f3 to f4,maths questions.20pts

✔ 最佳答案

1. AB = AC = 30cm (definition of rhombus)

OB = (30 - 9) cm = 21 cm (given)

Since AO2 + OB2 = AB2 (Pyth. theorem)

AO2 + 212 = 302

AO = 3√51 cm

So area of the rhombus

= (4 x 1/2 x 21 x 3√51) cm2

= 126√51 cm2

= 900 cm2 (corr. to 3 sig. fig.)

2. OC = AO cos 30 = 4√3 cm

Area of the shaded region

= [ 8(8)(π)(30/360) - (1/2)(8)(4√3)(sin30) ] cm2

= 2.90 cm2 (corr. to 3 sig. fig.)

3. ∠AOB + ∠BOC + ∠COD = 180 (adj. ∠s on st. line)

Since ∠AOB = ∠BOC = ∠COD (given)

∠AOB = ∠BOC = ∠COD = 180/3 = 60

OE = OC cos 60 = 6 cm

CE = OC sin 60 = 6√3 cm

Area of the shaded region

= Area of semicircle - Area of rectangle BCEF

= [ (1/2)(12)(12)(π) - (2)(6)(6√3) ] cm2

= 101 cm2 (corr. to 3 sig. fig.)

4a) Area of ΔBED = (1/2)(10)(4) cm2 = 20 cm2

BE = 20 / (8/2) cm = 5 cm

b) Since BE2 + AE2

= 52 + 122 = 169

= 132 = AB2

∠AEB = 90 (Converse of Pyth. theorem)

Area of ΔABE = (1/2)(5)(12) cm2 = 30 cm2

5a) ΔABE ~ ΔCDE (given)

AE: EC = AE: CE = AB: CD = 1 : 3 (corr. sides, ~Δs)

BE: DE = 1 : 3 (corr. sides, ~Δs)

b) Area of ΔABE : Area of ΔBEC =

= AE: EC (same height)

= 1 : 3 (proved)

Area of ΔBEC = 36 cm2

Area of ΔABE : Area of ΔADE =

= BE: DE (same height)

= 1 : 3 (proved)

Area of ΔADE = 36 cm2

c)Area of ΔABE : Area of ΔCDE

= (AE : CE)2

= 1 : 9

Area of ΔCDE = 108 cm2

d)Area of trapezium ABCD = (12 + 36 + 36 + 108) cm2 = 192 cm2

6. x(5%)(n) + x = 2x

x(0.05n + 1) = 2x

0.05n + 1 = 2

n = 20

The time required is 20 years.

7. (I use my F3 textbook to answer this question. It may be out-dated.)

Tax on the 1st $35000 = $35000 x 2% = $700

Tax on the 2nd $35000 = $35000 x 7% = $2450

Tax on the 3rd $35000 = $35000 x 12% = $4200

Since the tax the man had to pay is just 4325, his net chargable income is less than 105000. So he only need to pay the tax on the first two $35000.

(35000)(2%) + (35000)(7%) +(y - 70000)(17%) = 4325

3150 + 0.17y - 11900 = 4325

0.17y = 13075

y = 76911.76471

His net chargable income is $76912. (corr. to nearest dollar)

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