## 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) **

3^{2} = 9, 5^{2} = 25, 4^{2} = 16

∴ 3^{2} + 4^{2} = 9 + 16 = 25

∴ 3^{2} + 4^{2} = 5^{2}

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) **

4^{2} = 16, 9^{2} = 81, 12^{2} = 144

∴ 4^{2} + 9^{2} = 16 + 81 = 97

∴ 4^{2} + 9^{2} ≠ 12^{2}

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)**

5^{2} = 25, 12^{2} = 144, 13^{2} = 169

∴ 5^{2} + 12^{2} = 25 + 144 = 169

∴ 5^{2} + 12^{2} = 13^{2}

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) **

24^{2} = 576, 70^{2} = 4900, 74^{2} = 5476

**∴ **24^{2} + 70^{2} = 576 + 4900 = 5476

**∴** 24^{2} + 70^{2} = 74^{2}

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) **

10^{2} = 100, 24^{2} = 576, 27^{2} = 729

**∴ **10^{2} + 24^{2} = 100 + 576 = 676

**∴** 10^{2} + 24^{2} ≠ 27^{2}

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)**

11^{2} = 121, 60^{2} = 3600, 61^{2} = 3721

**∴ **11^{2} + 60^{2} = 121 + 3600 = 3721

**∴** 11^{2} + 60^{2} = 61^{2}

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.**

In Δ MNP,

A MNP = 90° …..(Given)

seg NQ **⊥** hypotenuse MP

by theorem of geometric mean.

NQ^{2} = MQ × PQ

**∴ **NQ^{2} = 9 × 4

**∴ **NQ^{2} = 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.**

In Δ QPR,

**∠** QPR = 90° …..(Given)

seg PM ⊥ hypotenuse QR …..(Given)

by theorem of geometric mean,

PM^{2} = QM × MR

**∴** 10^{2} = 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.**

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.**

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.**

##### 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,

AC^{2 }= AB^{2} + BC^{2}

10^{2} = x^{2} + x^{2}

100 = 2x^{2}

x^{2} = 100/2 = 50

**∴ **x^{2} = 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**

**(1) **In Δ DFE,

∠ DFE = 90° …..(Given)

seg FG ⊥ hypotenuse DE ….(Given)

**∴** by theorem of geometric mean,

FG^{2} = DG × EG

12^{2} = 8 × EG

**∴** EG = 144/8 = 18

**∴ **EG = 18

**(2)** In Δ DGF,

∠ DGF = 90° …..(Given)

**∴** by Pythagoras theorem,

FD^{2} = DG^{2} + GF^{2}

**∴** FD^{2} = 8^{2} + 12^{2}

**∴** FD^{2} = 64 + 144 = 208

**∴** FD^{2} = 208

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

**(3)** In Δ EGF,

∠ EFG = 90° …..(Given)

**∴** by Pythagoras theorem,

EF^{2} = EG^{2} + GF^{2}

**∴ **EF^{2} = 18^{2} +12^{2}

**∴ **EF^{2} = 324 + 144

**∴ **EF^{2} = 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.**

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,

QS^{2} = QR^{2} + RS^{2}

∴ QS^{2} = 35^{2} + 12^{2}

∴ QS^{2} = 1225 + 144

∴ QS^{2} = 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, ****PQ ^{2} = 4PM^{2} — 3PR^{2}**

**Proof :** In Δ PRQ,

∠ PRQ = 90° …..(Given)

∴ by Pythagoras theorem,

PQ^{2} = PR^{2} + QR^{2} …..(1)

In Δ PRM,

∠ PRM = 90° …..(Given)

∠ by Pythagoras theorem,

PM^{2} = PR^{2} + RM^{2} …..(2)

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

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

PM^{2} = PR^{2} + RQ^{2}

Multiplying each term with 4, we get,

4PM^{2} = 4PR^{2} + \(\frac{1}{4}\)RQ^{2}

4PM^{2} = 3PR^{2} + (PR^{2} + RQ^{2})

4PM^{2} = 3PR^{2} + PQ^{2} ….[From (1)]

4PM^{2} — 3PR^{2} = PQ^{2} OR

**PQ ^{2} = 4PM^{2} — 3PR^{2} …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.**

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,

AC^{2} = AB^{2} + BC^{2}

5.8^{2} = 4^{2} + BC^{2}

33.64 = 16 + BC^{2}

BC^{2} = 33.64 — 16

∴ BC^{2} = 17.64

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

In Δ EDC, ∠ EDC = 90°.

