(a)

∠ QHF = 90°

QF is the hypothenuse of the right-angled triangle QHF

QH is the adjacent side to ∠ EQH

QE = QH = Radius of the circle = 8$\sqrt{3}$

QF = QE + EF

EF = QF - QE

CAH

cos(θ) = $\frac{\mathrm{Adjacent}}{\mathrm{Hypothenuse}}$

cos 60° = $\frac{8\sqrt{3}}{\mathrm{|QF|}}$

Multiply both sides by the denominator QF

QF x cos 60° = 8$\sqrt{3}$

Divide both sides by cos 60°

QF = $\frac{8\sqrt{3}}{\mathrm{cos\; 60°}}$

QF = 27.71 cm

EF = QF - QE

EF = 27.71 cm - 8$\sqrt{3}$ cm

EF = 13.856 cm ≈ 13.9 cm

(b)

Shaded portion = Area of ∆QHF - Area of the sector of the circular shape

Area of ∆ QHF = $\frac{\mathrm{|QH|\; x\; |HF|}}{2}$

TOA

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

tan 60° = $\frac{\mathrm{|HF|}}{8\sqrt{3}}$

Multiply both sides by the denominator 8$\sqrt{3}$

|HF| = tan 60° x 8$\sqrt{3}$

|HF| = 24 cm

Area of ∆ QHF = $\frac{8\sqrt{3}\mathrm{x\; 24}}{2}$

Area of ∆ QHF = 166.28 cm²

Area of the sector of the circular shape = $\frac{60}{360}$ x π x QH x QH

Area of the sector of the circular shape = $\frac{60}{360}$ x $\frac{22}{7}$ x 8$\sqrt{3}$ x 8$\sqrt{3}$

Area of the sector of the circular shape = 100.57 cm²

Shaded portion = 166.28 cm² - 100.57 cm²

Shaded portion = 65.71 cm² ≈ 65.7 cm² (three significant figures)

(a)

The angle (θ) at the top of the cone when calculated at the two different parts is the same.

Since the opposite and adjacent sides of the angle (θ) is known, tan θ is used.

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

The total height of the cone = h + 12

tan θ = $\frac{\mathrm{h}}{\mathrm{r}}$ = $\frac{\mathrm{3r}}{\mathrm{h\; +\; 12}}$

$\frac{\mathrm{r}}{\mathrm{h}}$ = $\frac{\mathrm{3r}}{\mathrm{h\; +\; 12}}$

Cross multiply and solve for h

(h + 12) x r = h x 3r

Divide both sides by r

h + 12 = 3h

12 = 3h - h

12 = 2h

Divide both sides by 2

h = $\frac{12}{2}$ = 6

Height of the whole cone = 6 cm + 12 cm = 18 cm

(b)

Let r = radius of the cone.

Volume = 39,600 cm³

Height = 18 cm

π = $\frac{22}{7}$

39600 = $\frac{22}{7}$ x r² x $\frac{18}{3}$

39600 = $\frac{22}{7}$ x r² x 6

Get rid of the fraction by multiplying both sides by 7

39600 x 7 = 22 x r² x 6

277200 = 132 x r²

Divide both sides by 132

r² = $\frac{277200}{132}$

r² = 2100

Take square root of both sides.

r = $\sqrt{\mathrm{2100}}$

r = 45.825 ≈ 45.83 cm

(a)

Volume of a cube = Length x Length x Length

Length of the cubic tank = 1.2 m

Volume of the cubic tank = 1.2 m x 1.2 m x 1.2 m = 1.728 m³

1 m³ = 1000 litres

1.728 m³ = 1.728 x 1000 litres = 1,728 litres

(b)

The number of containers of water that can fill the tank = $\frac{\mathrm{Volume\; of\; the\; tank}}{\mathrm{Volume\; of\; the\; container}}$

Volume of tank = 1,728 litres

Volume of container = 54 litres

The number of containers of water that can fill the tank = $\frac{\mathrm{1728\; litres}}{\mathrm{54\; litres}}$ = 32

∴ 32 containers will fill the tank

(c)

If 1 container = GH₵ 1.75

32 containers = GH₵ 1.75 x 32 = GH₵ 56

∴ the amount of money paid to the labourer to fill the tank with water was GH₵ 56

(a)

Let h = height of the pyramid and the cone.

