Active Maths 1 – Strands 1–5 – Ch 24 Solutions 1

Chapter 24 Exercise 24.1

Q. 1. (i) (AB, EF), (BC, DF), (AC, ED)

(ii) (VW, SQ), (WU, QR), (UV, RS)

(iii) (XY, VW), (YZ, WU), (XZ, VU)

(iv) (AB, LM), (BC, MK), (AC, LK)

(v) (DE, ST), (EF, TR), (DF, SR)

Q. 2. (i) x ___ 10

= 14 ___ 16

⇒ x = 140 ____ 16 = 8.75

(ii) x __ 6 = 15 ___ 5

⇒ x = 90 ___ 5 = 18

(iii) x ___ 15 = 16 ___ 20

⇒ x = 240 ____ 20 = 12

(iv) x ___ 20 = 8 ___ 20

⇒ x = 8

(v) x ___ 35 = 12 ___ 28

⇒ x = 420 ____ 28 = 15

Q. 3. (i) y ___ 10 = 16 ___ 12

⇒ y = 160 ____ 12 = 13 1 __ 3

(ii) y __ 9 = 20 ___ 15

⇒ y = 180 ____ 15 = 12

(iii) y ___ 10 = 18 ___ 16

⇒ y = 180 ____ 16 = 11.25

(iv) y ___ 11 = 12.5 ____ 10

⇒ y = 137.5 ______ 10 = 13.75

(v) y ___ 17 = 40 ___ 12

⇒ y = 680 ____ 12 = 56 2 __ 3

Q. 4. (i) x __ 6

= 10.5 _____ 7

y ___ 10 = 10.5 ____ 7

⇒ x = 63 ___ 7

= 9 ⇒ y = 105 ____ 7

= 15

(ii) x ______ 14.75 = 8 ___ 10 y ____ 9.25 = 8 ___ 10

⇒ x = 118 ____ 10 = 11.8 ⇒ y = 74 ___ 10

= 7.4

(iii) x ___ 7.5 = 42 ___ 18 y ___ 17 = 42 ___ 18

⇒ x = 315 ____ 18 = 17.5 ⇒ y = 714 ____ 18

= 39 2 __ 3

(iv) x ___ 40 = 14.75 ______ 73.75 y ______ 10.25 = 40 ___ 8 = 5

⇒ x = 590 ______ 73.75 = 8 ⇒ y = 51.25

(v) x ___ 6.3 = 20 ___ 12 y ______ 18.25 = 12 ___ 20

⇒ x = 126 ____ 12 = 10.5 ⇒ y = 219 ____ 20

= 10.95

Q. 5.

x

6 25

7

Q

Q 4 (i) x 10.5 y 10.5

x __ 7 = 6 ___ 25 ⇒ 25x = 42

x = 1.68 m

2 Active Maths 1 – Strands 1–5 – Ch 24 Solutions

Q. 6.

1.6 m

4 10

x

x ___ 1.6 = 10 ___ 4 ⇒ 4x = 16 x = 4 m

Q. 7. 20

100 m

15

x

x ____ 100 = 15 ___ 20 ⇒ 20x = 1500

x = 75 m

100 – 75 = 25 m apart

Exercise 24.2

Q. 1. (i) Yes (SSS) (v) No

(ii) No (vi) Yes (ASA)

(iii) Yes (SSS) (vii) No

(iv) Yes (RHS) (viii) No

Q. 2. (i) Yes

|AC| = |CD| S

|∠ACB| = |∠BCD| A

BC is a common side S

(ii) Yes (SSS)

(iii) Yes (RHS)

(iv) No ∠SRT is not the included angle

(v) |∠LNH| = 180° – (45° + 65°) = 70°

|LN| = |BC| S

|∠LNH| = |∠ABC| A

|NH| = |AB| S

∴ Yes (SAS)

(vi) Yes ASA

(vii) Yes (SSS)

(viii) No

Q. 3. (i) 8 cm, 10 cm

(ii) Hypotenuse /diagonal

(iii) |AB| = |DC| ... rectangle

|AD| = |BC| ... rectangle

|BD| is a common side congruent (SSS)

Q. 4. (i) 8 cm, 7 cm

(ii) 60° … parallelogram

(iv) |ED| = |FG| … parallelogram

|∠EDG| = |∠EFG| = 60°

|DG| = |EF| = 8 cm

∴ ΔEDG ≡ ΔEFG (SAS)

