Proved)
BD = BC
(  Construction)
AB is common

ABD
ABC
(  SSS)
 ABD = ABC
(  CPCT)
 ABD = ABC = 90°
QED
Pythagoras Theorem
283
Comparision of Pythagoras theorem and its converse.
Pythagoras theorem
Converse of Pythagoras theorem
"If the square on the longest side of a
triangle is equal to the sum of the
squares on the other two sides, then
those two sides contain a right angle."
"In a right angled triangle, the square
on the hypotenuse is equal to the sum
of the squares on the other two sides.
A
A
B
Data: In
C
ABC,
B
C
Data: In ABC, AC2 = AB2 + BC2
ABC = 90°
To prove:
To prove: AC2 = AB2 + BC2
ABC = 90°
Now observe the following table. Study the relationship between the areas of three
sides of a triangle.
A
c
b
c
b
a
B

ABC,
 = 90°
ABC
AC2 = AB2 + BC2
b2 = c2 + a2
b
c
*
In
A
A
a
B
*
B
C
C
In ABC,
 < 90°
ABC
AC2< AB2 + BC2
b 2< c2 + a2

a
C
In ABC,

 > 90°
ABC
AC2> AB2 + BC2
b2> c2 + a2
Converse
2
If b2= c2 + a2
then
ABC = 90°
If b < c + a2
then
2
ABC < 90°
If b2> c2 + a2
then
ABC > 90°
Discuss
The Pythagoras theorem has two fundamental aspects, where one is about areas and the
other is about lengths. This landmark theorem connects two main branches of
mathematics - Geometry and Algebra.
284
UNIT-11
ILLUSTRATIVE EXAMPLES
Example1 : Verify whether the following measures represent the sides of a right angled
triangle.
(a) 6, 8, 10
Sol. Sides are : 6, 8, 10
Consider the areas of square on the sides : 62, 82, 102
i.e., 36, 64, 100
Consider the sum of areas of squares on the two smaller sides : 36 + 64 = 100
62 + 82 = 102
We observe that, square on the longest side of the triangle is equal to the sum of
squares on the other two sides.
By converse of Pythagoras theorem, those two smaller sides must contain a right
angle.
Conclusion: The sides 6, 8 and 10 form the sides of a right angled triangle with
hypotenuse 10 units and 6 and 8 units as the sides containing the right angle.
Note: Without actually constructing the triangle for the given measurements of
sides it is now possible to say whether the sides represent the sides of a right
angled triangle using converse of Pythagoras theorem.
(b) 1, 2,
3
Sides are
: 1,
3,2
: 1 2,
3
Consider the areas of
squares on the sides
2
, 22
i.e.,
: 1, 3, 4
Consider the sum of the areas of squares on the two smaller sides
: 1+3=4
12 +
3
2
= 22
We observe that the square on the longest side (2 units) is equal to the sum of the
squares on the other two sides.
By converse of pythagoras Theorem, these two smaller sides must contain a right
angle.
Conclusion: 1, 2,
3 forms the sides of a right angled triangle with 2 units as
hypotenuse and 1 and
3 units as sides containing the right angle.
(c) 4, 5, 6
Sides are
:
4, 5, 6
Areas of squares on the sides
:
42, 52, 62
i.e.,
:
16, 25, 36
Pythagoras Theorem
285
Sum of areas of squares on
the two smaller sides
:
16 + 25 = 41
42 + 52  62

We observe that the square on the longest side of the triangle is not equal to the
sum of the squares on the other two sides.
By converse of Pythagoras theorem, these two sides cannot contain a right angle.
Hence, 4, 5, and 6 cannot form the sides of right angled triangle.
ABC 90 and
Example 2 : In the quadrilateral ABCD,
AD2 = (AB2 + BC2 + CD2). Prove that :
ACD
90
D
A
C
B

ABC, ABC
90 [ Data]
Proof : In
[ Pythagoras theorem]
AC2 = AB2 + BC2

2
2
2
But
AD = (AB + BC )+CD

AD2 = AC2 + CD2
ACD = 90°

2
[ Data]
[ by data AB2 + BC2 = AC2]
[ converse of Pythagoras theorem]
EXERCISE 11.2
1. Verify whether the following measures represent the sides of a right angled triangle.
(a)
(c) n , n
(e)
(b) 6 3, 12, 6
2, 3, 5
x
2
1, 2n
1 x
,
1
2
1
, x
(d) x2 – 1, 2x , x2 + 1
(f) m2 – n2, 2mn, m2 + n2
2. In ABC, a + b = 18 units, b + c = 25 units and c + a = 17
units. What type of triangle is ABC ? Give reason.
A
b
c
B
a
C
286
UNIT-11
3. In ABC, CD  AB, CA = 2AD and BD = 3AD, Prove that
BCA
4. The shortest distance AP from a point 'A' to QR is 12 cm.
Q and R are respectively 15 cm and 20cm from 'A' and on
90
20
cm
12 cm
QAR
A
15
cm
opposite sides of AP. Prove that
90 .
R
P
A
12 cm
5. In the isosceles ABC, AB = AC, BC = 18 cm, AD  BC,
AD = 12 cm, BC is produced to 'E' and AE = 20cm.
Prove that
BAE
20
cm
90
D
B
E
C
18 cm
ADC
BC = AD = 6cm and CD = 3cm, Prove that
ACB
3 cm C
D
90 , AB = 9cm,
90
6 cm
B
0
6 cm
6. In the quadrilateral ABCD,
m
9c
A
P
7. ABCD is a rectangle. 'P' is any point outside it such that
PA2 + PC2 = BA2 + AD2. Prove that
APC 90
A
D
B
C
Pythagoras theorem
Statement of Pythagoras
theorem
Converse of Pythagoras
theorem
Logical proof
Pythagorean
triplets
Riders
ANSWERS
EXERCISE 11.1
1] 13 cm
2]12 2 cm
3] 100m
4] 24cm
5] 3.25m
6] 3 : 4 : 5
7] 12 ft