Practice Set 2.1 Pythagoras Theorem Class 10th Mathematics Part 2 MHB Solution

Practice Set 2.1
  1. Identify, with reason, which of the following are Pythagorean triplets. (i) (3, 5, 4)…
  2. In figure 2.17, ∠MNP = 90°, seg NQ ⊥ seg MP, MQ = 9,QP = 4, find NQ.…
  3. In figure 2.18, ∠QPR = 90°, seg PM ⊥ seg QR and Q - M - R,PM = 10, QM = 8, find QR.…
  4. See figure 2.19. Find RP and PSusing the information given in ΔPSR. Ans. RP = 12, PS =…
  5. For finding AB and BC with the help of information given in figure 2.20, complete…
  6. Find the side and perimeter of a square whose diagonal is 10 cm.
  7. In figure 2.21, DFE = 90, FG ED, If GD = 8, FG = 12,find (1) EG (2) FD and (3) EF…
  8. Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.…
  9. In the figure 2.22, M is themidpoint of QR. ∠PRQ = 90°. Prove that, PQ^2 = 4PM^2 -…
  10. Walls of two buildings on either side of a street are parellel to each other. A ladder…
Practice Set 2.1
Question 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:

For three numbers a, b and c if a2 + b2 = c2. then (a, b, c) is known as a Phythagorean Triplet. Also, order here doesn't matter and we take c > a, b.

1st case: 32 + 42 = 52. Thus this a triplet.


2nd case: 42 + 92 ≠ 122


3rd case: 52 + 122 = 132 . Thus this is a triplet.


4th case: 242 + 702 = 742. Thus this is a triplet.


5th case: 102 + 242 ≠ 272


6th case: 112 + 602 = 612. Thus this is a triplet..


Question 2.

In figure 2.17, ∠MNP = 90°, seg NQ ⊥ seg MP, MQ = 9,QP = 4, find NQ.



Answer:

In ∆MNP, ∠ MNP = 900,

By Pythagoras theorem,

MN2 + NP2 = MP2


⇒ MN2 + NP2 = (MQ + QP)2


⇒ MN2 + NP2 = (13)2


⇒ MN2 + NP2 = 169 … (1)


In ∆MQN, ∠MQN = 900,


QN2 + MQ2 = MN2


⇒ QN2 + 92 = MN2


⇒ QN2 + 81 = MN2 …(2)


In ∆PQN, ∠PQN = 900,


QN2 + PQ2 = PN2


⇒ QN2 + 42 = PN2


⇒ QN2 + 16 = PN2 … (3)


Now (2) + (3)


⇒ QN2 + 81 + QN2 + 16 = MN2 + PN2


⇒ 2QN2 + 97 = 169 [from(1)]


⇒ 2QN2 = 72


⇒ QN2 = 36


Thus NQ = 6.


Question 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 = 900


So PM2 + QM2 = PQ2

⇒ 102 + 82 = PQ2

⇒ 100 + 64 =PQ2

PQ2 = 164 …(1)


In ∆PQR, ∠RPQ = 900


So PQ2 + PR2 = QR2


⇒ 164 + PR2 = QR2


⇒ PR2 = QR2 – 164 …(2)


In ∆PMR, ∠PMR = 900


So PM2 + MR2 = PR2

⇒ 102 + (QR – QM)2 = QR2 – 164

⇒ 100 + (QR – QM)2 = QR2 – 164


⇒ 100 + QR2 – 2.QR.QM + QM2 = QR2 – 164


⇒ 100 – 2.QR.8 + 64 = – 164


⇒ 16QR = 2×164


⇒ QR = 20.5


Thus QR = 20.5


Question 4.

See figure 2.19. Find RP and PSusing the information given in ΔPSR.


Ans. RP = 12, PS = 6√3


Answer:

In ∆PSR, ∠PSR = 900


So PS2 + SR2 = RP2


⇒ 62 + (RP cos(30°))2 = RP2


⇒ 62 + RP = RP2


⇒ 62 = 


⇒ RP2 = 4×36


Thus RP = 12.


PS = RP cos(300)

⇒ PS = 12×

PS = 6√3.


Question 5.

For finding AB and BC with the help of information given in figure 2.20, complete following activity.





Answer:

In ∆ABC, ∠ABC = 900


So AB2 + BC2 = AC2


⇒ 2AB2 = 5


⇒ AB2 = 


⇒ AB = √ = X(Say)


AB = BC = √


∠BAC = 450 Since AB = BC


NOW √ = X√5


X = 


Similarly, X2√2 = √


X = 


Similarly, X√8 = √


X = 



Question 6.

Find the side and perimeter of a square whose diagonal is 10 cm.


Answer:

In a square of side say a cm, any diagonal divide the square into two right triangles of equal dimensions.



Thus a2 + a2 = 102


⇒ 2a2 = 100


⇒ a2 = 50


a = 5 cm


perimeter = 4a


= 4×5


= 20


Perimeter of square = 20cm


Question 7.

In figure 2.21, ∠DFE = 90°, FG ⊥ ED, If GD = 8, FG = 12,find (1) EG (2) FD and (3) EF


Answer:

In ∆DGF,DGF = 900


FD2 = DG2 + GF2


FD2 = 64 + 144


FD2 = 208


FD =


In ∆DEF,DFE = 900


ED2 = DF2 + EF2


⇒ (EG + 8)2 = 208 + EF2 … (1)


In ∆EGF,FGE = 900


EF2 = EG2 + GF2


 (EG + 8)2 – 208 = EG2 + 144


 EG2 + 2.EG.8 + 64 – 208 = EG2 + 144 (As we know (a+b)2= a2+b2+2ab


EG = 18


From (1)


⇒ (EG + 8)2 = 208 + EF2


EF = 6√13.


Question 8.

Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.


Answer:

The diagonal = √[length2 + breadth2]


= √(352 + 122)
= √(1225 + 144)

= √1369
= 37

Thus the diagonal is 37 cm.


Question 9.

In the figure 2.22, M is themidpoint of QR. ∠PRQ = 90°. Prove that, PQ2 = 4PM2 – 3PR2



Answer:

In ∆PRQ, ∠PRQ = 900


PQ2 = PR2 + QR2 – – – 1


In ∆PRM, ∠PRM = 900


PM2 = PR2 + MR2


⇒ PM2 = PR2 + 2 [ M is midpoint]


⇒ 4(PM2 – PR2) = QR2 – – – 2


1 And 2 implies


PQ2 = PR2 + 4(PM2 – PR2)


⇒ PQ2 = 4PM2 – 3PR2


PROVED.



Question 10.

Walls of two buildings on either side of a street are parellel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street , its top touches the window of the other building at a height 4.2 m. Find the width of the street.


Answer:

Let us consider a distance x m on the street from one building and a distance y m from the other one.


Now according to question,


In the 1st case,


5.82 = 42 + x2


⇒ x2 = 17.64


⇒ x = 4.2


Similarly for the second building,


5.82 = 4.22 + y2


⇒ y2 = 16


⇒ y = 4


Total width = x + y


= 4 + 4.2


= 8.2


Thus the total width is 8.2m.

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