Page 1
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
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