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