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Page 1
30
· Pythagorean triplet · Similarity and right angled triangles
· Theorem of geometric mean · Pythagoras theorem
· Application of Pythagoras theorem · Apollonius theorem
Pythagoras theorem :
In a right angled triangle, the square of the hypotenuse is equal to the sum
of the squares of remaning two sides.
In D PQR Ð PQR = 90°
l(PR)
2
= l(PQ)
2
+ l(QR)
2
We will write this as,
PR
2
= PQ
2
+ QR
2
The lengths PQ, QR and PR of D PQR can also be shown by letters r, p and q. With
this convention, refering to figure 2.1, Pythagoras theorem can also be stated as
q
2
= p
2
+ r
2
.
Pythagorean Triplet :
In a triplet of natural numbers, if the square of the largest number is equal to the
sum of the squares of the remaining two numbers then the triplet is called Pythagorean
triplet.
For Example: In the triplet ( 11, 60, 61 ) ,
11
2
= 121, 60
2
= 3600, 61
2
= 3721 and 121 + 3600 = 3721
The square of the largest number is equal to the sum of the squares of the other
two numbers.
\ 11, 60, 61 is a Pythagorean triplet.
Verify that (3, 4, 5), (5, 12, 13), (8, 15, 17), (24, 25, 7) are Pythagorean
triplets.
Numbers in Pythagorean triplet can be written in any order.
Fig. 2.1
P Q
R
2
Pythagoras Theorem
Let’s study.
Let’s recall.
Page 2
30
· Pythagorean triplet · Similarity and right angled triangles
· Theorem of geometric mean · Pythagoras theorem
· Application of Pythagoras theorem · Apollonius theorem
Pythagoras theorem :
In a right angled triangle, the square of the hypotenuse is equal to the sum
of the squares of remaning two sides.
In D PQR Ð PQR = 90°
l(PR)
2
= l(PQ)
2
+ l(QR)
2
We will write this as,
PR
2
= PQ
2
+ QR
2
The lengths PQ, QR and PR of D PQR can also be shown by letters r, p and q. With
this convention, refering to figure 2.1, Pythagoras theorem can also be stated as
q
2
= p
2
+ r
2
.
Pythagorean Triplet :
In a triplet of natural numbers, if the square of the largest number is equal to the
sum of the squares of the remaining two numbers then the triplet is called Pythagorean
triplet.
For Example: In the triplet ( 11, 60, 61 ) ,
11
2
= 121, 60
2
= 3600, 61
2
= 3721 and 121 + 3600 = 3721
The square of the largest number is equal to the sum of the squares of the other
two numbers.
\ 11, 60, 61 is a Pythagorean triplet.
Verify that (3, 4, 5), (5, 12, 13), (8, 15, 17), (24, 25, 7) are Pythagorean
triplets.
Numbers in Pythagorean triplet can be written in any order.
Fig. 2.1
P Q
R
2
Pythagoras Theorem
Let’s study.
Let’s recall.
31
For more information
Formula for Pythagorean triplet:
If a, b, c are natural numbers and a > b, then [(a
2
+ b
2
),(a
2
- b
2
),(2ab)] is
Pythagorean triplet.
\
(a
2
+ b
2
)
2
= a
4
+ 2a
2
b
2
+ b
4
.......... (I)
(a
2
- b
2
) = a
4
- 2a
2
b
2
+ b
4
.......... (II)
(2ab)
2
= 4a
2
b
2
.......... (III)
\by (I), (II) and (III) , (a
2
+ b
2
)
2
= (a
2
- b
2
)
2
+ (2ab)
2
\[(a
2
+ b
2
), (a
2
- b
2
), (2ab)] is Pythagorean Triplet.
This formula can be used to get various Pythagorean triplets.
For example, if we take a = 5 and b = 3,
a
2
+ b
2
= 34, a
2
- b
2
= 16 , 2ab = 30.
Check that (34, 16, 30) is a Pythagorean triplet.
Assign different values to a and b and obtain 5 Pythagorean triplet.
Last year we have studied the properties of right angled triangle with the angles
30° - 60° - 90° and 45° - 45° - 90°.
(I)Property of 30°-60°-90° triangle.
