Question 25 of 42advanced🔍 AnalyzeLong Answer5 marks

A chord of a circle of radius rr subtends an angle of 60°60° 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(\frac{1}{6} - \frac{\sqrt{3}}{4}\right).

Correct Answer

r²(π/6 - 3\sqrt{3}/4)

Exercise: Exercise Set 6.3 | Q: 7 | (Page 31)
For More Understanding

Explanation

The problem requires calculating the area of a minor segment by subtracting the area of the triangle from the area of the sector. The textbook context provides the formula for the area of a sector and defines a segment. It also establishes that a triangle with all three sides equal to the radius rr is an equilateral triangle, which applies here since the two radii and the chord form a triangle with an included angle of 60°60°.

Solution Steps

  1. Step 1: Understand that a segment is the region bounded by an arc and its chord. The area of the minor segment = Area of the sector − Area of the triangle.

  2. Step 2: Calculate the area of the sector. Given radius rr and angle θ=60°\theta = 60°,

    Area of sector=πr2×θ°360°=πr2×60°360°=πr26\text{Area of sector} = \pi r^2 \times \frac{\theta°}{360°} = \pi r^2 \times \frac{60°}{360°} = \frac{\pi r^2}{6}

  3. Step 3: Calculate the area of the triangle. The triangle is formed by two radii of length rr and the chord. Since the included angle is 60°60°, the triangle is an equilateral triangle with side rr. The area of an equilateral triangle with side rr is 34r2\frac{\sqrt{3}}{4}r^2.

  4. Step 4: Subtract the area of the triangle from the area of the sector.

    Area of minor segment=πr26−34r2\text{Area of minor segment} = \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2

  5. Step 5: Factor out r2r^2 from the expression to get the final form.

    Area of minor segment=r2(π6−34)\text{Area of minor segment} = r^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)
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