Std 10 Mathematics Part 2 · Chapter 2

2. Pythagoras Theorem

Solutions to Practice Sets 2.1, 2.2 and Problem Set 2, with figures and proofs. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 2 (Geometry).

Practice Set 2.1

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

A triplet is Pythagorean if the square of the largest number equals the sum of the squares of the other two.

(i) (3, 5, 4)

3² + 4² = 9 + 16 = 25 = 5²

∴ (3, 5, 4) is a Pythagorean triplet.

(ii) (4, 9, 12)

4² + 9² = 16 + 81 = 97, but 12² = 144

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

(iii) (5, 12, 13)

5² + 12² = 25 + 144 = 169 = 13²

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

(iv) (24, 70, 74)

24² + 70² = 576 + 4900 = 5476 = 74²

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

(v) (10, 24, 27)

10² + 24² = 100 + 576 = 676, but 27² = 729

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

(vi) (11, 60, 61)

11² + 60² = 121 + 3600 = 3721 = 61²

∴ (11, 60, 61) is a Pythagorean triplet.

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

Fig. 2.17

In Δ MNP, ∠ MNP = 90° and seg NQ ⊥ hypotenuse MP.

NQ² = MQ × QP … theorem of geometric mean

= 9 × 4 = 36

∴ NQ = 6

3In figure 2.18, ∠ QPR = 90°, seg PM ⊥ seg QR and Q − M − R, PM = 10, QM = 8, find QR.

Fig. 2.18

In Δ QPR, ∠ QPR = 90° and seg PM ⊥ hypotenuse QR.

PM² = QM × MR … theorem of geometric mean

∴ 10² = 8 × MR

∴ MR = 1008 = 12.5

QR = QM + MR … Q − M − R

= 8 + 12.5 = 20.5

∴ QR = 20.5

4See figure 2.19. Find RP and PS using the information given in Δ PSR.

Fig. 2.19

In Δ PSR, ∠ S = 90°, ∠ P = 30°, so ∠ R = 60°.

By the 30° − 60° − 90° theorem,

RS = 12 × RP … side opposite 30°

∴ 6 = 12 × RP

∴ RP = 12

PS = √32 × RP … side opposite 60°

= √32 × 12 = 6√3

∴ RP = 12 and PS = 6√3

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

Fig. 2.20

AB = BC … given

∴ ∠ BAC = 45°

∴ AB = BC = 1√2 × AC

= 1√2 × √8

= 1√2 × 2√2

= 2

In Δ ABC, ∠ B = 90° and AB = BC, so ∠ BAC = ∠ BCA = 45°. By the 45° − 45° − 90° theorem, each side making the right angle is 1√2 times the hypotenuse.

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

Let the side of the square be x cm.

A diagonal divides the square into two right angled triangles.

x² + x² = 10² … Pythagoras theorem

∴ 2x² = 100

∴ x² = 50

∴ x = √50 = 5√2

Perimeter = 4 × 5√2 = 20√2

∴ Side = 5√2 cm and perimeter = 20√2 cm

7In figure 2.21, ∠ DFE = 90°, FG ⊥ ED. If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF.

Fig. 2.21

(1) EG

In Δ DFE, ∠ DFE = 90° and seg FG ⊥ hypotenuse ED.

FG² = EG × GD … theorem of geometric mean

∴ 12² = EG × 8

∴ EG = 1448 = 18

∴ EG = 18

(2) FD

In Δ FGD, ∠ FGD = 90°.

FD² = FG² + GD² … Pythagoras theorem

= 144 + 64 = 208

∴ FD = √208 = √16 × 13 = 4√13

∴ FD = 4√13

(3) EF

In Δ FGE, ∠ FGE = 90°.

EF² = EG² + FG² … Pythagoras theorem

= 324 + 144 = 468

∴ EF = √468 = √36 × 13 = 6√13

∴ EF = 6√13

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

Each angle of a rectangle is 90°. A diagonal is the hypotenuse of a right angled triangle with sides 35 cm and 12 cm.

Diagonal² = 35² + 12² … Pythagoras theorem

= 1225 + 144 = 1369

∴ Diagonal = 37

∴ The diagonal is 37 cm

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

Fig. 2.22

Proof: In Δ PRQ, ∠ PRQ = 90°.