If acute angles of a right angled triangle are 30° and 60°, then the side opposite
30°angle is half of the hypotenuse and the side opposite to 60° angle is
3
2
times the
hypotenuse.
See figure 2.2. In D LMN, Ð L = 30°, Ð N = 60°, Ð M = 90°
30°
60°
90°
M
L
N
Fig. 2.2
\side opposite 30°angle = MN =
1
2
´ LN
side opposite 60°angle = LM =
3
2
´ LN
If LN = 6 cm, we will find MN and LM.
MN =
1
2
´ LN LM =
3
2
´ LN
=
1
2
´ 6 =
3
2
´ 6
= 3 cm = 3 3 cm
Page 3
30
· Pythagorean triplet · Similarity and right angled triangles
· Theorem of geometric mean · Pythagoras theorem
· Application of Pythagoras theorem · Apollonius theorem
Pythagoras theorem :
In a right angled triangle, the square of the hypotenuse is equal to the sum
of the squares of remaning two sides.
In D PQR Ð PQR = 90°
l(PR)
2
= l(PQ)
2
+ l(QR)
2
We will write this as,
PR
2
= PQ
2
+ QR
2
The lengths PQ, QR and PR of D PQR can also be shown by letters r, p and q. With
this convention, refering to figure 2.1, Pythagoras theorem can also be stated as
q
2
= p
2
+ r
2
.
Pythagorean Triplet :
In a triplet of natural numbers, if the square of the largest number is equal to the
sum of the squares of the remaining two numbers then the triplet is called Pythagorean
triplet.
For Example: In the triplet ( 11, 60, 61 ) ,
11
2
= 121, 60
2
= 3600, 61
2
= 3721 and 121 + 3600 = 3721
The square of the largest number is equal to the sum of the squares of the other
two numbers.
\ 11, 60, 61 is a Pythagorean triplet.
Verify that (3, 4, 5), (5, 12, 13), (8, 15, 17), (24, 25, 7) are Pythagorean
triplets.
Numbers in Pythagorean triplet can be written in any order.
Fig. 2.1
P Q
R
2
Pythagoras Theorem
Let’s study.
Let’s recall.
31
For more information
Formula for Pythagorean triplet:
If a, b, c are natural numbers and a > b, then [(a
2
+ b
2
),(a
2
- b
2
),(2ab)] is
Pythagorean triplet.
\
(a
2
+ b
2
)
2
= a
4
+ 2a
2
b
2
+ b
4
.......... (I)
(a
2
- b
2
) = a
4
- 2a
2
b
2
+ b
4
.......... (II)
(2ab)
2
= 4a
2
b
2
.......... (III)
\by (I), (II) and (III) , (a
2
+ b
2
)
2
= (a
2
- b
2
)
2
+ (2ab)
2
\[(a
2
+ b
2
), (a
2
- b
2
), (2ab)] is Pythagorean Triplet.
This formula can be used to get various Pythagorean triplets.
For example, if we take a = 5 and b = 3,
a
2
+ b
2
= 34, a
2
- b
2
= 16 , 2ab = 30.
Check that (34, 16, 30) is a Pythagorean triplet.
Assign different values to a and b and obtain 5 Pythagorean triplet.
Last year we have studied the properties of right angled triangle with the angles
30° - 60° - 90° and 45° - 45° - 90°.
(I)Property of 30°-60°-90° triangle.
If acute angles of a right angled triangle are 30° and 60°, then the side opposite
30°angle is half of the hypotenuse and the side opposite to 60° angle is
3
2
times the
hypotenuse.
See figure 2.2. In D LMN, Ð L = 30°, Ð N = 60°, Ð M = 90°
30°
60°
90°
M
L
N
Fig. 2.2
\side opposite 30°angle = MN =
1
2
´ LN
side opposite 60°angle = LM =
3
2
´ LN
If LN = 6 cm, we will find MN and LM.
MN =
1
2
´ LN LM =
3
2
´ LN
=
1
2
´ 6 =
3
2
´ 6
= 3 cm = 3 3 cm
32
(II ) Property of 45°-45°-90°
If the acute angles of a right angled triangle are 45° and 45°, then each of the
perpendicular sides is
1
2
times the hypotenuse.