Q. 5. |AB| = |CD| = 3 cm

|∠BAE| = |∠CDE| ... alternate

|∠ABE| = |∠ECD| ... alternate

⇒ ΔABE ≡ ΔCDE (ASA)

Exercise 24.3

Q. 1. (i) a = 8, b = 6, h = ?

h2 = a2 + b2

⇒ h2 = 82 + 62

⇒ h2 = 64 + 36

⇒ h2 = 100

⇒ h = √____

100

⇒ h = 10

(ii) a = 8, b = 15, h = ?

h2 = a2 + b2

⇒ h2 = 82 + 152

⇒ h2 = 64 + 225

⇒ h2 = 289

⇒ h = √____

289

⇒ h = 17

3Active Maths 1 – Strands 1–5 – Ch 24 Solutions

(iii) a = 7, b = 24, h = ?

h2 = a2 + b2

⇒ h2 = 72 + 242

⇒ h2 = 49 + 576

⇒ h2 = 625

⇒ h = √____

625

⇒ h = 25

(iv) a = 9, b = 12, h = ?

h2 = a2 + b2

⇒ h2 = 92 + 122

⇒ h2 = 81 + 144 = 225

⇒ h = 15

(v) a = 16, b = 63, h = ?

h2 = a2 + b2

⇒ h2 = 162 + 632

⇒ h2 = 256 + 3969 = 4225

⇒ h = 65

(vi) √_________

442 + 332 = 55 m

(vii) √_________

202 + 212 = 29 cm

(viii) √_________

392 + 522 = 65 mm

(ix) √_________

652 + 722 = 97 cm

(x) √_________

452 + 282 = 53 m

Q. 2. (i) a = 10, b = ?, h = 26

h2 = a2 + b2 ⇒ b2 = h2 – a2 ← use this

⇒ b2 = 262 – 102

⇒ b2 = 676 – 100

⇒ b2 = 576

⇒ b = √____

576

⇒ b = 24

(ii) a = 24, b = ?, h = 25

b2 = h2 – a2

⇒ b2 = 252 – 242

⇒ b2 = 625 – 576 = 49

⇒ b = 7

(iii) a = 12, b = ?, h =37

b2 = h2 – a2

⇒ b2 = 372 – 122

⇒ b2 = 1369 – 144

⇒ b2 = 1225

⇒ b = √_____

1225

⇒ b = 35

(iv) a = 10, b = ?, h = 12.5

b2 = h2 – a2

⇒ b2 = 12.52 – 102

⇒ b2 = 156.25 – 100

⇒ b2 = 56.25

⇒ b = 7.5

(v) a = 36, b = ?, h = 39

b2 = h2 – a2

⇒ b2 = 392 – 362

⇒ b2 = 1521 – 1296

⇒ b2 = 225

⇒ b = 15

(vi) √__________

1652 – 992 = 132 cm

(vii) √___________

1702 – 1502 = 80 m

(viii) √_________

912 – 842 = 35 m

(ix) √__________

1002 – 802 = 60 mm

(x) √__________

1702 – 802 = 150 cm

Q. 3. (Longest side)2 (Side)2 + (Other side)2 Right-angled (i) 902 = 8,100 652 + 722 = 9,409 No (ii) 1032 = 10,609 402 + 702 = 6,500 No(iii) 852 = 7,225 362 + 772 = 7,225 Yes(iv) 1012 = 10,201 202 + 992 = 10,201 Yes (v) 822 = 6,724 802 + 142 = 6,596 No

4 Active Maths 1 – Strands 1–5 – Ch 24 Solutions

Q. 4. (i) a = 3, b = 9, h = ?

h2 = a2 + b2

⇒ h2 = 32 + 92

⇒ h2 = 9 + 81 = 90

⇒ h = 9.49

(ii) a = 7, b = ?, h = 8

b2 = h2 – a2

⇒ b2 = 82 – 72

⇒ b2 = 64 – 49 = 15

⇒ b = 3.87

(iii) a = 2, b = 2, h = ?