∴ PQ² = PR² + RQ² …… (I) … Pythagoras theorem

RQ = 2RM … M is the midpoint of QR

∴ RQ² = 4RM² …… (II)

In Δ PRM, ∠ PRM = 90°.

∴ PM² = PR² + RM² … Pythagoras theorem

∴ RM² = PM² − PR² …… (III)

∴ PQ² = PR² + 4RM² … from (I) and (II)

= PR² + 4(PM² − PR²) … from (III)

= PR² + 4PM² − 4PR²

∴ PQ² = 4PM² − 3PR²

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.

O is the foot of the ladder. A and D are the windows.

In Δ ABO, ∠ ABO = 90°.

OB² = OA² − AB² … Pythagoras theorem

= 5.8² − 4² = 33.64 − 16 = 17.64

∴ OB = 4.2 m

In Δ DCO, ∠ DCO = 90°.

OC² = OD² − DC² … Pythagoras theorem

= 5.8² − 4.2² = 33.64 − 17.64 = 16

∴ OC = 4 m

Width of the street = BC = OB + OC = 4.2 + 4 = 8.2

∴ The width of the street is 8.2 m

Practice Set 2.2

1In 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 a median.

PQ² + PR² = 2PS² + 2QS² … Apollonius theorem

∴ 11² + 17² = 2 × 13² + 2QS²

∴ 121 + 289 = 338 + 2QS²

∴ 2QS² = 410 − 338 = 72

∴ QS² = 36

∴ QS = 6

QR = 2 × QS = 2 × 6 = 12

∴ QR = 12

2In 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. M is the midpoint of AB, so AM = 102 = 5.

CA² + CB² = 2CM² + 2AM² … Apollonius theorem

∴ 7² + 9² = 2CM² + 2 × 5²

∴ 49 + 81 = 2CM² + 50

∴ 2CM² = 80

∴ CM² = 40

∴ CM = √40 = 2√10

∴ The median from C is 2√10

3In the figure 2.28, seg PS is the median of Δ PQR and PT ⊥ QR. Prove that, (1) PR² = PS² + QR × ST + (QR2)² (2) PQ² = PS² − QR × ST + (QR2)²

Fig. 2.28

QS = SR = QR2 … S is the midpoint of QR

In Δ PTS, ∠ PTS = 90°. ∴ PT² + ST² = PS² …… (I) … Pythagoras theorem

(1)

In Δ PTR, ∠ PTR = 90°.

PR² = PT² + TR² … Pythagoras theorem

= PT² + (ST + SR)² … T − S − R

= PT² + ST² + 2ST × SR + SR²

= PS² + 2ST × SR + SR² … from (I)

= PS² + 2ST × QR2 + (QR2)²

∴ PR² = PS² + QR × ST + (QR2)²

(2)

In Δ PTQ, ∠ PTQ = 90°.

PQ² = PT² + QT² … Pythagoras theorem

= PT² + (QS − ST)² … Q − T − S

= PT² + ST² − 2ST × QS + QS²

= PS² − 2ST × QS + QS² … from (I)

= PS² − 2ST × QR2 + (QR2)²

∴ PQ² = PS² − QR × ST + (QR2)²

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

Fig. 2.29

In Δ ABC, seg AM is a median.

AB² + AC² = 2AM² + 2BM² … Apollonius theorem

∴ 290 = 2 × 8² + 2BM²

∴ 290 = 128 + 2BM²

∴ 2BM² = 162

∴ BM² = 81

∴ BM = 9

BC = 2 × BM = 2 × 9 = 18

∴ BC = 18 cm

5★In figure 2.30, 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.)

Fig. 2.30

Construction: Draw seg AB ∥ side SR through T, with A on side PS and B on side QR.

Proof: □ PQRS is a rectangle and seg AB ∥ side SR.

∴ seg AB ⊥ side PS and seg AB ⊥ side QR.

∴ □ APQB and □ ASRB are rectangles.

∴ AP = BQ and AS = BR …… (I) … opposite sides of rectangles

In right angled Δ TAS and Δ TBQ,

TS² = TA² + AS² and TQ² = TB² + BQ² … Pythagoras theorem

∴ TS² + TQ² = TA² + AS² + TB² + BQ² …… (II)

In right angled Δ TAP and Δ TBR,

TP² = TA² + AP² and TR² = TB² + BR² … Pythagoras theorem

∴ TP² + TR² = TA² + AP² + TB² + BR²

= TA² + BQ² + TB² + AS² …… (III) … from (I)

∴ TS² + TQ² = TP² + TR² … from (II) and (III)

Problem Set 2

1Some questions and their alternative answers are given. Select the correct alternative.