See Figure 2.3. In D XYZ,
XY =
1
2
´ ZY
XZ =
1
2
´ ZY
\ XY = XZ =
1
2
´ ZY
If ZY = 32 cm then we will find XY and
ZX
XY = XZ =
1
2
´ 32
XY = XZ = 3cm
In 7
th
standard we have studied theorem of Pythagoras using areas of four
right angled triangles and a square. We can prove the theorem by an alternative
method.
Activity:
Take two congruent right angled triangles. Take another isosceles right angled
triangle whose congruent sides are equal to the hypotenuse of the two congruent right
angled triangles. Join these triangles to form a trapezium
Area of the trapezium =
1
2
´ (sum of the lengths of parallel sides) ´ height
Using this formula, equating the area of trapezium with the sum of areas of the
three right angled triangles we can prove the theorem of Pythagoras.
Fig. 2.4
x
y
y
x
z
z
Fig. 2.3
45°
45°
Z
X Y
Page 4
30
· Pythagorean triplet · Similarity and right angled triangles
· Theorem of geometric mean · Pythagoras theorem
· Application of Pythagoras theorem · Apollonius theorem
Pythagoras theorem :
In a right angled triangle, the square of the hypotenuse is equal to the sum
of the squares of remaning two sides.
In D PQR Ð PQR = 90°
l(PR)
2
= l(PQ)
2
+ l(QR)
2
We will write this as,
PR
2
= PQ
2
+ QR
2
The lengths PQ, QR and PR of D PQR can also be shown by letters r, p and q. With
this convention, refering to figure 2.1, Pythagoras theorem can also be stated as
q
2
= p
2
+ r
2
.
Pythagorean Triplet :
In a triplet of natural numbers, if the square of the largest number is equal to the
sum of the squares of the remaining two numbers then the triplet is called Pythagorean
triplet.
For Example: In the triplet ( 11, 60, 61 ) ,
11
2
= 121, 60
2
= 3600, 61
2
= 3721 and 121 + 3600 = 3721
The square of the largest number is equal to the sum of the squares of the other
two numbers.
\ 11, 60, 61 is a Pythagorean triplet.
Verify that (3, 4, 5), (5, 12, 13), (8, 15, 17), (24, 25, 7) are Pythagorean
triplets.
Numbers in Pythagorean triplet can be written in any order.
Fig. 2.1
P Q
R
2
Pythagoras Theorem
Let’s study.
Let’s recall.
31
For more information
Formula for Pythagorean triplet:
If a, b, c are natural numbers and a > b, then [(a
2
+ b
2
),(a
2
- b
2
),(2ab)] is
Pythagorean triplet.
\
(a
2
+ b
2
)
2
= a
4
+ 2a
2
b
2
+ b
4
.......... (I)
(a
2
- b
2
) = a
4
- 2a
2
b
2
+ b
4
.......... (II)
(2ab)
2
= 4a
2
b
2
.......... (III)
\by (I), (II) and (III) , (a
2
+ b
2
)
2
= (a
2
- b
2
)
2
+ (2ab)
2
\[(a
2
+ b
2
), (a
2
- b
2
), (2ab)] is Pythagorean Triplet.
This formula can be used to get various Pythagorean triplets.
For example, if we take a = 5 and b = 3,
a
2
+ b
2
= 34, a
2
- b
2
= 16 , 2ab = 30.
Check that (34, 16, 30) is a Pythagorean triplet.
Assign different values to a and b and obtain 5 Pythagorean triplet.
Last year we have studied the properties of right angled triangle with the angles
30° - 60° - 90° and 45° - 45° - 90°.
(I)Property of 30°-60°-90° triangle.
If acute angles of a right angled triangle are 30° and 60°, then the side opposite
30°angle is half of the hypotenuse and the side opposite to 60° angle is
3
2
times the
hypotenuse.
See figure 2.2. In D LMN, Ð L = 30°, Ð N = 60°, Ð M = 90°
30°
60°
90°
M
L
N
Fig. 2.2
\side opposite 30°angle = MN =
1
2
´ LN
side opposite 60°angle = LM =
3
2
´ LN
If LN = 6 cm, we will find MN and LM.
