Solutions-Class-10-Mathematics-2-Chapter-2-Pythagoras Theorem-Maharashtra Board

Pythagoras Theorem

Class-10-Mathematics-2-Chapter-2-Maharashtra Board

Solutions

Practice set 2.1

Question 1 Identify, with reason, which of the following are Pythagorean triplets.

(i) (3, 5, 4)

Answer :

32 = 9, 52 = 25, 42 = 16

∴ 32 + 42 = 9 + 16 = 25

∴ 32 + 42 = 52

The square of the largest number is equal to the sum of the squares of the other two numbers.

(3, 5, 4) is a Pythagorean triplet.

(ii) (4, 9, 12)

Answer :

42 = 16, 92 = 81, 122 = 144

∴ 42 + 92 = 16 + 81 = 97

∴ 42 + 92 ≠ 122

The square of the largest number is not equal to the sum of the squares of the other two numbers.

(4, 9, 12) is not a Pythagorean triplet.

(iii) (5, 12, 13)

Answer :

52 = 25, 122 = 144, 132 = 169

∴ 52 + 122 = 25 + 144 = 169

∴ 52 + 122 = 132

The square of the largest number is equal to the sum of the squares of the other two numbers.

(5, 12, 13) is a Pythagorean triplet.

(iv) (24, 70, 74)

Answer :

242 = 576, 702 = 4900, 742 = 5476

242 + 702 = 576 + 4900 = 5476

242 + 702 = 742

The square of the largest number is equal to the sum of the squares of the other two numbers.

(24, 70, 74) is a Pythagorean triplet.

(v) (10, 24, 27)

Answer :

102 = 100, 242 = 576, 272 = 729

102 + 242 = 100 + 576 = 676

102 + 242 ≠ 272

The square of the largest number is not equal to the sum of the squares of the other two numbers.

(10, 24, 27) is not a Pythagorean triplet.

(vi) (11, 60, 61)

Answer :

112 = 121, 602 = 3600, 612 = 3721

112 + 602 = 121 + 3600 = 3721

112 + 602 = 612

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.

Question 2. In figure, MNP = 90°, seg NQ seg MP, MQ = 9, QP = 4, find NQ.

Answer :

In Δ MNP,

A MNP = 90°          …..(Given)

seg NQ hypotenuse MP

by theorem of geometric mean.

NQ2 = MQ × PQ

NQ2 = 9 × 4

NQ2 = 36

NQ = 6    ….  (Taking square roots of both the sides)

Answer is : NQ = 6.

Question 3. In figure, QPR = 90°, seg PM seg QR and Q—M—R, PM = 10, QM = 8, find QR.

Answer :

In Δ QPR,

QPR = 90°       …..(Given)

seg PM ⊥ hypotenuse QR     …..(Given)

by theorem of geometric mean,

PM2 = QM × MR

102 = 8 × MR

MR = 100/8

MR = 12.5

QR = QM + MR  (Q—M—R)

QR = 8 + 12.5  = 20.5

Answer is : QR = 20.5.

Question 4. See figure. Find RP and PS using the information given in Δ PSR.

Answer :

In Δ PSR,

PSR = 90°     ……(Given)

SPR = 30°

SRP = 60°     ……(Remaining angle of a triangle)

Δ PSR is a 30°— 60°— 90° triangle.

by 30°— 60°— 90° triangle theorem,

SR = \(\frac{1}{2}\)RP              …..(Side opposite to 30°)

6 = \(\frac{1}{2}\) × RP

RP = 6 × 2  = 12

PS = \(\frac{\sqrt{3}}{2}\)RP                 ……(Slde opposite to 60°)

PS = \(\frac{\sqrt{3}}{2}\) × 12
∴ PS = \(6\sqrt{3}\)
Answer is : RP = 12 and PS = \(6\sqrt{3}\)

Question 5. For finding AB and BC with the help of information given in figure, complete following activity.