(1) Out of the following which is the Pythagorean triplet? — (B) (3, 4, 5)
3² + 4² = 25 = 5².
(2) In a right angled triangle, if sum of the squares of the sides making right angle is 169 then what is the length of the hypotenuse? — (B) 13
Hypotenuse² = 169, so hypotenuse = 13.
(3) Out of the dates given below which date constitutes a Pythagorean triplet? — (A) 15/08/17
8² + 15² = 64 + 225 = 289 = 17².
(4) If a, b, c are sides of a triangle and a² + b² = c², name the type of triangle. — (C) Right angled triangle
By the converse of Pythagoras theorem.
(5) Find perimeter of a square if its diagonal is 10√2 cm. — (D) 40 cm
Side = 10√2√2 = 10 cm, so perimeter = 4 × 10 = 40 cm.
(6) Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude. — (C) 6 cm
Altitude² = 4 × 9 = 36 (theorem of geometric mean), so altitude = 6 cm.
(7) Height and base of a right angled triangle are 24 cm and 18 cm, find the length of its hypotenuse. — (B) 30 cm
Hypotenuse² = 24² + 18² = 576 + 324 = 900, so hypotenuse = 30 cm.
(8) In Δ ABC, AB = 6√3 cm, AC = 12 cm, BC = 6 cm. Find measure of ∠ A. — (A) 30°
AB² + BC² = 108 + 36 = 144 = AC², so ∠ B = 90°. BC is half of hypotenuse AC, so the angle opposite BC, ∠ A, is 30°.

2Solve the following examples.

(1) Find the height of an equilateral triangle having side 2a.

Let Δ ABC be equilateral with side 2a. Draw AD ⊥ BC. D is the midpoint of BC, so BD = a.

In Δ ADB, AD² = AB² − BD² … Pythagoras theorem

= (2a)² − a² = 4a² − a² = 3a²

∴ AD = √3a

∴ Height = √3a

(2) Do sides 7 cm, 24 cm, 25 cm form a right angled triangle? Give reason.

7² + 24² = 49 + 576 = 625 and 25² = 625

∴ 7² + 24² = 25²

∴ Yes, they form a right angled triangle. … converse of Pythagoras theorem

(3) Find the length of a diagonal of a rectangle having sides 11 cm and 60 cm.

Diagonal² = 11² + 60² … Pythagoras theorem

= 121 + 3600 = 3721

∴ Diagonal = 61 cm

(4) Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.

Hypotenuse² = 9² + 12² … Pythagoras theorem

= 81 + 144 = 225

∴ Hypotenuse = 15 cm

(5) A side of an isosceles right angled triangle is x. Find its hypotenuse.

Both sides making the right angle are x.

Hypotenuse² = x² + x² = 2x² … Pythagoras theorem

∴ Hypotenuse = √2x

(6) In Δ PQR; PQ = √8, QR = √5, PR = √3. Is Δ PQR a right angled triangle? If yes, which angle is of 90°?

PQ² = 8, QR² = 5, PR² = 3

QR² + PR² = 5 + 3 = 8 = PQ²

∴ Δ PQR is a right angled triangle. … converse of Pythagoras theorem

PQ is the hypotenuse, so the angle opposite PQ is 90°.

∴ Yes, ∠ R = 90°

3In Δ RST, ∠ S = 90°, ∠ T = 30°, RT = 12 cm then find RS and ST.

In Δ RST, ∠ S = 90°, ∠ T = 30°, so ∠ R = 60°.

By the 30° − 60° − 90° theorem,

RS = 12 × RT … side opposite 30°

= 12 × 12 = 6

ST = √32 × RT … side opposite 60°

= √32 × 12 = 6√3

∴ RS = 6 cm and ST = 6√3 cm

4Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm.

Area of rectangle = length × breadth

∴ 192 = 16 × breadth

∴ Breadth = 19216 = 12 cm

Diagonal² = 16² + 12² … Pythagoras theorem

= 256 + 144 = 400

∴ Diagonal = 20 cm

5★Find the length of the side and perimeter of an equilateral triangle whose height is √3 cm.