MN =
1
2
´ LN LM =
3
2
´ LN
=
1
2
´ 6 =
3
2
´ 6
= 3 cm = 3 3 cm
32
(II ) Property of 45°-45°-90°
If the acute angles of a right angled triangle are 45° and 45°, then each of the
perpendicular sides is
1
2
times the hypotenuse.
See Figure 2.3. In D XYZ,
XY =
1
2
´ ZY
XZ =
1
2
´ ZY
\ XY = XZ =
1
2
´ ZY
If ZY = 32 cm then we will find XY and
ZX
XY = XZ =
1
2
´ 32
XY = XZ = 3cm
In 7
th
standard we have studied theorem of Pythagoras using areas of four
right angled triangles and a square. We can prove the theorem by an alternative
method.
Activity:
Take two congruent right angled triangles. Take another isosceles right angled
triangle whose congruent sides are equal to the hypotenuse of the two congruent right
angled triangles. Join these triangles to form a trapezium
Area of the trapezium =
1
2
´ (sum of the lengths of parallel sides) ´ height
Using this formula, equating the area of trapezium with the sum of areas of the
three right angled triangles we can prove the theorem of Pythagoras.
Fig. 2.4
x
y
y
x
z
z
Fig. 2.3
45°
45°
Z
X Y
33
Proof : In D ADB and D ABC
Ð DAB @ Ð BAC ...(common angle)
Ð ADB @ Ð ABC ... (each 90°)
D ADB ~ D ABC ... (AA test)... (I)
In D BDC and D ABC
Ð BCD @ Ð ACB .....(common angle)
Ð BDC @ Ð ABC ..... (each 90°)
D BDC ~ D ABC ..... (AA test) ... (II)
Fig. 2.5
A
B
D
C
´
´
Let’s learn.
Now we will give the proof of Pythagoras theorem based on properties of similar
triangles. For this, we will study right angled similar triangles.
Similarity and right angled triangle
Theorem : In a right angled triangle, if the altitude is drawn to the hypotenuse, then
the two triangles formed are similar to the original triangle and to each
other.
Given : In D ABC, Ð ABC = 90°,
seg BD ^ seg AC, A-D-C
T o prove : D ADB ~ D ABC
D BDC ~ D ABC
D ADB ~ D BDC
\ D ADB ~ D BDC from (I) and (II) ........(III)
\ from (I), (II) and (III), D ADB ~ D BDC ~ D ABC ....(transitivity)
Theorem of geometric mean
In a right angled triangle, the perpendicular segment to the hypotenuse
from the opposite vertex, is the geometric mean of the segments into
which the hypotenuse is divided.
Proof ? In right angled triangle PQR, seg QS ^ hypotenuse PR
D QSR ~ D PSQ .......... ( similarity of right triangles )
QS
PS
=
SR
SQ
QS
PS
=
SR
QS
QS
2
= PS ´ SR
\ seg QS is the ‘geometric mean’ of seg PS and SR.
Fig. 2.6
R
P Q
S
Page 5
30
· Pythagorean triplet · Similarity and right angled triangles
· Theorem of geometric mean · Pythagoras theorem
· Application of Pythagoras theorem · Apollonius theorem
Pythagoras theorem :
In a right angled triangle, the square of the hypotenuse is equal to the sum
of the squares of remaning two sides.
In D PQR Ð PQR = 90°
l(PR)
2
= l(PQ)
2
+ l(QR)
2
We will write this as,
PR
2
= PQ
2
+ QR
2
The lengths PQ, QR and PR of D PQR can also be shown by letters r, p and q. With
this convention, refering to figure 2.1, Pythagoras theorem can also be stated as
q
2
= p
2
+ r
2
.
Pythagorean Triplet :
In a triplet of natural numbers, if the square of the largest number is equal to the
sum of the squares of the remaining two numbers then the triplet is called Pythagorean
triplet.
For Example: In the triplet ( 11, 60, 61 ) ,
11
2
= 121, 60
2
= 3600, 61
2
= 3721 and 121 + 3600 = 3721
The square of the largest number is equal to the sum of the squares of the other
two numbers.
\ 11, 60, 61 is a Pythagorean triplet.
