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Pythagoras Theorem 
Practice Set 2.1 
Q. 1. Identify, with reason, which of the following are Pythagorean triplets. 
(i) (3, 5, 4) 
(ii) (4, 9, 12) 
(iii) (5, 12, 13) 
(iv) (24, 70, 74) 
(v) (10, 24, 27) 
(vi) (11, 60, 61) 
Answer : In a triangle with sides (a,b,c), the Pythagoran’s theorem states that 
a
2
 + b
2
 = c
2
. If this condition is satisfied then (a,b,c)are Pythagorean triplets. 
1
st
 case: 3
2
 + 4
2
 = 5
2
. Thus this a triplet. 
2
nd
 case: 4
2
 + 9
2
 ? 12
2
 
3
rd
 case: 5
2
 + 12
2
 = 13
2
. Thus this is a triplet. 
4
th
 case: 24
2 +
 70
2
 = 74
2
. Thus this is a triplet. 
5
th
 case: 10
2
 + 24
2
 ? 27
2
 
6
th
 case: 11
2
 + 60
2
 = 61
2
. Thus this is a triplet.
 
Q. 2. In figure 2.17, ?MNP = 90°, seg NQ ? seg MP, MQ = 9, QP = 4, find NQ. 
 
 
Answer : In ?MNP, ? MNP = 90
0
, 
MN
2
 + NP
2
 = MP
2
 
Page 2


Pythagoras Theorem 
Practice Set 2.1 
Q. 1. Identify, with reason, which of the following are Pythagorean triplets. 
(i) (3, 5, 4) 
(ii) (4, 9, 12) 
(iii) (5, 12, 13) 
(iv) (24, 70, 74) 
(v) (10, 24, 27) 
(vi) (11, 60, 61) 
Answer : In a triangle with sides (a,b,c), the Pythagoran’s theorem states that 
a
2
 + b
2
 = c
2
. If this condition is satisfied then (a,b,c)are Pythagorean triplets. 
1
st
 case: 3
2
 + 4
2
 = 5
2
. Thus this a triplet. 
2
nd
 case: 4
2
 + 9
2
 ? 12
2
 
3
rd
 case: 5
2
 + 12
2
 = 13
2
. Thus this is a triplet. 
4
th
 case: 24
2 +
 70
2
 = 74
2
. Thus this is a triplet. 
5
th
 case: 10
2
 + 24
2
 ? 27
2
 
6
th
 case: 11
2
 + 60
2
 = 61
2
. Thus this is a triplet.
 
Q. 2. In figure 2.17, ?MNP = 90°, seg NQ ? seg MP, MQ = 9, QP = 4, find NQ. 
 
 
Answer : In ?MNP, ? MNP = 90
0
, 
MN
2
 + NP
2
 = MP
2
 
? MN
2
 + NP
2
 = (MQ + QP)
2
 
? MN
2
 + NP
2
 = (13)
2
 
? MN
2
 + NP
2
 = 169 … (1) 
In ?MQN, ?MQN = 90
0
, 
QN
2
 + MQ
2
 = MN
2
 
? QN
2
 + 9
2
 = MN
2
 
? QN
2
 + 81 = MN
2
 …(2) 
In ?PQN, ?PQN = 90
0
, 
QN
2
 + PQ
2
 = PN
2
 
? QN
2
 + 4
2
 = PN
2
 
? QN
2
 + 16 = PN
2
 … (3) 
Now (2) + (3) 
? QN
2
 + 81 + QN
2
 + 16 = MN
2
 + PN
2
 
? 2QN
2
 + 97 = 169 [from (1)] 
? 2QN
2
 = 72 
? QN
2
 = 36 
Thus NQ = 6. 
Q. 3. In figure 2.18, ?QPR = 90°, seg PM ? seg QR and Q – M – R,PM = 10, QM = 8, 
find QR. 
Page 3


Pythagoras Theorem 
Practice Set 2.1 
Q. 1. Identify, with reason, which of the following are Pythagorean triplets. 
(i) (3, 5, 4) 
(ii) (4, 9, 12) 
(iii) (5, 12, 13) 
(iv) (24, 70, 74) 
(v) (10, 24, 27) 
(vi) (11, 60, 61) 
Answer : In a triangle with sides (a,b,c), the Pythagoran’s theorem states that 
a
2
 + b
2
 = c
2
. If this condition is satisfied then (a,b,c)are Pythagorean triplets. 
1
st
 case: 3
2
 + 4
2
 = 5
2
. Thus this a triplet. 
2
nd
 case: 4
2
 + 9
2
 ? 12
2
 
3
rd
 case: 5
2
 + 12
2
 = 13
2
. Thus this is a triplet. 
4
th
 case: 24
2 +
 70
2
 = 74
2
. Thus this is a triplet. 
5
th
 case: 10
2
 + 24
2
 ? 27
2
 
6
th
 case: 11
2
 + 60
2
 = 61
2
. Thus this is a triplet.
 