Answer :

AB = BC ……. (given )

BAC = 45°

AB = BC  = \(\frac{1}{\sqrt{2}}\) × AC

                  = \(\frac{1}{\sqrt{2}}\) × \(\sqrt{8}\)

                  = \(\frac{1}{\sqrt{2}}\) × \(2\sqrt{2}\)

                  = 2

Question 6. Find the side and perimeter of a square whose diagonal is 10 cm.

Answer :

Let □ ABCD be the given square and AC = diagonal = 10 cm

Let the side of the square be x cm.

AB = BC = x cm.

In Δ ABC,

∠ ABC = 90°     ……(Angle of a square)

by Pythagoras theorem,

AC2 = AB2 + BC2

102 = x2 + x2

100 = 2x2

x2 = 100/2 = 50

x2 = 50

x = \(5\sqrt{2}\)

AB = \(5\sqrt{2}\) cm.

side of square is \(5\sqrt{2}\) cm.

Perimeter of a square = 4 × side

= 4 × \(5\sqrt{2}\)

= \(20\sqrt{2}\) cm

Answer is :  Side of a square is \(20\sqrt{2}\)  cm and its perimeter is  cm.

Question 7. In figure, DFE = 90°, FG ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF

Answer :

(1) In Δ DFE,

∠ DFE = 90°       …..(Given)

seg FG ⊥ hypotenuse DE     ….(Given)

by theorem of geometric mean,

FG2 = DG × EG

122 = 8 × EG

EG = 144/8 = 18

EG = 18

(2) In Δ DGF,

∠ DGF = 90°      …..(Given)

by Pythagoras theorem,

FD2 = DG2 + GF2

FD2 = 82 + 122

FD2 = 64 + 144 = 208

FD2 = 208

FD = \(4\sqrt{13}\)    ….  (Taking square roots of both the sides)

(3) In Δ EGF,

∠ EFG = 90°       …..(Given)

by Pythagoras theorem,

EF2 = EG2 + GF2

EF2 = 182 +122

EF2 = 324 + 144

EF2 = 468

EF = \(6\sqrt{13}\)     …..  (Taking square roots of both the sides)

Answer is :  (1) EG = 18 (2) FD = \(4\sqrt{13}\)  and (3) EF = \(6\sqrt{13}\) .

Question 8. Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.

Answer :

Let □ PQRS be the given rectangle (fig.). QR = 35 cm, RS = 12 cm.

In Δ QRS,

∠ QRS = 90°       …..(Angle of a rectangle)

∴ by Pythagoras theorem,

QS2 = QR2 + RS2

∴ QS2 = 352 + 122

∴ QS2 = 1225 + 144

∴ QS2 = 1369

∴ QS = 37 cm       ….. (Taking square roots of both the sides)

Answer is : The diagonal of rectangle is 37 cm.

Question 9. In the figure, M is the midpoint of QR. PRQ = 90°.  Prove that, PQ2 = 4PM2 — 3PR2

Answer :

Proof : In Δ PRQ,

∠ PRQ = 90°      …..(Given)

∴ by Pythagoras theorem,

PQ2 = PR2 + QR2     …..(1)

In Δ PRM,

∠ PRM = 90°     …..(Given)

∠ by Pythagoras theorem,

PM2 = PR2 + RM2          …..(2)

RM = \(\frac{1}{2}\)RQ            ….(M is the midpoint of seg RQ)  ….(3)

∴ PM2 = PR2 + \((\frac{1}{2}RQ)^2\)          ….[From (2) and (3)]

PM2 = PR2 +  RQ2

Multiplying each term with 4, we get,

4PM2 = 4PR2 + \(\frac{1}{4}\)RQ2

4PM2 = 3PR2 + (PR2 + RQ2)

4PM2 = 3PR2 + PQ2          ….[From (1)]

4PM2 — 3PR2 = PQ2          OR

PQ2 = 4PM2 — 3PR2  …proved.

