Selina Concise Class 7 Math Chapter 16 Pythagoras Theorem Exercise 16 Solution

Selina Concise Class 7 Math Chapter 16 Pythagoras Theorem Exercise 16 Solution

EXERCISE 16

(1) Applying Pythagoras theorem,

BC2 = AB2 + AC2

⇒ BC2 = (24)2 + (18)2

⇒ BC2 = 576 + 324 = 900

⇒ BC = √900 = 30 cm

(2) Applying Pythagoras theorem,

XY2 = XZ2 + ZY2

⇒ XZ2 + ZY2 = XY2

⇒ (12)2 + ZY2 = (13)2

⇒ ZY2 = (13)2 – (12)2

⇒ ZY2 = 169 – 144 = 25

⇒ ZY = √25 = 5 cm

(3) Applying Pythagoras theorem,

PQ2 = QR2 + PR2

⇒ QR2 + PR2 = PQ2

⇒ (33.6)2 + PR2 = (34)2

⇒ PR2 = (34)2 – (33.6)2

⇒ PR2 = 1156 – 1128.96 = 27.04

⇒ PR = √27.04 = 5.2 cm

(4) The given triangle will be a right-angled triangle if square on its largest side is equal to the sum of the squares on the other two sides.

(i) (20)2 = (16)2 + (12)2

⇒ 400 = 256 + 144

⇒ 400 = 400

So, the triangle is a right-angled triangle.

(ii) (13)2 = (9)2 + (6)2

⇒ 169 = 81 + 36

⇒ 169 ≠ 117

So, the triangle is not a right-angled triangle.

(5) Applying Pythagoras theorem,

BC2 = AB2 + AC2

⇒ BC2 = (400)2 + (300)2

⇒ BC2 = 160000 + 90000

⇒ BC2 = 250,000

⇒ BC = √250,000 = 500

(6) (i) In ∆PAC,

Applying Pythagoras Theorem,

PA2 = AC2 + CP2

⇒ AC2 + CP2 = PA2

⇒ (12)2 + CP2 = (15)2

⇒ CP2 = (15)2 – (12)2

⇒ CP2 = 225 – 144 = 81

⇒ CP = √81 = 9 m

(ii) In ∆PDB,

Applying Pythagoras Theorem,

PB2 = PD2 + BD2

⇒ PD2 + BD2 = PB2

⇒ PD2 = (15)2 – (9)2

⇒ PD2 = 225 – 81 = 144

⇒ PD = √144 = 12 m

(iii) CD = CP + PD = (9 + 12) = 21 m

(7) (i) If, PQ = 8 cm and QR = 6 cm

Then, PR2 = PQ2 + QR2

⇒ PR2 = (8)2 + (6)2

⇒ PR2 = 64 + 36 = 100

⇒ PR = √100 = 10 cm

(ii) If PR = 34 cm and QR = 30 cm

Then, PR2 = PQ2 + QR2

⇒ (34)2 = PQ2 + (30)2

⇒ PQ2 = (34)2 – (30)2

⇒ PQ2 = 1156 – 900 = 256

⇒ PQ = √256 = 16 cm

(8) The given triangle will be a right-angled triangle if square on its largest side is equal to the sum of the squares on the other two sides.

Then, (41)2 = (40)2 + (9)2

⇒ 1681 = 1600 + 81

⇒ 1681 = 1681

So, the triangle is a right-angled triangle.

(9) (i) In ∆ABC,

AB2 = AC2 + BC2

⇒ (10)2 = AC2 + (6)2

⇒ AC2 = (10)2 – (6)2

⇒ AC2 = 100 – 36 = 64

⇒ AC = √64 = 8 cm

(ii) In ∆ADC,

AD2 = AC2 + CD2

⇒ (17)2 = (8)2 + CD2

⇒ CD2 = (17)2 – (8)2

⇒ CD2 = 289 – 64 = 225

⇒ CD = √225 = 15 cm

(10) In ∆ACD,

AC2 = AD2 + CD2

⇒ (26)2 = (24)2 + CD2

⇒ CD2 = (26)2 – (24)2

⇒ CD2 = 676 – 576 = 100

⇒ CD = √100 = 10 cm

Given, BD = DC = 10 cm

Therefore, BC = BD + DC = 10 + 10 = 20 cm

(11) In ∆ABC,

AC2 = AB2 + BC2

⇒ AC2 = (3)2 + (12)2

⇒ AC2 = 9 + 144 = 153

⇒ AC = √153 cm

Now, in ∆ADC

AD2 = DC2 + AC2

⇒ (13)2 = DC2 + (√153)2

⇒ DC2 = (13)2 – (√153)2

⇒ DC2 = 169 – 153 = 16

⇒ DC = √16 = 4 cm

(12) BC2 = AB2 + AC2

⇒ (6.5)2 = (2.5)2 + AC2

⇒ AC2 = (6.5)2 – (2.5)2

⇒ AC2 = 42.25 – 5 = 36

⇒ AC = √36 = 6 m

(13) BC2 = AB2 + AC2

⇒ BC2 = (12)2 + (5)2

⇒ BC2 = 144 + 25 = 169

⇒ BC = √169 = 13 m

(14) Here, DO = BC = 24 cm

And, DC = OB = 10 cm

Then, AO = AB – OB = 20 – 10 = 10 cm

Now, in ∆AOD,

AD2 = OD2 + OA2

⇒ AD2 = (24)2 + (10)2

⇒ AD2 = 576 + 100 = 676

⇒ AD = √676 = 26 cm

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