CBSE Class 10 Mathematics Area Relates to Circles Assignment Set 07

X - Mathematics Assignment No. 02 - Area related to circle

Q1. Calculate the area of a sector of \(60^\circ\) cut off from the circle of radius 21 cm.
Answer: Radius of the circle (\(r\)) = 21 cm
Angle of the sector (\(\theta\)) = \(60^\circ\)
Area of the sector = \(\frac{\theta}{360^\circ} \times \pi r^2\)
Area of the sector = \(\frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21\)
Area of the sector = \(\frac{1}{6} \times 22 \times 3 \times 21\)
Area of the sector = \(11 \times 21 = 231 \text{ cm}^2\)
In simple words: The area of the sector is found by multiplying the total circle area by the fraction of the angle out of \(360^\circ\).

Exam Tip: Always use \(\pi = \frac{22}{7}\) when the radius is a multiple of 7 to simplify calculations easily.

Q2. Calculate the area of a sector of \(50^\circ\) cut off from the circle of diameter 12 cm (use \(\pi = 3.14\))
Answer: Diameter (\(d\)) = 12 cm
Radius (\(r\)) = \(\frac{12}{2} = 6\) cm
Angle of the sector (\(\theta\)) = \(50^\circ\)
Area of the sector = \(\frac{\theta}{360^\circ} \times \pi r^2\)
Area of the sector = \(\frac{50^\circ}{360^\circ} \times 3.14 \times 6 \times 6\)
Area of the sector = \(\frac{5}{36} \times 3.14 \times 36\)
Area of the sector = \(5 \times 3.14 = 15.70 \text{ cm}^2\)
In simple words: First find the radius by dividing the diameter by 2, then apply the sector area formula using \(\pi = 3.14\).

Exam Tip: Be careful to use radius, not diameter, in the area formula.

Q3. A sector is cut off from a circle of radius 21 cm. The angle of the sector is \(120^\circ\). Find the length of its arc and the area.
Answer: Radius (\(r\)) = 21 cm
Angle (\(\theta\)) = \(120^\circ\)
Length of the arc = \(\frac{\theta}{360^\circ} \times 2\pi r\)
Length of the arc = \(\frac{120^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21\)
Length of the arc = \(\frac{1}{3} \times 2 \times 22 \times 3 = 44 \text{ cm}\)
Area of the sector = \(\frac{\theta}{360^\circ} \times \pi r^2\)
Area of the sector = \(\frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21\)
Area of the sector = \(\frac{1}{3} \times 22 \times 3 \times 21 = 462 \text{ cm}^2\)
In simple words: Use the fraction of the angle to find both the arc length (using circumference formula) and the area.

Exam Tip: Remember that arc length uses \(2\pi r\) while area uses \(\pi r^2\).

Q4. The minute hand of a clock is \(2\sqrt{3}\) cm long. Find the area described by the minute hand on the face of the clock between 7.00 am and 7.05 am.
Answer: Length of minute hand (\(r\)) = \(2\sqrt{3}\) cm
Time elapsed = 5 minutes
In 60 minutes, the minute hand sweeps \(360^\circ\).
Angle swept in 1 minute = \(\frac{360^\circ}{60} = 6^\circ\)
Angle swept in 5 minutes (\(\theta\)) = \(5 \times 6^\circ = 30^\circ\)
Area described = \(\frac{\theta}{360^\circ} \times \pi r^2\)
Area described = \(\frac{30^\circ}{360^\circ} \times \frac{22}{7} \times (2\sqrt{3})^2\)
Area described = \(\frac{1}{12} \times \frac{22}{7} \times (4 \times 3)\)
Area described = \(\frac{1}{12} \times \frac{22}{7} \times 12\)
Area described = \(\frac{22}{7} = 3.14 \text{ cm}^2\)
In simple words: A minute hand covers \(360^\circ\) in one hour, so in 5 minutes it covers \(30^\circ\). Calculate the area of that sector.

Exam Tip: For clock problems, always compute the angle per minute as \(6^\circ\) first.

Q5. A sector of \(100^\circ\) cut off from a circle contains 70.65 sq. cm. Find the radius of the circle (use \(\pi = 3.14\))
Answer: Area of sector = 70.65 cm\(^2\)
Angle (\(\theta\)) = \(100^\circ\)
\(\frac{\theta}{360^\circ} \times \pi r^2 = 70.65\)
\(\frac{100^\circ}{360^\circ} \times 3.14 \times r^2 = 70.65\)
\(\frac{10}{36} \times 3.14 \times r^2 = 70.65\)
\(r^2 = \frac{70.65 \times 36}{10 \times 3.14}\)
\(r^2 = \frac{2543.4}{31.4} = 81\)
\(r = \sqrt{81} = 9 \text{ cm}\)
In simple words: Set the sector area formula equal to the given area and solve algebraically for the radius \(r\).