Q. 2. In figure 2.17, ?MNP = 90°, seg NQ ? seg MP, MQ = 9, QP = 4, find NQ. 
 
 
Answer : In ?MNP, ? MNP = 90
0
, 
MN
2
 + NP
2
 = MP
2
 
? MN
2
 + NP
2
 = (MQ + QP)
2
 
? MN
2
 + NP
2
 = (13)
2
 
? MN
2
 + NP
2
 = 169 … (1) 
In ?MQN, ?MQN = 90
0
, 
QN
2
 + MQ
2
 = MN
2
 
? QN
2
 + 9
2
 = MN
2
 
? QN
2
 + 81 = MN
2
 …(2) 
In ?PQN, ?PQN = 90
0
, 
QN
2
 + PQ
2
 = PN
2
 
? QN
2
 + 4
2
 = PN
2
 
? QN
2
 + 16 = PN
2
 … (3) 
Now (2) + (3) 
? QN
2
 + 81 + QN
2
 + 16 = MN
2
 + PN
2
 
? 2QN
2
 + 97 = 169 [from (1)] 
? 2QN
2
 = 72 
? QN
2
 = 36 
Thus NQ = 6. 
Q. 3. In figure 2.18, ?QPR = 90°, seg PM ? seg QR and Q – M – R,PM = 10, QM = 8, 
find QR. 
 
 
Answer : 
 
 
In ?PMQ, ?PMQ = 90
0
 
So PM
2
 + QM
2
 = PQ
2
 
?
 
10
2
 + 8
2
 = PQ
2
 
?
 
100 + 64 =
 
PQ
2
 
PQ
2
 = 164 …(1) 
In ?PQR, ?RPQ = 90
0
 
So PQ
2
 + PR
2
 = QR
2
 
? 164 + PR
2
 = QR
2
 
? PR
2
 = QR
2
 – 164 …(2) 
In ?PMR, ?PMR = 90
0
 
So PM
2
 + MR
2
 = PR
2
 
Page 4


Pythagoras Theorem 
Practice Set 2.1 
Q. 1. Identify, with reason, which of the following are Pythagorean triplets. 
(i) (3, 5, 4) 
(ii) (4, 9, 12) 
(iii) (5, 12, 13) 
(iv) (24, 70, 74) 
(v) (10, 24, 27) 
(vi) (11, 60, 61) 
Answer : In a triangle with sides (a,b,c), the Pythagoran’s theorem states that 
a
2
 + b
2
 = c
2
. If this condition is satisfied then (a,b,c)are Pythagorean triplets. 
1
st
 case: 3
2
 + 4
2
 = 5
2
. Thus this a triplet. 
2
nd
 case: 4
2
 + 9
2
 ? 12
2
 
3
rd
 case: 5
2
 + 12
2
 = 13
2
. Thus this is a triplet. 
4
th
 case: 24
2 +
 70
2
 = 74
2
. Thus this is a triplet. 
5
th
 case: 10
2
 + 24
2
 ? 27
2
 
6
th
 case: 11
2
 + 60
2
 = 61
2
. Thus this is a triplet.
 
Q. 2. In figure 2.17, ?MNP = 90°, seg NQ ? seg MP, MQ = 9, QP = 4, find NQ. 
 
 
Answer : In ?MNP, ? MNP = 90
0
, 
MN
2
 + NP
2
 = MP
2
 
? MN
2
 + NP
2
 = (MQ + QP)
2
 
? MN
2
 + NP
2
 = (13)
2
 
? MN
2
 + NP
2
 = 169 … (1) 
In ?MQN, ?MQN = 90
0
, 
QN
2
 + MQ
2
 = MN
2
 
? QN
2
 + 9
2
 = MN
2
 
? QN
2
 + 81 = MN
2
 …(2) 
In ?PQN, ?PQN = 90
0
, 
QN
2
 + PQ
2
 = PN
2
 
? QN
2
 + 4
2
 = PN
2
 
? QN
2
 + 16 = PN
2
 … (3) 
Now (2) + (3) 
? QN
2
 + 81 + QN
2
 + 16 = MN
2
 + PN
2
 
? 2QN
2
 + 97 = 169 [from (1)] 
? 2QN
2
 = 72 
? QN
2
 = 36 
Thus NQ = 6. 
Q. 3. In figure 2.18, ?QPR = 90°, seg PM ? seg QR and Q – M – R,PM = 10, QM = 8, 
find QR. 
 