Question 10. Walls of two buildings on either side of a street are parallel 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 :

In the figure, seg XB and seg YD represent the walls of two buildings on either side of a street BD.

Seg AC represents the first position of the ladder and seg CE represents the second position of the ladder.

AC = CE = 5.8 m, AB = 4 m and DE = 4.2 m.

In Δ ABC, ∠ ABC = 90°.

∴ by Pythagoras theorem,

AC2 = AB2 + BC2

5.82 = 42 + BC2

33.64 = 16 + BC2

BC2 = 33.64 — 16

∴ BC2 = 17.64

BC = 4.2 m      … (Taking square roots of both the sides)

In Δ EDC, ∠ EDC = 90°.

∴ by Pythagoras theorem,

CE2 = DE2 + CD2

5.82 = 4.22 + CD2

CD2 = 5.82 — 4.22

CD2 = 33.64 — 17.64

∴ CD2 = 16

∴  CD = 4 m     …. (Taking square roots of both the sides)

BD = BC + CD  (B—C—D)

BD = 4.2 + 4

 BD = 8.2 m
Answer is : Width of the street is 8.2 m

Practice set 2.2

Question 1. In Δ PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

Answer :

In Δ PQR, seg PS is the median  …..(By definition)

∴ by Apollonius theorem,

PQ2 + PR2 = 2PS2 + 2QS2

∴ 112 + 172 = 2(13)2 + 2QS2

∴ 121 + 289 = 2 × 169 + 2QS2

∴ 410 = 388 + 2QS2

∴ 2QS2 = 410 — 388

∴ 2QS2 = 72

∴ QS2 = 36

∴ QS = 6    ….  (Taking square roots of both the sides)

QS = \(\frac{1}{2}\)QR  (S is the midpoint of seg QR)

∴ 6 = \(\frac{1}{2}\)QR

∴ QR = 6 × 2

∴ QR = 12

Answer is : QR = 12

Question 2. In Δ ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB

Answer :

Let seg CM be the median drawn from the vertex C to side AB.

∴ M is the midpoint of side AB     …..(By definition of a median)

∴ AM = \(\frac{1}{2}\)AB = \(\frac{1}{2}\) × 10 = 5

In Δ ACB, seg CM is the median,

∴ by Apollonius theorem,

AC2 + BC2 = 2CM2 + 2AM2

∴ 72 + 92 = 2CM2 + 2(5) 2

∴ 49 + 81 = 2CM2 + 50

∴ 130 – 50 = 2CM2

∴ CM2 = 80/2 = 40

∴ CM =  \(2\sqrt{10}\)      …. (Taking square roots of both the sides)

Answer is : Length of the median drawn from point C to side AB is \(2\sqrt{10}\) 

Question 3. In the figure seg PS is the median of Δ PQR and PT QR. Prove that,

(1) PR2 = PS2 + QR × ST + \((\frac{QR}{2})^2\)

(ii) PQ2 = PS2 — QR × ST + \((\frac{QR}{2})^2\)

Answer :

Proof :  Seg PS is the median of Δ PQR      …..(Given)

QS = SR = \(\frac{1}{2}\)QR     ….. (S is the midpoint of side QR)  ….(1)

In Δ PTS, ∠ PTS = 90°       …..(Given)

∴ by Pythagoras theorem,

PS2 = PT2 + TS2      …..(2)

(1) In Δ PTR, ∠ PTR = 90°     …..(Given)

∴ by Pythagoras theorem,

PR2 = PT2 + TR2

∴ PR2 = PT2 + (TS + SR)2       …..(T—S—R)

∴ PR2 = PT2 + TS2 + 2ST.SR + SR2      …. [(a + b)2 = a2 + 2ab + b2]

∴ PR2 = (PT2 + TS2) + 2ST.SR + SR2

∴ PR2 = PS2 + 2ST.\((\frac{QR}{2})\)+\((\frac{QR}{2})^2\)   ….. [From (1) and (2)]