Exam Tip: Simplify fractions carefully before cross-multiplying to avoid calculation errors.

Q6. The perimeter of a certain sector of a circle of radius 5.7 metres is 27.2 m. Find the area of the sector.
Answer: Radius (\(r\)) = 5.7 m
Perimeter of sector = \(\text{Arc length} + 2 \times \text{radius} = 27.2\)
\(\text{Arc length} + 2(5.7) = 27.2\)
\(\text{Arc length} + 11.4 = 27.2\)
\(\text{Arc length} = 27.2 - 11.4 = 15.8 \text{ m}\)
Area of sector = \(\frac{1}{2} \times \text{Arc length} \times \text{radius}\)
Area of sector = \(\frac{1}{2} \times 15.8 \times 5.7\)
Area of sector = \(7.9 \times 5.7 = 45.03 \text{ m}^2\)
In simple words: The perimeter of a sector includes the arc and two radii. Subtract the two radii from the perimeter to get the arc length, then use the formula Area = \(\frac{1}{2} \times \text{arc} \times \text{radius}\).

Exam Tip: Memorize the alternative sector area formula \(\frac{1}{2} \times l \times r\) when arc length (\(l\)) is known.

Q7. PQ is a chord of a circle of radius 10 cm. The chord subtends a right angle at the centre of the circle. Find the area of the minor segment (use \(\pi = 3.14\))
Answer: Radius (\(r\)) = 10 cm
Angle (\(\theta\)) = \(90^\circ\)
Area of minor segment = Area of sector - Area of right-angled triangle
Area of sector = \(\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 10 \times 10 = \frac{314}{4} = 78.5 \text{ cm}^2\)
Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2\)
Area of minor segment = \(78.5 - 50 = 28.5 \text{ cm}^2\)
In simple words: Subtract the area of the right-angled triangle formed by the radii and chord from the area of the sector.

Exam Tip: For a \(90^\circ\) sector, the base and height of the triangle are both equal to the radius \(r\).

Q8. A chord of a circle subtends an angle of \(60^\circ\) at the centre. If the length of the chord is 100 cm, find the areas of the two segments into which the chord divides the circle (use \(\pi = 3.14, \sqrt{3} = 1.73\))
Answer: Angle (\(\theta\)) = \(60^\circ\)
Chord length = 100 cm
Since the chord subtends \(60^\circ\) and the two radii are equal, the triangle formed is equilateral. Therefore, radius (\(r\)) = 100 cm.
Area of sector = \(\frac{60^\circ}{360^\circ} \times 3.14 \times 100 \times 100 = \frac{1}{6} \times 31400 = 5233.33 \text{ cm}^2\)
Area of equilateral triangle = \(\frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 100 \times 100 = 1.73 \times 2500 = 4325 \text{ cm}^2\)
Area of minor segment = Area of sector - Area of triangle
Area of minor segment = \(5233.33 - 4325 = 908.33 \approx 908.3 \text{ cm}^2\)
Total area of the circle = \(\pi r^2 = 3.14 \times 100 \times 100 = 31400 \text{ cm}^2\)
Area of major segment = Total circle area - Area of minor segment
Area of major segment = \(31400 - 908.3 = 30491.7 \text{ cm}^2\)
In simple words: Find the minor segment by subtracting the equilateral triangle area from the sector area, then subtract the minor segment from the total circle area to get the major segment.

Exam Tip: When a chord equals the radius in length, it forms an equilateral triangle with the center.

Q9. The radius of a circle is 5 cm. A chord of length \(\sqrt{50}\) cm is drawn in the circle. Find the area of the major segment (use \(\pi = 3.14\))
Answer: Radius (\(r\)) = 5 cm
Chord length (\(a\)) = \(\sqrt{50} = 5\sqrt{2}\) cm
Check triangle sides: \(r^2 + r^2 = 5^2 + 5^2 = 25 + 25 = 50 = (\sqrt{50})^2\)
Since \(r^2 + r^2 = a^2\), the triangle formed by the chord and radii is a right-angled isosceles triangle with \(\theta = 90^\circ\).
Area of minor segment = Area of sector (\(90^\circ\)) - Area of right triangle
Area of sector = \(\frac{90^\circ}{360^\circ} \times 3.14 \times 5^2 = \frac{1}{4} \times 3.14 \times 25 = 19.625 \text{ cm}^2\)
Area of triangle = \(\frac{1}{2} \times 5 \times 5 = 12.5 \text{ cm}^2\)
Area of minor segment = \(19.625 - 12.5 = 7.125 \text{ cm}^2\)
Total area of circle = \(\pi r^2 = 3.14 \times 25 = 78.5 \text{ cm}^2\)
Area of major segment = Total area - Area of minor segment
Area of major segment = \(78.5 - 7.125 = 71.375 \text{ cm}^2\)
In simple words: Verify that the triangle is right-angled using Pythagoras theorem, find the minor segment, and subtract it from the total circle area to get the major segment.