Let the side of the equilateral triangle be x cm. The height divides it into two 30° − 60° − 90° triangles.

Height = √32 × side … side opposite 60°

∴ √3 = √32 × x

∴ x = 2

Perimeter = 3 × 2 = 6 cm

∴ Side = 2 cm and perimeter = 6 cm

6In Δ ABC, seg AP is a median. If BC = 18, AB² + AC² = 260, find AP.

P is the midpoint of BC, so BP = 182 = 9.

AB² + AC² = 2AP² + 2BP² … Apollonius theorem

∴ 260 = 2AP² + 2 × 81

∴ 2AP² = 260 − 162 = 98

∴ AP² = 49

∴ AP = 7

7★Δ ABC is an equilateral triangle. Point P is on base BC such that PC = 13 BC. If AB = 6 cm find AP.

AM ⊥ BC is drawn.

Construction: Draw AM ⊥ BC.

BC = AB = 6 cm. M is the midpoint of BC, so BM = MC = 3 cm.

PC = 13 × 6 = 2 cm. ∴ MP = MC − PC = 3 − 2 = 1 cm

AM = √32 × AB = √32 × 6 = 3√3 cm … 30° − 60° − 90° theorem

In Δ AMP, ∠ AMP = 90°.

AP² = AM² + MP² … Pythagoras theorem

= (3√3)² + 1² = 27 + 1 = 28

∴ AP = √28 = 2√7

∴ AP = 2√7 cm

8From the information given in the figure 2.31, prove that PM = PN = √3 × a.

Fig. 2.31: PQ = PR = QR = a, MQ = RN = a

Proof: PQ = PR = QR = a, so Δ PQR is equilateral.

PS ⊥ QR, so S is the midpoint of QR. ∴ QS = SR = a2

PS = √32 × PQ = √3a2 … 30° − 60° − 90° theorem

MS = MQ + QS = a + a2 = 3a2

In Δ PSM, ∠ PSM = 90°.

PM² = PS² + MS² … Pythagoras theorem

= 3a²4 + 9a²4 = 12a²4 = 3a²

∴ PM = √3 × a

Similarly, NS = NR + RS = 3a2, so PN = √3 × a.

∴ PM = PN = √3 × a

9Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

□ ABCD is a parallelogram. Its diagonals meet at O.

Given: □ ABCD is a parallelogram. Diagonals AC and BD intersect at O.

To prove: AB² + BC² + CD² + DA² = AC² + BD²

Proof: Diagonals of a parallelogram bisect each other.

∴ AO = OC = AC2 and BO = OD = BD2

In Δ ABD, seg AO is a median.

∴ AB² + AD² = 2AO² + 2BO² …… (I) … Apollonius theorem

In Δ CBD, seg CO is a median.

∴ CB² + CD² = 2CO² + 2BO² …… (II) … Apollonius theorem

Adding (I) and (II),

AB² + BC² + CD² + DA² = 2AO² + 2CO² + 4BO²

= 4AO² + 4BO² … AO = CO

= (2AO)² + (2BO)²

= AC² + BD²

10Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 15√2 km. Find their speed per hour.

They walk at the same speed, so each walks the same distance, say x km, in 2 hours.

East and North are at right angles.

x² + x² = (15√2)² … Pythagoras theorem

∴ 2x² = 450

∴ x² = 225

∴ x = 15 km

Speed = 152 = 7.5 km per hour

∴ The speed of each is 7.5 km/hr

11★In Δ ABC, ∠ BAC = 90°, seg BL and seg CM are medians of Δ ABC. Then prove that: 4(BL² + CM²) = 5BC².

Fig. 2.32

Proof: L is the midpoint of AC and M is the midpoint of AB.

∴ AL = AC2 and AM = AB2

In Δ BAL, ∠ BAL = 90°.

∴ BL² = AB² + AL² = AB² + AC²4 … Pythagoras theorem

∴ 4BL² = 4AB² + AC² …… (I)

In Δ CAM, ∠ CAM = 90°.

∴ CM² = AC² + AM² = AC² + AB²4 … Pythagoras theorem

∴ 4CM² = 4AC² + AB² …… (II)

Adding (I) and (II),

4(BL² + CM²) = 5AB² + 5AC² = 5(AB² + AC²)

But AB² + AC² = BC² … Pythagoras theorem in Δ BAC

∴ 4(BL² + CM²) = 5BC²

12Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.