h2 = a2 + b2

⇒ h2 = 22 + 22

⇒ h2 = 4 + 4 = 8

⇒ h = 2.83

(iv) a = 13, b = ?, h = 25

b2 = h2 – a2

⇒ b2 = 252 – 132

⇒ b2 = 625 – 169

⇒ b2 = 456

⇒ b = 21.35

(v) a = 14, b = ?, h = 30

b2 = h2 – a2

⇒ b2 = 302 – 142

⇒ b2 = 900 – 196

⇒ b2 = 704

⇒ b = 26.53

(vi) √__________

1002 + 402 = 107.70 m

(vii) √__________

1032 – 452 = 92.65 cm

(viii) √_________

702 – 502 = 48.99 km

(ix) √_____________

5.722 + 4.552 = 7.31 cm

(x) √______________

80.782 – 40.152 = 70.10 cm

Q. 5. (Longest side)2 (Side)2 + (Other side)2 Right-angled (i) 612 = 3,721 112 + 602 = 3,721 Yes (ii) 302 = 900 202 + 242 = 976 No(iii) 802 = 6,400 362 + 772 = 7,225 No(iv) 4332 = 187,489 1452 + 4082 = 187,489 Yes (v) 2772 = 76,729 1152 + 2522 = 76,729 Yes

Q. 6. (i) We have 2 sides in the bottom Δ

a = 12, b = 16, h = x

h2 = a2 + b2

⇒ x2 = 122 + 162

⇒ x2 = 144 + 256

⇒ x2 = 400

⇒ x = 20

Top Δ:

a = x = 20, b = y, h = 29

b2 = h2 – a2

⇒ y2 = 292 – 202

⇒ y2 = 841 – 400

⇒ y2 = 441

⇒ y = 21

(ii) a = 4, b = 3, h = x

h2 = a2 + b2

⇒ x2 = 42 + 32

⇒ x2 = 16 + 9 = 25

⇒ x = 5

a = 5, b = y, h = 13

b2 = h2 + a2

⇒ y2 = 132 – 52

⇒ y2 = 169 – 25 = 144

⇒ y = 12

(iii) X = √__________

1202 +272 = 123 m

Y = √___________

1252 – 1232 = 22.27 m

(iv) X = √_________

212 +722 = 75 units

Y = √_________

402 + 752 = 85 units

Z = √__________

1572 – 852 = 132 units

Q 5

Q 6 (i) We have 2 sides in the bottom Δ (ii) a = 4 b = 3 h = x

5Active Maths 1 – Strands 1–5 – Ch 24 Solutions

Q. 7. (i) 22 = 1.62 + x2

⇒ 4 = 2.56 + x2

⇒ x2 = 1.44

⇒ x = 1.2 m

(ii) y2 = 22 – 12

⇒ y2 = 3

⇒ y = √__

3

⇒ y = 1.73 m

Q. 8. Road = (48 – 2) + (14 – 2) = 58 m

Laneway ⇒ x2 = 482 + 142

⇒ x2 = 2304 + 196

⇒ x2 = 2500

⇒ x = 50 m

Total distance using laneway = 50 m + 2 m + 2 m = 54 m

∴ Laneway is shorter

Q. 9. (i) Let d = diameter

d2 = 182 + 242

⇒ d2 = 324 + 576

⇒ d2 = 900

⇒ d = 30

⇒ radius = 15 m

(ii) x2 + x2 = 302

⇒ 2x2 = 900

⇒ x2 = 450

⇒ x = 21.21 m

Perimeter = 18 + 24 + 21.21 + 21.21= 84.42 m

Cost = €9.50 × 84.42 = €802

Exercise 24.4

Q. 1. (i) |∠A| = 90° … angle in a semicircle (iii) |∠A| = 90° … angle in a semicircle

BC is the diameter BC is the diameter

(ii) |∠B| = 90° … angle in a semicircle (iv) |∠B| = 90° AC is the diameter

AC is the diameter

Q. 2. (i) |∠A| = 90° … angle in a semicircle

|∠B| = 180° – (25° +90°) = 65° … angle in a Δ

(ii) |∠B| = 90° … angle in a semicircle

|∠A| = 180° – (90° + 72°) = 18° … angles in a Δ

(iii) |∠A| = |∠B| = 180° – 90° __________ 2 = 45°

Isosceles ∆, angle in a semicircle, angles in a Δ.