 
Answer : 
 
 
In ?PMQ, ?PMQ = 90
0
 
So PM
2
 + QM
2
 = PQ
2
 
?
 
10
2
 + 8
2
 = PQ
2
 
?
 
100 + 64 =
 
PQ
2
 
PQ
2
 = 164 …(1) 
In ?PQR, ?RPQ = 90
0
 
So PQ
2
 + PR
2
 = QR
2
 
? 164 + PR
2
 = QR
2
 
? PR
2
 = QR
2
 – 164 …(2) 
In ?PMR, ?PMR = 90
0
 
So PM
2
 + MR
2
 = PR
2
 
? 10
2
 + (QR – QM)
2
 = QR
2
 – 164 
? 100 + (QR – QM)
2
 = QR
2
 – 164 
? 100 + QR
2
 – 2.QR.QM + QM
2
 = QR
2
 – 164 
? 100 – 2.QR.8 + 64 = – 164 
? 16QR = 2 × 164 
? QR = 20.5 
Thus QR = 20.5 
Q. 4. See figure 2.19. Find RP and PS using the information given in ?PSR. 
 
 
Ans. RP = 12, PS = 6 
Answer : In ?PSR, ?PSR = 90
0
 
So PS
2
 + SR
2
 = RP
2
 
? 6
2
 + (RP cos(30
0
))
2
 = RP
2
 
? 6
2
 + RP
2
 = RP
2
 
? 6
2
 =  
? RP
2
 = 4 × 36 
Thus RP = 12. 
PS = RP cos(30
0
) 
? PS = 12 ×  
PS = 6v3. 
Page 5


Pythagoras Theorem 
Practice Set 2.1 
Q. 1. Identify, with reason, which of the following are Pythagorean triplets. 
(i) (3, 5, 4) 
(ii) (4, 9, 12) 
(iii) (5, 12, 13) 
(iv) (24, 70, 74) 
(v) (10, 24, 27) 
(vi) (11, 60, 61) 
Answer : In a triangle with sides (a,b,c), the Pythagoran’s theorem states that 
a
2
 + b
2
 = c
2
. If this condition is satisfied then (a,b,c)are Pythagorean triplets. 
1
st
 case: 3
2
 + 4
2
 = 5
2
. Thus this a triplet. 
2
nd
 case: 4
2
 + 9
2
 ? 12
2
 
3
rd
 case: 5
2
 + 12
2
 = 13
2
. Thus this is a triplet. 
4
th
 case: 24
2 +
 70
2
 = 74
2
. Thus this is a triplet. 
5
th
 case: 10
2
 + 24
2
 ? 27
2
 
6
th
 case: 11
2
 + 60
2
 = 61
2
. Thus this is a triplet.
 
Q. 2. In figure 2.17, ?MNP = 90°, seg NQ ? seg MP, MQ = 9, QP = 4, find NQ. 
 
 
Answer : In ?MNP, ? MNP = 90
0
, 
MN
2
 + NP
2
 = MP
2
 
? MN
2
 + NP
2
 = (MQ + QP)
2
 
? MN
2
 + NP
2
 = (13)
2
 
? MN
2
 + NP
2
 = 169 … (1) 
In ?MQN, ?MQN = 90
0
, 
QN
2
 + MQ
2
 = MN
2
 
? QN
2
 + 9
2
 = MN
2
 
? QN
2
 + 81 = MN
2
 …(2) 
In ?PQN, ?PQN = 90
0
, 
QN
2
 + PQ
2
 = PN
2
 
? QN
2
 + 4
2
 = PN
2
 
? QN
2
 + 16 = PN
2
 … (3) 
Now (2) + (3) 
? QN
2
 + 81 + QN
2
 + 16 = MN
2
 + PN
2
 
? 2QN
2
 + 97 = 169 [from (1)] 
? 2QN
2
 = 72 
? QN
2
 = 36 
Thus NQ = 6. 
Q. 3. In figure 2.18, ?QPR = 90°, seg PM ? seg QR and Q – M – R,PM = 10, QM = 8, 
find QR. 
 