Since the volumes are the same, the two formula can be equated.

π x r² x $\frac{\mathrm{h}}{3}$ = $\frac{1}{3}$ x Base area x h

Base area of the pyramid = 154

π x r² x $\frac{\mathrm{h}}{3}$ = $\frac{1}{3}$ x 154 x h

Divide both sides by h to get rid of the h

π x r² x $\frac{1}{3}$ = $\frac{1}{3}$ x 154

Get rid of the fractions by multiplying both sides by the denominator 3

π x r² = 154

π = $\frac{22}{7}$

$\frac{22}{7}$ x r² = 154

Get rid of the fraction by multiplying both sides by the denominator 7

22 x r² = 154 x 7

Divide both sides by 22

r² = $\frac{\mathrm{154\; x\; 7}}{22}$

r² = 49

Take square roots of both sides.

r = $\sqrt{\mathrm{49}}$ = 7

The radius of the cone is 7 cm.

(b)

Volume of water poured = 6 litres = 6 x 1000 cm³ = 6000 cm³

The water is a quarter full. Hence the volume of water is $\frac{1}{4}$ of the full volume of the sphere

π = $\frac{22}{7}$

$\frac{1}{4}$ x $\frac{4}{3}$ x $\frac{22}{7}$ x r³ = 6000

$\frac{\mathrm{4\; x\; 22}}{\mathrm{4\; x\; 3\; x\; 7}}$ x r³ = 6000

$\frac{88}{84}$ x r³ = 6000

Get rid of the fraction by multiplying both sides by the denominator 84.

88 x r³ = 6000 x 84

Divide both sides by 88

r³ = $\frac{\mathrm{6000\; x\; 84}}{88}$

r³ = 5727.27

Take cube root of both sides.

r = ${5727.27}^{\frac{1}{3}}$

r = 17.8916044472 cm

Diameter is 2 times the radius

Diameter = 2 x 17.8916044472 cm

Diameter = 35.7832088944 cm ≈ 35.8 cm(three significant figures)

(i)

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

sin 25° = $\frac{\mathrm{|BO|}}{8}$

|BO| = 8sin 25° = 3.381 cm

|BD| = 2 x |BO|

|BD| = 2 x 3.381 cm

|BD| = 6.762 cm ≈ 7 cm (nearest whole number)

(ii)

Area of ∆BCD = $\frac{1}{2}$ x |CD| x |BC|

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

|BC|² + |CD|² = |BD|²

|BC|² + 5² = 6.762²

|BC|² = 6.762² - 5²

Take square root of both sides.

|BC| = $\sqrt{{\mathrm{6.762}}^2-5^2}$

|BC| = 4.552 cm

Area of ∆BCD = $\frac{1}{2}$ x 5 cm x 4.552 cm

Area of ∆BCD = 11.38 cm² ≈ 11 cm² (nearest whole number)

(a)

The length of the arc of the sector becomes the circumference of the base of the cone.

Let R = radius of the circle of the sector

r = radius of the cone.

Circumference = 2 x π x radius

The circumference of the circular base of the cone = 2 x π x r

The length of the arc of the sector = the circumference of the circular base of the cone.

$\frac{\mathrm{\theta }}{360}$ x 2 x π x R = 2 x π x r

Divide both sides by 2 x π to cancel out.

$\frac{\mathrm{\theta }}{360}$ x R = r

r = $\frac{\mathrm{\theta }}{360}$ x R

θ = 60

R = 22 cm

r = $\frac{60}{360}$ x 22 cm = 3.6667 cm

∴ The radius of the cone is 3.7 cm (one decimal place)

(b)

l² = h² + r²

h² = l² - r²

Take square root of both sides.

h = $\sqrt{{\mathrm{l}}^2-{\mathrm{r}}^2}$

l = 22 cm

r = 3.6667 cm

h = $\sqrt{{22}^2-{3.6667}^2}$

h = 21.6923 cm

∴ The height of the cone is 21.7 cm (one decimal place)

(c)

h = 21.6923 cm

r = 3.6667 cm

π = $\frac{22}{7}$

Volume of the cone = $\frac{22}{7}$ x 3.6667² x $\frac{21.6923}{3}$ = 305.5341 cm³

∴ Volume of the cone = 305.5 cm³ (one decimal place)