Verify that (3, 4, 5), (5, 12, 13), (8, 15, 17), (24, 25, 7) are Pythagorean
triplets.
Numbers in Pythagorean triplet can be written in any order.
Fig. 2.1
P Q
R
2
Pythagoras Theorem
Let’s study.
Let’s recall.
31
For more information
Formula for Pythagorean triplet:
If a, b, c are natural numbers and a > b, then [(a
2
+ b
2
),(a
2
- b
2
),(2ab)] is
Pythagorean triplet.
\
(a
2
+ b
2
)
2
= a
4
+ 2a
2
b
2
+ b
4
.......... (I)
(a
2
- b
2
) = a
4
- 2a
2
b
2
+ b
4
.......... (II)
(2ab)
2
= 4a
2
b
2
.......... (III)
\by (I), (II) and (III) , (a
2
+ b
2
)
2
= (a
2
- b
2
)
2
+ (2ab)
2
\[(a
2
+ b
2
), (a
2
- b
2
), (2ab)] is Pythagorean Triplet.
This formula can be used to get various Pythagorean triplets.
For example, if we take a = 5 and b = 3,
a
2
+ b
2
= 34, a
2
- b
2
= 16 , 2ab = 30.
Check that (34, 16, 30) is a Pythagorean triplet.
Assign different values to a and b and obtain 5 Pythagorean triplet.
Last year we have studied the properties of right angled triangle with the angles
30° - 60° - 90° and 45° - 45° - 90°.
(I)Property of 30°-60°-90° triangle.
If acute angles of a right angled triangle are 30° and 60°, then the side opposite
30°angle is half of the hypotenuse and the side opposite to 60° angle is
3
2
times the
hypotenuse.
See figure 2.2. In D LMN, Ð L = 30°, Ð N = 60°, Ð M = 90°
30°
60°
90°
M
L
N
Fig. 2.2
\side opposite 30°angle = MN =
1
2
´ LN
side opposite 60°angle = LM =
3
2
´ LN
If LN = 6 cm, we will find MN and LM.
MN =
1
2
´ LN LM =
3
2
´ LN
=
1
2
´ 6 =
3
2
´ 6
= 3 cm = 3 3 cm
32
(II ) Property of 45°-45°-90°
If the acute angles of a right angled triangle are 45° and 45°, then each of the
perpendicular sides is
1
2
times the hypotenuse.
See Figure 2.3. In D XYZ,
XY =
1
2
´ ZY
XZ =
1
2
´ ZY
\ XY = XZ =
1
2
´ ZY
If ZY = 32 cm then we will find XY and
ZX
XY = XZ =
1
2
´ 32
XY = XZ = 3cm
In 7
th
standard we have studied theorem of Pythagoras using areas of four
right angled triangles and a square. We can prove the theorem by an alternative
method.
Activity:
Take two congruent right angled triangles. Take another isosceles right angled
triangle whose congruent sides are equal to the hypotenuse of the two congruent right
angled triangles. Join these triangles to form a trapezium
Area of the trapezium =
1
2
´ (sum of the lengths of parallel sides) ´ height
Using this formula, equating the area of trapezium with the sum of areas of the
three right angled triangles we can prove the theorem of Pythagoras.
Fig. 2.4
x
y
y
x
z
z
Fig. 2.3
45°
45°
Z
X Y
33
Proof : In D ADB and D ABC
Ð DAB @ Ð BAC ...(common angle)
Ð ADB @ Ð ABC ... (each 90°)
D ADB ~ D ABC ... (AA test)... (I)
In D BDC and D ABC
Ð BCD @ Ð ACB .....(common angle)
Ð BDC @ Ð ABC ..... (each 90°)
D BDC ~ D ABC ..... (AA test) ... (II)
Fig. 2.5
A
B
D
C
´
´
Let’s learn.
Now we will give the proof of Pythagoras theorem based on properties of similar
triangles. For this, we will study right angled similar triangles.
Similarity and right angled triangle
Theorem : In a right angled triangle, if the altitude is drawn to the hypotenuse, then
the two triangles formed are similar to the original triangle and to each
other.