(iv) |∠A| = |∠B| = 90° … angles in a semicircle

Q. 3. (i) |∠C| = 90° … angle in a semicircle

|∠D| = 90° + 50° = 140° … external angle

(ii) |∠C| = 90° – 28° = 62° … angle in a semicircle

|∠D| = 180° – (90° + 62°) = 28° … angles in a Δ

(iii) |∠C| = 180° – (90° + 30°) = 60° … angles in a Δ

|∠D| = 180° – (90° + 48°) = 42° … angles in a Δ

(iv) |∠D| = 180° – 114° = 66° … straight angle

|∠C| = 180° – (57° + 66°) = 57° … angles in a Δ and vertically opposite angle

Q

Exercise 24 4

6 Active Maths 1 – Strands 1–5 – Ch 24 Solutionsti th St d Ch 24 S l ti

Q. 4. (i) |∠1| = 1 __ 2

(180° – 60°) = 60° … isosceles Δ

|∠2| = 1 __ 2 (60°) = 30° … exterior angle and isosceles Δ

(ii) |∠1| (90° – 31°) = 59° … angle in a semicircle and isosceles Δ

|∠2| = 180° – (59° + 59°) = 62° … vertically opposite and angles in a Δ

(iii) |∠1| = 40° … isosceles Δ

|∠2| = 40° + 40° = 80° … exterior angle

(iv) |∠1| = 1 __ 2 (180° – 110°) = 35° … angles in a Δ and vertically opposite angle

|∠2| = 90° – 35° = 55° … angle in a semicircle + isosceles Δ

Q. 5. (i) |∠ACE| = 90° … angle in a semicircle

⇒ |∠CAE| = 180° – (90° + 66°) = 24° … angles in a Δ

(ii) |∠ADB| = 90° … angle in a semicircle

|∠BED| = 66° … vertically opposite

⇒ |∠DBE| = 180° – (90° + 66°) = 24° … angles in a Δ

(iii) |∠EAB| = 180° – (90° + 40° + 24°) = 26° … angles in a Δ

(iv) |∠AEB| = 180° – (40° + 26°) = 114° … angles in a Δ

Yes |∠CEA| = |∠BED| = 66°

|∠ACE| = |∠EDB| = 90°

|∠EAC| = |∠EBD| = 24°

Q. 6. (i) |∠A| = 180° – (53° + 53°) = 74° … isosceles Δ

(ii) |∠B| = 74° – 17° = 57° … external angle

(iii) The two angles in the small triangle containing C are 90° – 37° = 53° (angle in a semicircle) and 57° (vertically opposite)

⇒ |∠C| = 180° – (53°+ 57°) = 70° … angle in a Δ

(iv) |∠D| = 180° – (90°+ 70°) = 20° … angle in a ∆

Q. 7. (i) a = 8, b = 6, h = ?

h2 = a2 + b2

⇒ h2 = 82 + 62

⇒ h2 = 64 + 36 = 100

⇒ h = 10

(ii) radius = 10

⇒ diameter = 20

(iii) a = 24, b = |OT|, h = 25

b2 = h2 – a2

⇒ |OT|2 = 252 – 242

⇒ |OT|2 = 625 – 576 = 49

⇒ |OT| = 7

⇒ diameter = 14

7Active Maths 1 – Strands 1–5 – Ch 24 Solutions

Revision Exercises

Q. 1. (a) (i) Yes (RHS)

|∠B| =|∠E| = 90°, |AC| = |DF|, |AB|= |EF|

(ii) Yes (ASA)

|∠Q| =|∠N|, |QR| = |LN|, |∠R| = |∠L|

(iii) Yes (SSS)

|IJ| = |HK|, |JK| = |HI| and IK is a common side

(iv) No 3 equal angles implies similar but not congruent

(v) No YZ is not the included side

(vi) Yes (ASA)

|∠Q| = |∠V|, |PQ| = |UV|, |∠P| = |U| (remaining angle)

(b) (i)

A D B

C

(ii) |AD| = |BD| … D is midpoint

|∠ADC| = |∠CDB| = 90° perpendicular bisector (C is directly above D)

CD is a common side

⇒ Δ ADC ≡ ΔCDB (SAS)

(c)

A D

CB

Q. 2. (a) (i) Equilateral triangle

⇒ x = y = 9

(ii) x __ 5 = 11.875 _______ 9.5 y __ 9 = 11.875 _______ 9.5

⇒ x = 59.375 ________ 95

= 6.25 ⇒ y = 106.875 _________ 9.5

= 11.25

(iii) x ___ 20 = 37.5 ____ 12 y ___ 17 = 37.5 ____ 12

⇒ x = 750 ____ 12

= 62.5 ⇒ y = 637.5 ______ 12

= 53.125

(iv) x ___ 24 = 8 ___ 16 y __ 6 = 16 ___ 8

⇒ x = 192 ____ 16

= 12 ⇒ y = 96 ___ 8

= 12

(v) x ___ 20 = 21 ___ 15 y ___ 28 = 21 ___ 15

⇒ x = 420 ____ 15

= 28 ⇒ y = 588 ____ 15

= 39.2

(i) |AB| = |CD| … opposite side

|AD| = |BC| … opposite side

BD is a common side

⇒ Δ ABD ≡ Δ BCD (SSS)

(ii) The areas of each of the triangles are equal.

