(a) The scale of a map is 1 : 20,000. Calculate the area, in square centimetres, on the map of a forest reserve which covers 85\(km^{2}\). (b) A rectangular playing field is 18m wide. It is surrounded by a path 6m wide such that its area is equal to the perimeterof the path. Calculate the length of the field. (c) The diagram shows a circle centre O. If < POQ = x°, the diameter of the circle is 7 cm and the area of the shaded portion is 27.5\(cm^{2}\). Find, correct to the nearest degree, the value of x. [Take \(\pi = \frac{22}{7}\)].

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(a) The scale of a map is 1 : 20,000. Calculate the area, in square centimetres, on the map of a forest reserve which covers 85

k m^{2}

.
(b) A rectangular playing field is 18m wide. It is surrounded by a path 6m wide such that its area is equal to the perimeter of the path. Calculate the length of the field.
(c) The diagram shows a circle centre O. If < POQ = x°, the diameter of the circle is 7 cm and the area of the shaded portion is 27.5

c m^{2}

. Find, correct to the nearest degree, the value of x. [Take

π = \frac{22}{7}

Explanation

(a) 1km = 100,000 cm.
Area =

\frac{85 \times 100, 000 \times 100, 000}{20, 000 \times 20, 000}

85 \times 5 \times 5

2125 c m^{2}

(b)
Perimeter of path =

2 (l e n g t h + w i d t h)

2 (30 + (12 + a)) = 2 (42 + a)

84 + 2 a

Area of field =

l e n g t h \times w i d t h

18 \times a

= 18a

∴ 18 a = 84 + 2 a

18 a - 2 a = 84 ⟹ 16 a = 84

a = \frac{84}{16}

5.25 c m

\frac{360 - x}{360} \times π r^{2}

27.5 = \frac{360 - x}{360} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}

27.5 = \frac{360 - x}{360} \times \frac{77}{2}

360 - x = \frac{27.5 \times 360 \times 2}{77}

360 - x = 257.14

x = 360 ° - 257.14 ° = 102.86 °

x ≊ 103 °

(to the nearest degree).

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(a) The scale of a map is 1 : 20,000. Calculate the area, in square centimetres, on the ...

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