 
Answer : 
 
 
In ?PMQ, ?PMQ = 90
0
 
So PM
2
 + QM
2
 = PQ
2
 
?
 
10
2
 + 8
2
 = PQ
2
 
?
 
100 + 64 =
 
PQ
2
 
PQ
2
 = 164 …(1) 
In ?PQR, ?RPQ = 90
0
 
So PQ
2
 + PR
2
 = QR
2
 
? 164 + PR
2
 = QR
2
 
? PR
2
 = QR
2
 – 164 …(2) 
In ?PMR, ?PMR = 90
0
 
So PM
2
 + MR
2
 = PR
2
 
? 10
2
 + (QR – QM)
2
 = QR
2
 – 164 
? 100 + (QR – QM)
2
 = QR
2
 – 164 
? 100 + QR
2
 – 2.QR.QM + QM
2
 = QR
2
 – 164 
? 100 – 2.QR.8 + 64 = – 164 
? 16QR = 2 × 164 
? QR = 20.5 
Thus QR = 20.5 
Q. 4. See figure 2.19. Find RP and PS using the information given in ?PSR. 
 
 
Ans. RP = 12, PS = 6 
Answer : In ?PSR, ?PSR = 90
0
 
So PS
2
 + SR
2
 = RP
2
 
? 6
2
 + (RP cos(30
0
))
2
 = RP
2
 
? 6
2
 + RP
2
 = RP
2
 
? 6
2
 =  
? RP
2
 = 4 × 36 
Thus RP = 12. 
PS = RP cos(30
0
) 
? PS = 12 ×  
PS = 6v3. 
Q. 5. For finding AB and BC with the help of information given in figure 2.20, 
complete following activity. 
 
 
 
Answer : In ?ABC, ?ABC = 90
0
 
So AB
2
 + BC
2
 = AC
2
 
? 2AB
2
 = 5 
? AB
2
 =  
? AB = v = X(Say) 
AB = BC = v 
?BAC = 45
0
 Since AB = BC 
Now v = Xv5

	Mathematics Class 10 (Maharashtra SSC Board) 26 videos\|208 docs\|38 tests

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FAQs on Textbook Solutions: Pythagoras Theorem - Mathematics Class 10 (Maharashtra SSC Board)

1. What is the Pythagorean theorem and how is it used in geometry?

Ans. The Pythagorean theorem states that in a right-angled 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. This can be expressed as a² + b² = c², where c is the hypotenuse and a and b are the other two sides. It is used in geometry to determine the lengths of sides in right triangles, to calculate distances, and in various real-world applications like construction and navigation.

2. Can the Pythagorean theorem be applied in real-life situations?

Ans. Yes, the Pythagorean theorem has numerous real-life applications. It can be used in fields such as architecture, engineering, and computer graphics. For example, it helps architects determine the proper lengths and angles when designing buildings, ensures safe construction practices, and is used in navigation to find the shortest paths or distances between points on a map.

3. How do you solve for the length of a side using the Pythagorean theorem?

Ans. To solve for the length of a side using the Pythagorean theorem, you need to identify which side of the triangle you are solving for. If you have the lengths of the other two sides, you can rearrange the theorem's formula. For example, if you need to find the length of the hypotenuse (c), you would use c = √(a² + b²). If you need to find one of the other sides, say a, you would rearrange it to a = √(c² - b²).

4. What are some common misconceptions about the Pythagorean theorem?

Ans. A common misconception is that the Pythagorean theorem applies to all triangles when it only applies to right-angled triangles. Another misconception is that the theorem can be used in non-Euclidean geometries without modification, which is not accurate. Additionally, some students may confuse the squares of the sides with the sides themselves, forgetting to square the lengths when applying the theorem.

5. Are there any proofs or historical significance associated with the Pythagorean theorem?

Ans. Yes, the Pythagorean theorem has several proofs, including geometric proofs, algebraic proofs, and even proofs using calculus. Historically, it is attributed to the ancient Greek mathematician Pythagoras, who lived around the 6th century BCE. Its significance lies not only in its mathematical applications but also in its foundational role in the development of mathematics and geometry, influencing various fields and cultures throughout history.

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