Given : In D ABC, Ð ABC = 90°,
seg BD ^ seg AC, A-D-C
T o prove : D ADB ~ D ABC
D BDC ~ D ABC
D ADB ~ D BDC
\ D ADB ~ D BDC from (I) and (II) ........(III)
\ from (I), (II) and (III), D ADB ~ D BDC ~ D ABC ....(transitivity)
Theorem of geometric mean
In a right angled triangle, the perpendicular segment to the hypotenuse
from the opposite vertex, is the geometric mean of the segments into
which the hypotenuse is divided.
Proof ? In right angled triangle PQR, seg QS ^ hypotenuse PR
D QSR ~ D PSQ .......... ( similarity of right triangles )
QS
PS
=
SR
SQ
QS
PS
=
SR
QS
QS
2
= PS ´ SR
\ seg QS is the ‘geometric mean’ of seg PS and SR.
Fig. 2.6
R
P Q
S
34
Fig. 2.9 Fig. 2.8
A
B C
P
Q R
Pythagoras Theorem
In a right angled triangle, the square of the hypotenuse is equal to the sum of
the squares of remaining two sides.
Given : In D ABC, ÐABC = 90°
To prove : AC
2
= AB
2
+ BC
2
Construction :Draw perpendicular seg BD on side AC.
A-D-C.
Proof : In right angled D ABC, seg BD ^ hypotenuse AC ..... (construction)
\ D ABC ~ D ADB ~ D BDC ..... (similarity of right angled triangles)
D ABC ~ D ADB Similarly, D ABC ~ D BDC
AB
AD
=
BC
DB
=
AC
AB
- corresponding
AB
BD
=
BC
DC
=
AC
BC
-corresponding
sides sides
AB
AD
=
AC
AB
BC
DC
=
AC
BC
AB
2
= AD ´ AC .......... (I) BC
2
= DC ´ AC .......... (II)
Adding (I) and (II)
AB
2
+ BC
2
= AD ´ AC + DC ´ AC
= AC (AD + DC)
= AC ´ AC .......... (A-D-C)
\ AB
2
+ BC
2
= AC
2
\ AC
2
= AB
2
+ BC
2
Converse of Pythagoras theorem
In a triangle if the square of one side is equal to the sum of the squares of the
remaining two sides, then the triangle is a right angled triangle.
Given ? In D ABC, AC
2
= AB
2
+ BC
2
To prove ? Ð ABC = 90°
A
B C
D
Fig. 2.7
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FAQs on Textbook: Pythagoras Theorem - Mathematics Class 10 (Maharashtra SSC Board)
| 1. What is the Pythagorean Theorem? |  |
Ans. The Pythagorean Theorem is a fundamental principle in geometry that relates the lengths of the sides of a right triangle. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. The formula is expressed as a² + b² = c², where c represents the hypotenuse, and a and b represent the other two sides.
| 2. How can the Pythagorean Theorem be applied in real-life situations? | |
Ans. The Pythagorean Theorem can be applied in various real-life situations, such as in construction, navigation, and design. For example, it is used to determine the correct lengths of ladders, to calculate distances in maps, or to create right angles in building projects. By applying the theorem, one can ensure that structures are built accurately and safely.
| 3. What are some examples of problems that can be solved using the Pythagorean Theorem? | |
Ans. Some examples of problems that can be solved using the Pythagorean Theorem include finding the length of one side of a right triangle when the lengths of the other two sides are known, determining the distance between two points on a coordinate plane, and calculating the height of an object when the distance and angle are known. These problems often arise in various fields such as engineering, architecture, and even everyday situations.
| 4. Can the Pythagorean Theorem be used for triangles that are not right triangles? | |
Ans. No, the Pythagorean Theorem specifically applies to right triangles only. For triangles that are not right triangles, other methods such as the Law of Cosines or the Law of Sines must be used to find unknown side lengths or angles. The Pythagorean Theorem relies on the unique properties of right triangles, which do not hold true for other types of triangles.
| 5. What are the limitations of the Pythagorean Theorem? | |
Ans. The primary limitation of the Pythagorean Theorem is that it only applies to right triangles. Additionally, it does not provide information about the angles of the triangle or the relationships between non-right triangles. Furthermore, the theorem assumes a flat, Euclidean plane, so it does not apply in non-Euclidean geometries such as spherical or hyperbolic geometry.
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