Measuring Space: Perimeter and Area | Exercise 6.3

Question 4

A chord of a circle of radius 10 cm10\text{ cm} subtends 90∘90^\circ at the centre. Find the area of the corresponding:

(i) minor sector (that subtends 90∘90^\circ at the centre), and (ii) major sector (that subtends 270∘270^\circ at the centre). (Use π≈3.14\pi \approx 3.14.)

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Solution
Understand the Question
  • Given a circle of radius r=10 cmr = 10\text{ cm}.
  • The area of a sector with central angle θ\theta is given by: Area of sector=θ360∘×πr2\text{Area of sector} = \dfrac{\theta}{360^\circ} \times \pi r^2
  • For the minor sector, the central angle is θ=90∘\theta = 90^\circ.
  • For the major sector, the central angle is θ=360∘−90∘=270∘\theta = 360^\circ - 90^\circ = 270^\circ.
  • We use the given approximation π≈3.14\pi \approx 3.14.

(i) minor sector (that subtends 90∘90^\circ at the centre)

Step 1 · Calculate Area of Minor Sector

Given r=10 cmr = 10\text{ cm} and θ=90∘\theta = 90^\circ.Diagram 1

Area of minor sector=90∘360∘×πr2=90360×3.14×(10)2=14×3.14×100=14×314=78.5 cm2\begin{aligned} \text{Area of minor sector} &= \dfrac{90^\circ}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{90}{360} \times 3.14 \times (10)^2 \\[0.6em] &= \dfrac{1}{4} \times 3.14 \times 100 \\[0.6em] &= \dfrac{1}{4} \times 314 \\[0.6em] &= 78.5\text{ cm}^2 \end{aligned}
Answer

(i) 78.5 cm278.5\text{ cm}^2

(ii) major sector (that subtends 270∘270^\circ at the centre)

Step 1 · Calculate Area of Major Sector

For the major sector, the central angle is θ=270∘\theta = 270^\circ.

Area of major sector=270∘360∘×πr2=270360×3.14×(10)2=34×3.14×100=34×314=235.5 cm2\begin{aligned} \text{Area of major sector} &= \dfrac{270^\circ}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{270}{360} \times 3.14 \times (10)^2 \\[0.6em] &= \dfrac{3}{4} \times 3.14 \times 100 \\[0.6em] &= \dfrac{3}{4} \times 314 \\[0.6em] &= 235.5\text{ cm}^2 \end{aligned}
Answer

(ii) 235.5 cm2235.5\text{ cm}^2

Common Mistakes
  • Value of π\pi: Using π=227\pi = \dfrac{22}{7} instead of the specified π=3.14\pi = 3.14, which leads to slight rounding differences.
  • Sector vs. Segment: Confusing the area of a sector (pie slice) with the area of a segment (region bounded by the chord and the arc).
  • Angle Selection: Mixing up the central angle of the minor sector (90∘90^\circ) with that of the major sector (360∘−90∘=270∘360^\circ - 90^\circ = 270^\circ).

More questions in Exercise 6.3

Q1

Unless stated otherwise, use the approximation 227\dfrac{22}{7} for π\pi.

Find the area of a sector of a circle with radius 7 cm7\text{ cm} if the angle of the sector is 60∘60^\circ.

Q2

Find the area of a quadrant of a circle whose circumference is 44 cm44\text{ cm}.

Q3

The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

Q4

A chord of a circle of radius 10 cm10\text{ cm} subtends 90∘90^\circ at the centre. Find the area of the corresponding:

(i) minor sector (that subtends 90∘90^\circ at the centre), and (ii) major sector (that subtends 270∘270^\circ at the centre). (Use π≈3.14\pi \approx 3.14.)

Q5

A chord of a circle of radius 15 cm15\text{ cm} subtends an angle of 60∘60^\circ at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π≈3.14\pi \approx 3.14 and 3≈1.73\sqrt{3} \approx 1.73.)

Q6

A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm28\text{ cm} and sweeps through an angle of 120∘120^\circ. Find the total area cleaned at each sweep of the blades.

Q7

A chord of a circle of radius rr subtends an angle of 60∘60^\circ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to πr2(16−34)\pi r^2 \left( \dfrac{1}{6} - \dfrac{\sqrt{3}}{4} \right).

Q8

An equilateral triangle is inscribed in a circle of radius rr. Show that the ratio of the area of the triangle to the area of the circle is equal to 334π≈0.413\dfrac{3\sqrt{3}}{4\pi} \approx 0.413.

Q9

A square is inscribed in a circle of radius rr. Show that the ratio of the area of the square to the area of the circle is equal to 2π≈0.637\dfrac{2}{\pi} \approx 0.637.

Q10

A hexagon is inscribed in a circle of radius rr. Show that the ratio of the area of the hexagon to the area of the circle is equal to 332π≈0.827\dfrac{3\sqrt{3}}{2\pi} \approx 0.827. Can you see why the answer is exactly twice the answer to Question 8?

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