(iii) It bisects the area of a parallelogram.

8 Active Maths 1 – Strands 1–5 – Ch 24 SolutionsA ti M th 1 St d 1 5 Ch 24 S l ti

(b) (i) A square

(ii) E F

G

M

D

(c) The three angles are equal (vertically opposite and alternate angles) but the three sides are not the same.

∴ (i) Triangles are similar. (ii) Triangles are not congruent.

Q. 3. (a) (i) a = 21, b = ?, h = 29

b2 = h2 – a2

⇒ b2 = 292 – 212

⇒ b2 = 841 – 441

⇒ b2 = 400

⇒ b = 20

(ii) a = 33, b = 56, h = ?

h2 = a2 + b2

⇒ h2 = 332 + 562

⇒ h2 = 1089 + 3136

⇒ h2 = 4225

⇒ h = 65

(iii) a = 24, b = ?, h = 25

b2 = h2 – a2

⇒ b2 = 252 – 242

⇒ b2 = 625 – 576 = 49

⇒ b = 7

(iv) a = 60, b = ?, h = 61

b2 = h2 – a2

⇒ b2 = 612 – 602

⇒ b2 = 3721 – 3600 = 121

⇒ b = 11

(v) a = √__

2 , b = √__

2 , h = ?

h2 = a2 + b2

⇒ h2 = ( √__

2 ) 2 + ( √__

2 ) 2

⇒ h2 = 2 + 2 = 4 ⇒ h = 2

(iii) |∠D| = |∠F| = 90°

EG is a common side and the hypotenuse. Also, |ED| = |FG|.

⇒ ΔEDG ≡ ΔEFG (RHS)

(v) ΔFMG, ΔDMG, ΔDEM

Q. 3. (a) (i) a = 21, b = ?, h = 29 (iii) a = 24, b = ?, h = 25

(b) (i) p ______

21.75 = 20 ___

30

q ___

14 = 21.75 ______

14.5

⇒ p = 435 ____ 30

= 14.5 ⇒ q = 304.5 ______ 14.5

= 21

(ii)

20 p

17.25

35

q

p + 18.75

q ______

17.25 = 35 ___

20

p ______

18.75 = 20 ___

15

⇒ q = 603.75 ________ 20

= 30.1875 ⇒ p = 375 ____ 15

= 25

9Active Maths 1 – Strands 1–5 – Ch 24 Solutions

(iii) p ____ 34.5 = 6 ____ 20.7

q ____ 19.6 = 10 ____ 24.5

⇒ p = 207 _____ 20.7

= 10 ⇒ q = 196 _____ 24.5

= 8

(c) (i) a = 120, b = h, h = 200

⇒ h2 = 2002 – 1202

⇒ h2 = 40,000 – 14,400 = 25,600

⇒ h = 160 cm

⇒ h = 1 __ 3

(w) ⇒ w = 3h

⇒ w = 480 cm

(ii) If you put one on top of the other like so, two ramps will have the height of one ramp.

hh

⇒ 10 ramps = 5h = 5(160 cm) = 800 cm= 8 m

(iii) volume = 800 × 120 × 480 = 46, 080, 000 cm3

Q. 4. (a) (i) Hypotenuse is the biggest side (if it is h)

a = 12, b = 35, h = 37

a2 + b2 = 122 + 352

= 144 + 1225 =1369

h2 = 372 = 1369

⇒ a2 + b2 = h2

∴ Yes it is a right-angled ∆

(ii) a = 84, b = 13, h = 85

a2 + b2 = 842 + 132 = 7056 + 169 = 7225

h2 = 852 = 7225

⇒ a2 + b2 = h2

∴ Yes it is a right-angled ∆

(iii) a = 32, b = 25, h = 40

a2 + b2 = 322 + 252 = 1024 + 625 = 1649

h2 = 402 = 1600

⇒ a2 + b2 ≠ h2

∴ Not right-angled

(iv) a = 60, b = 32, c = 68

a2 + b2 = 602 + 322 = 3600 + 1024 = 4624

h2 = 682 = 4624

⇒ a2 + b2 = h2

∴ Right-angled

(v) a = 24, b = 8, h = 26

a2 + b2 = 242 + 82 = 576 + 64 = 640

h2 = 262 = 676

⇒ a2 + b2 ≠ h2

∴ Not right-angled

∴ Ans Q.4. (a): (i), (ii), (iv)