Exam Tip: Always check if the triangle is right-angled when side lengths involve square roots like \(\sqrt{50}\).

Q10. A square of side 4 cm is inscribed in a circle. Find the area enclosed between the circle and the square.
Answer: Side of square (\(s\)) = 4 cm
Diagonal of the square = \(s\sqrt{2} = 4\sqrt{2}\) cm
Since the square is inscribed in the circle, the diagonal of the square is equal to the diameter of the circle.
Diameter (\(d\)) = \(4\sqrt{2}\) cm
Radius (\(r\)) = \(\frac{4\sqrt{2}}{2} = 2\sqrt{2}\) cm
Area of the circle = \(\pi r^2 = \frac{22}{7} \times (2\sqrt{2})^2 = \frac{22}{7} \times 8 = \frac{176}{7} = 25\frac{1}{7} \text{ cm}^2\)
Area of the square = \(s^2 = 4 \times 4 = 16 \text{ cm}^2\)
Area enclosed = Area of circle - Area of square
Area enclosed = \(\frac{176}{7} - 16 = \frac{176 - 112}{7} = \frac{64}{7} = 9\frac{1}{7} \text{ cm}^2\) (or \(9\frac{1}{7}\) cm)
In simple words: The circle's diameter is the square's diagonal. Subtract the square's area from the circle's area.

Exam Tip: For an inscribed square, diameter equals \(\sqrt{2} \times \text{side}\).

Q11. Four equal circles are described at the four corners of a square so that each touches two of the others. The shaded area, enclosed between the circles, being \(3\frac{3}{7}\text{ cm}^2\), find the radius of the circles.
Answer: Let the radius of each circle be \(r\).
The four circles are at the corners of a square of side length \(2r\).
Area of the square = \((2r)^2 = 4r^2\)
The four sectors inside the square combine to form a full circle of radius \(r\) (since the sum of the four corner angles is \(360^\circ\)).
Area of 4 sectors = Area of circle of radius \(r\) = \(\pi r^2\)
Shaded area = Area of square - Area of 4 sectors
\(3\frac{3}{7} = 4r^2 - \pi r^2\)
\(\frac{24}{7} = r^2 (4 - \frac{22}{7})\)
\(\frac{24}{7} = r^2 \left(\frac{28 - 22}{7}\right)\)
\(\frac{24}{7} = r^2 \left(\frac{6}{7}\right)\)
\(r^2 = \frac{24}{7} \times \frac{7}{6} = 4\)
\(r = \sqrt{4} = 2 \text{ cm}\)
In simple words: The enclosed area equals the square's area minus one full circle's area. Set up the equation and solve for \(r\).

Exam Tip: Four right-angled sectors at the corners of any polygon always add up to a full circle.

Q12. A chord AB of a circle of radius 7 cm makes an angle of \(30^\circ\) at the centre of the circle. Find the area of the major segment of the circle.
Answer: Radius (\(r\)) = 7 cm
Angle (\(\theta\)) = \(30^\circ\)
Area of minor sector = \(\frac{30^\circ}{360^\circ} \times \pi r^2 = \frac{1}{12} \times \frac{22}{7} \times 7 \times 7 = \frac{1}{12} \times 154 = \frac{77}{6} \text{ cm}^2\)
Area of minor triangle = \(\frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 7 \times 7 \times \sin 30^\circ = \frac{49}{2} \times \frac{1}{2} = \frac{49}{4} \text{ cm}^2\)
Area of minor segment = Area of sector - Area of triangle
Area of minor segment = \(\frac{77}{6} - \frac{49}{4} = \frac{154 - 147}{12} = \frac{7}{12} \text{ cm}^2\)
Total area of circle = \(\pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \text{ cm}^2\)
Area of major segment = Total circle area - Area of minor segment
Area of major segment = \(154 - \frac{7}{12} = \frac{1848 - 7}{12} = \frac{1841}{12} = 153\frac{5}{12} \text{ cm}^2\)
In simple words: Calculate the minor segment using sector area minus triangle area (\(\frac{1}{2}r^2\sin\theta\)), then subtract it from the total circle area.

Exam Tip: Remember that for non-standard angles like \(30^\circ\), use the general triangle area formula \(\frac{1}{2}r^2\sin\theta\).