Let the sides be a and b, and the other diagonal be d cm. Then a² + b² = 130.

Sum of squares of diagonals = sum of squares of the four sides … proved in Q 9

∴ 14² + d² = 2(a² + b²)

∴ 196 + d² = 2 × 130 = 260

∴ d² = 64

∴ d = 8

∴ The other diagonal is 8 cm

13In Δ ABC, seg AD ⊥ seg BC, DB = 3CD. Prove that: 2AB² = 2AC² + BC².

Fig. 2.33

Proof: BC = CD + DB = CD + 3CD = 4CD … C − D − B

∴ BC² = 16CD² …… (I)

In Δ ADB, AB² = AD² + DB² = AD² + 9CD² …… (II) … Pythagoras theorem

In Δ ADC, AC² = AD² + CD² …… (III) … Pythagoras theorem

Subtracting (III) from (II),

AB² − AC² = 8CD² = BC²2 … from (I)

∴ 2AB² − 2AC² = BC²

∴ 2AB² = 2AC² + BC²

14★In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.

G is the centroid of Δ ABC.

Let Δ ABC have AB = AC = 13 cm and BC = 10 cm. Let AD be the median to BC, and G the centroid.

In an isosceles triangle, the median to the base is perpendicular to it. BD = 5 cm.

In Δ ADB, AD² = AB² − BD² … Pythagoras theorem

= 169 − 25 = 144

∴ AD = 12 cm

The centroid divides a median in the ratio 2 : 1 from the vertex.

∴ AG = 23 × AD = 23 × 12 = 8

∴ The required distance is 8 cm

15In a trapezium ABCD, seg AB ∥ seg DC, seg BD ⊥ seg AD, seg AC ⊥ seg BC. If AD = 15, BC = 15 and AB = 25. Find A(□ ABCD).

Fig. 2.34: DM ⊥ AB and CN ⊥ AB are drawn.

Construction: Draw DM ⊥ AB and CN ⊥ AB.

In Δ ADB, ∠ ADB = 90°.

BD² = AB² − AD² = 625 − 225 = 400 … Pythagoras theorem

∴ BD = 20

A(Δ ADB) = 12 × AD × BD = 12 × AB × DM

∴ 15 × 20 = 25 × DM

∴ DM = 12

In Δ DMA, AM² = AD² − DM² = 225 − 144 = 81 … Pythagoras theorem

∴ AM = 9

In the same way, in Δ ACB, AC = 20, CN = 12 and BN = 9.

□ DMNC is a rectangle, so DC = MN = AB − AM − BN = 25 − 9 − 9 = 7

A(□ ABCD) = 12 × (AB + DC) × DM

= 12 × (25 + 7) × 12 = 192

∴ A(□ ABCD) = 192 sq. units

16★In the figure 2.35, Δ PQR is an equilateral triangle. Point S is on seg QR such that QS = 13 QR. Prove that: 9PS² = 7PQ².

Fig. 2.35

Proof: Let PQ = QR = PR = a. Draw PT ⊥ QR.

T is the midpoint of QR, so QT = a2. QS = a3.

∴ ST = QT − QS = a2 − a3 = a6

PT = √32a … 30° − 60° − 90° theorem

In Δ PTS, ∠ PTS = 90°.

PS² = PT² + ST² … Pythagoras theorem

= 3a²4 + a²36 = 27a² + a²36 = 28a²36 = 7a²9

∴ 9PS² = 7a²

∴ 9PS² = 7PQ²

17★Seg PM is a median of Δ PQR. If PQ = 40, PR = 42 and PM = 29, find QR.

M is the midpoint of QR.

PQ² + PR² = 2PM² + 2QM² … Apollonius theorem

∴ 1600 + 1764 = 2 × 841 + 2QM²

∴ 3364 = 1682 + 2QM²

∴ 2QM² = 1682

∴ QM² = 841

∴ QM = 29

QR = 2 × QM = 2 × 29 = 58

∴ QR = 58

18Seg AM is a median of Δ ABC. If AB = 22, AC = 34, BC = 24, find AM.

M is the midpoint of BC, so BM = 12.

AB² + AC² = 2AM² + 2BM² … Apollonius theorem

∴ 484 + 1156 = 2AM² + 2 × 144

∴ 1640 = 2AM² + 288

∴ 2AM² = 1352

∴ AM² = 676

∴ AM = 26

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