(b) (i) a = 5, b = x, h= 13

b2 = h2 – a2

⇒ x2 = 132 – 52

⇒ x2 = 169 – 25 = 144

⇒ x = 12

y2 = 152 – 122

⇒ y2 = 225 – 144 = 81

⇒ y = 9

Q. 4. (a) (i) Hypotenuse is the biggest side (iv) a = 60, b = 32, c = 68

10 Active Maths 1 – Strands 1–5 – Ch 24 Solutions

(ii) x2 = 292 – 202

⇒ x2 = 841 – 400

⇒ x2 = 441

⇒ x = 21

y2 = 352 – 212

⇒ y2 = 1225 – 441 = 784

⇒ y = 28

(iii) x2 = 242 + 322

⇒ x2 = 576 + 1024 = 1600

⇒ x = 40

y2 = 852 – 402

⇒ y2 = 7225 – 1600 = 5625

⇒ y = 75

(c) Let d be the diagonal length:

d2 = 82 + 62

⇒ d2 = 64 + 36 = 100

⇒ d = 10

∴ Total length = 6 + 6+ 10 + 10 = 32 m

Q. 5. (a) (i) |∠A| = 90° angle in a semicircle

|∠B| = 180° – (90° + 25°) = 65° … angles in a Δ

(ii) |∠A| = 90° angle in a semicircle

|∠B| = 90° + 53° = 143° exterior angle

(iii) |∠A| = 1 __ 2 (110°) = 55° external angle + isosceles Δ

|∠B| = 180° – (90° + 55°) = 35° angle in a Δ

(iv) |∠B| = 35° vertically opposite and isosceles Δ

|∠A| = 90° – 35° = 55° angle in a semicircle and isosceles Δ

(v) |∠A| = 25° alternate

|∠B| = 180° – (90° + 25°) = 65° angle in a Δ and semicircle

(b) (i) |∠ADB| = 180° – (90° + 30°) = 60° angle in a Δ + semicircle

|∠ADC| = 180° – (30° + 63°) = 87° straight angle

(ii) Tangent

(iii) No |∠A| and |∠C| are right angles but |∠D| = 87° (shown above) and |∠B| = 93°, so the opposite angles are not equal. It’s not a parallelogram.

(c) (i) |∠ACB| = 90° angle in a semicircle

(ii) |∠ADO| = 90° corresponding

(iii) |∠AOD| = 72° corresponding

⇒ |∠BOD| = 180° – 72° = 108° straight angle

(iv) |∠DAO| = 180° – (90° + 72°) = 18° angle in a Δ

11Active Maths 1 – Strands 1–5 – Ch 24 Solutions

Q. 6. (a) (i)

480

360

x

x2 = 4802 + 3602

x2 = 360,000

x2 = 600 km

(ii) Time=> 16:50

14:15

2:35

Speed = 600 ____ 2 7 ___ 12

= 232.258…

= 232 km/hr

(b)

100

x

x

1002 = x2 + x2

10,000 = 2x2

5,000 = x2 → x = √______

5,000

Perimeter = 4( √______

5,000 ) = 282.84…

(4l) = 283 m

(c) (i)

C

B

2.7 m

A

2.8 m

3.5 m

4.5

m

3.52 = x2 + 2.82 4.52 = y2 + 2.72

12.25 = x2 + 7.84 20.25 = y2 + 7.29

4.41 = x2 12.96 = y2

2.1 = x 3.6 = y

3.6 – 2.1 = 1.5 m

(ii)

z 3.6

A

C 2.8

(c) (i) A

z2 = 3.62 + 2.82

z2 = 20.8

z = 4.56 m

12 Active Maths 1 – Strands 1–5 – Ch 24 Solutions

Q. 7. (a) √_________

402 + 422 = 58 cm

√_________

332 + 442 = 55 cm

(b) 12.52 = 156.25 7.52 + 102 = 156.25 Has right angle

112 = 121 10.82 + 8.12 = 182.25 Does not have right angle

13.22 = 174.24 10.52 +8.62 = 184.21 Does not have right angle

The second and third designs fulfil the condition.

(c)

B

11 m

20 m

A

18.5 – 11 = 7.5

20 m

B: √_________

112 + 202 = 22.825…

Linesman: ≈ 23 m from ball

A: √__________

202 + 7.52 = 21.3600… m

Referee: ≈ 21 m from ball