PR2 = PS2 + QR × ST + \((\frac{QR}{2})^2\) 

(2) In Δ PTQ, ∠ PTQ = 90°  (Given)

∴ by Pythagoras theorem,

PQ2 = PT2 + TQ2

∴ PQ2 = PT2 + (QS — TS)2              …(Q—T—S)

∴ PQ2 = PT2 + QS2 — 2QS.TS + TS2    ….. [(a — b)2 = a2 — 2ab + b2]

∴ PQ2 = (PT2 + TS2) — 2QS.TS + QS2

∴ PQ2 = PS2 — 2\((\frac{QR}{2})\) × TS + \((\frac{QR}{2})^2\)      …..[From (1) and (2)]

PQ2 = PS2 — QR × ST + \((\frac{QR}{2})^2\)

Question 4. In Δ ABC, point M is the midpoint of side BC. If, AB2 + AC2 = 290 cm2, AM = 8 cm, find BC.

Answer :

In Δ ABC,  seg AM is the median.      ….(Given)

∴ by Apollonius theorem,

AB2 + AC2 = 2AM2 + 2BM2

∴ 290 = 2(8) 2 + 2BM2

290 = 128 + 2BM2

∴ 290 — 128 = 2BM2

∴ 2BM2 = 162

∴ BM2 = 162/2

∴ BM2 = 81

∴ BM = 9 cm     … (Taking square roots of both the sides)

BM = \(\frac{1}{2}\)BC   …….(M is the midpoint of side BC)

∴ 9 = \(\frac{1}{2}\)BC

∴ BC = 18 cm

Answer is : BC = 18 cm.

Question 5. In figure, point T is in the interior of rectangle PQRS. Prove that, TS2 + TQ2 = TP2 + TR2.

(As shown in the figure, draw seg AB || side SR and A—T—B)

Answer :

Construction : Through T, draw seg AB || side SR such that A—T—B, P—A—S and  Q—B—R

Proof :

seg PS || seg QR      ….(Opposite sides of rectangle)

seg AS || seg BR       ……(P—A—S and Q—B—R)

also seg AB || seg SR     …..(Construction)

∴ □ ASRB is a parallelogram  (By definition)

∠ ASR = 90°               ……(Angle of rectangle PSRQ)

∴ □ ASRB is a rectangle    …. (A parallelogram is a rectangle, if one of its angles is a

right angle.)

∠ SAB = ∠ ABR = 90°             …..(Angles of a rectangle)

∴ seg TA ⊥ side PS and seg TB ⊥ side QR     ….(1)

AS = BR     …. (Opposite sides of rectangle are equal)  ….(2)

Similarly, we can prove AP = BQ  …..(3)

In Δ TAS,

∠ TAS = 90°       …..[From (1)]

∴ by Pythagoras theorem,

TS2 = TA2 + AS2     …..(4)

In Δ TBQ,

∠ TBQ = 90°           ….[From (1)]

∴ by Pythagoras theorem,

TQ2 = TB2 + BQ2     ……(5)

Adding (4) and (5), we get,

TS2 + TQ2 = TA2 + AS2 + TB2 + BQ2      ….(6)

In Δ TAP,

∠ TAP = 90°  [From (1)]

∴ by Pythagoras theorem,

TP2 = TA2 + AP2     ….(7)

In Δ TBR,

∠ TBR = 90°           …..[From (1)]

∴ by Pythagoras theorem,

TR2 = TB2 + BR2       …..(8)

Adding (7) and (8), We get,

TP2 + TR2 = TA2 + AP2 + TB2 + BR2

TP2 + TR2 = TA2 + BQ2 + TB2 + AS2      ….[From (2) and (3)]  …(9)

from (6) and (9), we get,

TS2 + TQ2 = TP2 + TR2

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