# Maharashtra Board Class 10 Math Part 2 Solution Chapter 2 Practice Set 2.1 Pythagoras Theorem

## Maharashtra Board Class 10 Math Part – 2 Solution Chapter 2 Practice Set 2.1 – Pythagoras Theorem

Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 2: Pythagoras Theorem. Marathi or English Medium Students of Class 10 get here Pythagoras Theorem full Exercise Solution.

 Std Maharashtra Class 10 Subject Math Part 2 Solution Chapter Pythagoras Theorem Practice Set 2.1

#### Practice set 2.1

(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)

Solution:

(i) Given, (3, 5, 4)

∴ 32 = 9, 52 = 25, 42 = 16

32 + 42 = 16 + 9 = 25

∴ 52 = 32 + 42

∴ (3, 5, 4) are Pythagoras triplets.

(ii) (4, 9, 12)

42 = 16, 92 = 81, 122 = 144

∴ 122 ≠ 42 + 92

∴ (4, 9, 12) are not Pythagoras triplets.

(iii) (5, 12, 13)

∴ 52 = 25, 122 = 144, 132 = 169

∴ 132 = 52 + 122

∴ (5, 12, 13) are Pythagoras triplets.

(iv) (24, 70, 74)

242 = 576, 702 = 4900, 742 = 5476

242 + 702 = 576 + 4900 = 5476

∴ 742 = 242 + 702

∴ (29, 70, 74) are Pythagoras triplets.

(v) (10, 24, 27)

102 = 100, 242 = 576, 272 = 727

∴ 272≠ 102 + 242

∴ (10, 24, 7) are not Pythagoras triplets.

(vi) (11, 60, 61)

112 = 121, 602 = 3600, 612 = 3721

602 + 112 = 3600 + 121

= 3721

∴ 612 = 112 + 612

∴(11, 60, 61) are Pythagoras triplets.

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

Solution:

Given, ∠MNP = 90°

NS⊥MP, MQ = 9, BP = 4

NQ2 = MQ×QP [By theorem of geometric mean]

Or, NQ = √9×4

= √36

= 6

(3) In figure 2.18, ∠QPR = 90°, segPM⊥ segQR and Q-M-R, PM = 10, QM = 8, find QR.

Solution:

Given, ∠QPR = 90°

PM⊥QR, Q-M-R, PM = 10, QN = 8

By theorem of geometric mean,

PM2 = QN×MR

Or, 102 = 8×MR

Or, MR = 100/8 = 12.5

∴ QR = 9M+MR

= 8 + 12.5

= 20.5

(4) See figure 2.19. Find RP and PS using the information given in ∆PSR.

Solution:-

From the given into in adjacent fig.

∠PRS = 180° – (90° + 30°)

= 180° – 120° [Sum of angles of triangle is 180°]

= 60°

∴ SR = ½ RP [By 30° – 60° – 90° theorem]

Or, 6 = 1/2 × RP = 6×2 = 12

∴ PS = √3/2 × RP [By 30° – 60° – 90° theorem]

= √3/2 × 12 = 6√3

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

AB = BC …… —-

∠BAC =

AB = BC = —— × AC

= —- ×√8

= —–× 2√2

= —-

Solution:

AB = BC given,

∠BAC = 180°-90/2 = 90/2 = 45°

[∵ AB = BC so ∠BAC = ∠BCA opposite angles of equal sides are equal]

AB = BC = 1/√2 × AC [By 45° – 45° – 90° theorem]

= 1/√2 × √8

= 1/√2 × 2√2

= 2

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

Solution:

Given, diagonal of square = 10 cm.

We know, sides of square are equal & interior angles of square are all 90°.

∴ Let, side of square be x.

∠BAC = ∠BCA = 180°-90/2 = 90/2 = 45°

[∵ AB = BC sides of square. so opposite angles of equal sides are equal]

∴ AB = BC = 1/√2 × 10

= 1/√2 × 2 × 5 = 5√2 cm.

∴ Perimeter of square ABCD = 4 × x

= 4×5√2 = 20√2 cm

(7) In figure 2.21, √DFE = 90°,

FG⊥ED, If GD = 8, FG = 12, Find (1) EG (2) FD and (3) EF

Solution:

Given, ∠DEF = 90°

FG⊥ED, GD = 8, FG = 12

(1) FG2 = GE×GD [Theorem of geometric mean]

Or, 122 = GE × 8

Or, GE = 144/8 = 18

(2) In ∆FDG,

FD2 = FG2 + GD2 [by Pythagoras theorem]

Or, FD = √122 + 82 = √144+64 = √208 = 4√13

(3) In ∆FGE, EF2 = FG2 + GE2 [By Pythagoras theorem]

Or, EF = √122+182 = √144+324 = √468

= 6√13

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

Solution:

ABCD is a rectangle.

Length BC = 35cm [Given]

Breadth AB = 12 cm [Given]

∠ABC = 90° [Interior angle of rectangle]

∴ By Pythagoras theorem,

AC2 = AB2 + BC2

Or, AC = √122 + 352 = √144+1225 = √1369

= 37 cm

∴ Diagonal AC = 37 cm

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

Solution:

Given, M is the midpoint of QR ∠PRQ = 90°

∴ RM = MQ

In ∆PRM

By Pythagoras theorem,

PM2 = PR2 + RM2

Or, RM2 = PM2 – PR2 —- (i)

In ∆PQR,

By Pythagoras theorem

PQ2 = PR2 + QR2

= PR2 + (RM + MQ)2

∵ R – M – Q

= PR2 + (RM + RM)2

∵ RM = MQ

= PR2 + (2RM)2

= PR2 + 4RM2

Now, putting the value of RM2 from equation (i) we get.

PQ2 = PR2 + 4RM2

Or, PQ2 = PR2 + 4 (PM2 – PR2)

Or, PQ2 = PR2 + 4PM2 – 4PR2

Or, PQ2 = 4PM2 – 3PR2 Hence proved

(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.

Solution:

The adjacent figure represents the description given in the question AB & CD represents the two building AB||CD (given) A & D are point where ∴ AB & CD are ⊥ to BC ground.

∴ In ∆ABE

By Pythagoras theorem,

AE2 = AB2 + BE2

Or, 5.82 = 42 + BE2

Or BE = √33.64 – 16 = √17.64 = 4.2m

Now, In ∆DCE,

By Pythagoras theorem,

DE2 = CD2 + EC2

Or, 5.82 = 4.22 + EC2

Or, EC = √33.64 – 17.64 = √16 = 4m

∴ width of the street is BC

Now, BC = BE + EC

= 4.2 + 4

= 8.2 m

∴ The width of the street is 8.2 m.

Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 2 Pythagoras Theorem Practice Set 2.1

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Updated: April 2, 2022 — 5:02 pm