∴ by Pythagoras theorem,

CE^{2} = DE^{2} + CD^{2}

5.8^{2} = 4.2^{2} + CD^{2}

CD^{2} = 5.8^{2} — 4.2^{2}

CD^{2} = 33.64 — 17.64

∴ CD^{2} = 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.**

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

∴ by Apollonius theorem,

PQ^{2} + PR^{2} = 2PS^{2} + 2QS^{2}

∴ 11^{2} + 17^{2} = 2(13)^{2} + 2QS^{2}

∴ 121 + 289 = 2 × 169 + 2QS^{2}

∴ 410 = 388 + 2QS^{2}

∴ 2QS^{2} = 410 — 388

∴ 2QS^{2} = 72

∴ QS^{2} = 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**

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,

AC^{2} + BC^{2} = 2CM^{2} + 2AM^{2}

∴ 7^{2} + 9^{2} = 2CM^{2} + 2(5)^{ 2}

∴ 49 + 81 = 2CM^{2} + 50

∴ 130 – 50 = 2CM^{2}

∴ CM^{2} = 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) PR ^{2} = PS^{2} + QR **

**×**

**ST + \((\frac{QR}{2})^2\)**

**(ii) PQ ^{2} = PS^{2} — QR **

**×**

**ST + \((\frac{QR}{2})^2\)**

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,

PS^{2} = PT^{2} + TS^{2} …..(2)

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

∴ by Pythagoras theorem,

PR^{2} = PT^{2} + TR^{2}

∴ PR^{2} = PT^{2} + (TS + SR)^{2} …..(T—S—R)

∴ PR^{2} = PT^{2} + TS^{2} + 2ST.SR + SR^{2} …. [(a + b)^{2} = a^{2} + 2ab + b^{2}]

∴ PR^{2} = (PT^{2} + TS^{2}) + 2ST.SR + SR^{2}

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

**∴ ****PR ^{2} = PS^{2} + QR **

**×**

**ST + \((\frac{QR}{2})^2\)**

**(2) **In Δ PTQ, ∠ PTQ = 90° (Given)

∴ by Pythagoras theorem,

PQ^{2} = PT^{2} + TQ^{2}

∴ PQ^{2} = PT^{2} + (QS — TS)^{2} …(Q—T—S)

∴ PQ^{2} = PT^{2} + QS^{2} — 2QS.TS + TS^{2} ….. [(a — b)^{2} = a^{2} — 2ab + b^{2}]

∴ PQ^{2} = (PT^{2} + TS^{2}) — 2QS.TS + QS^{2}

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

**PQ ^{2} = PS^{2} — QR **

**×**

**ST + \((\frac{QR}{2})^2\)**

**Question 4. In ****Δ** **ABC, ****point ****M ****is the midpoint of side ****BC. If, AB ^{2} + AC^{2} = 290 **

**cm**

^{2}**, AM = 8**

**cm, find**

**BC.**

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

∴ by Apollonius theorem,

AB^{2} + AC^{2} = 2AM^{2} + 2BM^{2}

∴ 290 = 2(8)^{ 2} + 2BM^{2}

290 = 128 + 2BM^{2}

∴ 290 — 128 = 2BM^{2}

∴ 2BM^{2} = 162

∴ BM^{2} = 162/2

∴ BM^{2} = 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, ****TS ^{2} + TQ^{2} = TP^{2} + TR^{2}. **

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

**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,

TS^{2} = TA^{2} + AS^{2} …..(4)

In Δ TBQ,

∠ TBQ = 90° ….[From (1)]

∴ by Pythagoras theorem,

TQ^{2} = TB^{2} + BQ^{2} ……(5)

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

TS^{2} + TQ^{2} = TA^{2} + AS^{2} + TB^{2} + BQ^{2} ….(6)

In Δ TAP,

∠ TAP = 90° [From (1)]

∴ by Pythagoras theorem,

TP^{2} = TA^{2} + AP^{2} ….(7)

In Δ TBR,

∠ TBR = 90° …..[From (1)]

∴ by Pythagoras theorem,

TR^{2} = TB^{2} + BR^{2} …..(8)

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

TP^{2} + TR^{2} = TA^{2} + AP^{2 }+ TB^{2} + BR^{2}

TP^{2} + TR^{2} = TA^{2} + BQ^{2} + TB^{2} + AS^{2} ….[From (2) and (3)] …(9)

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

TS^{2} + TQ^{2} = TP^{2} + TR^{2}

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