(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 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\(cm^{2}\). Find, correct to the nearest degree, the value of x. [Take \(\pi = \frac{22}{7}\)].
(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\(cm^{2}\). Find, correct to the nearest degree, the value of x. [Take \(\pi = \frac{22}{7}\)].
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Correct Answer: Option n
Explanation:
(a) 1km = 100,000 cm.
Area = \(\frac{85 \times 100,000 \times 100,000}{20,000 \times 20,000}\)
= \(85 \times 5 \times 5\)
= \(2125 cm^{2}\)
(b)
Perimeter of path = \(2(length + width)\)
= \(2(30 + (12 + a)) = 2(42 + a)\)
= \(84 + 2a\)
Area of field = \(length \times width\)
= \(18 \times a\)
= 18a
\(\therefore 18a = 84 + 2a\)
\(18a - 2a = 84 \implies 16a = 84\)
\(a = \frac{84}{16}\)
= \(5.25 cm\)
(c) Area of major sector = \(\frac{360 - x}{360} \times \pi r^{2}\)
\(27.5 = \frac{360 - x}{360} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
\(27.5 = \frac{360 - x}{360} \times \frac{77}{2}\)
\(360 - x = \frac{27.5 \times 360 \times 2}{77}\)
\(360 - x = 257.14\)
\(x = 360° - 257.14° = 102.86°\)
\(x \approxeq 103°\) (to the nearest degree).
(a) 1km = 100,000 cm.
Area = \(\frac{85 \times 100,000 \times 100,000}{20,000 \times 20,000}\)
= \(85 \times 5 \times 5\)
= \(2125 cm^{2}\)
(b)
Perimeter of path = \(2(length + width)\)
= \(2(30 + (12 + a)) = 2(42 + a)\)
= \(84 + 2a\)
Area of field = \(length \times width\)
= \(18 \times a\)
= 18a
\(\therefore 18a = 84 + 2a\)
\(18a - 2a = 84 \implies 16a = 84\)
\(a = \frac{84}{16}\)
= \(5.25 cm\)
(c) Area of major sector = \(\frac{360 - x}{360} \times \pi r^{2}\)
\(27.5 = \frac{360 - x}{360} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
\(27.5 = \frac{360 - x}{360} \times \frac{77}{2}\)
\(360 - x = \frac{27.5 \times 360 \times 2}{77}\)
\(360 - x = 257.14\)
\(x = 360° - 257.14° = 102.86°\)
\(x \approxeq 103°\) (to the nearest degree).