CBSE Class 10 Mathematics Areas Related to Circles Assignment Set 14

Exercise-1: Subjective Questions on Perimeter and Area of a Circle

Question A-1. If the diameter of a semicircular protractor is 14 cm, then find its perimeter.
Answer:
Radius = 7 cm. Perimeter of a semicircle = πr + 2r = \( \dfrac{22}{7} \times 7 + 14 = 22 + 14 = \) 36 cm.
In simple words: The perimeter includes the curved edge and the straight diameter.

Teacher's Note:
1. Perimeter = arc + diameter.
2. Arc = πr.

Question A-2. Diameter of a tyre is 1.26 m. Find the distance covered by it in 500 revolutions.
Answer:
Distance in one revolution = circumference = \( \pi d = \dfrac{22}{7} \times 1.26 = 3.96 \) m.
Distance in 500 revolutions = 3.96 × 500 = 1980 m.
In simple words: One revolution covers one circumference of the tyre.

Teacher's Note:
1. Distance = revolutions × circumference.
2. Use π = 22/7.

Question A-3. Two circles touch internally. The sum of their areas is 116π cm² and the distance between their centres is 6 cm. Find the radii of the circles.
Answer:
Let the radii be R and r (R > r). Internal touching gives R − r = 6, and \( \pi R^2 + \pi r^2 = 116\pi \), so \( R^2 + r^2 = 116 \).
\( (R + r)^2 = 2(R^2 + r^2) - (R - r)^2 = 232 - 36 = 196 \), so R + r = 14.
Solving: R = 10 and r = 4. The radii are 10 cm and 4 cm.
In simple words: The distance between the centres of internally touching circles is the difference of the radii.

Teacher's Note:
1. Use R − r = 6 and R² + r² = 116.
2. Find R + r from the identity.

Question A-4. The inner circumference of a circular track is 220 m. The track is 7 m wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of Rs. 2 per metre [use π = 22/7].
Answer:
Inner circumference = 220 m, so 2πr = 220 and r = 35 m. Outer radius = 35 + 7 = 42 m.
Outer circumference = \( 2 \times \dfrac{22}{7} \times 42 = 264 \) m. Cost of fencing = 264 × 2 = Rs. 528.
In simple words: Find the inner radius, add the width and then the outer circumference.

Teacher's Note:
1. Outer radius = inner radius + 7.
2. Cost = length × rate.

Teacher's Note:
1. The shaded area is E − D − (one small semicircle).
2. 25 paise = Rs. 0.25.

Exercise-1: Subjective Questions on Sector and Segment of a Circle

Question B-1. A chord of a circle of radius 12 cm subtends an angle of 60° at the centre. Find the area of the corresponding shaded segment of the circle (use π = 3.14 and \( \sqrt{3} = 1.73 \)).
Answer:
Area of the sector = \( \dfrac{60}{360} \times 3.14 \times 12^2 = 75.36 \) cm². The triangle formed by the radii is equilateral with side 12 cm: area = \( \dfrac{\sqrt{3}}{4} \times 144 = 62.28 \) cm².
Area of the segment = 75.36 − 62.28 = 13.08 cm².
In simple words: Segment = sector − triangle.

Teacher's Note:
1. A 60° isosceles triangle is equilateral.
2. Use √3 = 1.73.

Question B-2. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in one minute [use π = 22/7].
Answer:
The minute hand turns through 6° in one minute. Area swept = \( \dfrac{6}{360} \times \dfrac{22}{7} \times 14^2 = \dfrac{154}{15} = \) 10.27 cm² (approximately).
In simple words: The hand sweeps a sector of 6° each minute.

Teacher's Note:
1. The minute hand covers 360° in 60 minutes.
2. Use the sector formula.

Question B-3. In a circle with centre O and radius 5 cm, AB is a chord of length \( 5\sqrt{3} \) cm. Find the area of the sector AOB.
Answer:
Draw OM ⊥ AB. Then \( AM = \dfrac{5\sqrt{3}}{2} \) and \( \sin\angle AOM = \dfrac{AM}{OA} = \dfrac{\sqrt{3}}{2} \), so ∠AOM = 60° and ∠AOB = 120°.
Area of the sector AOB = \( \dfrac{120}{360} \times \pi \times 5^2 = \) \( \dfrac{25\pi}{3} \) cm².
In simple words: The chord subtends 120° at the centre.

Teacher's Note:
1. The perpendicular from the centre bisects the chord.
2. Use sin 60° = √3/2.

Question B-4. A chord AB of a circle of radius 10 cm makes a right angle at the centre of the circle. Find the area of the major and minor segments (take π = 3.14).
Answer:
Minor segment = sector (90°) − △OAB = \( \dfrac{1}{4} \times 3.14 \times 100 - \dfrac{1}{2} \times 10 \times 10 = 78.5 - 50 = \) 28.5 cm².
Major segment = area of the circle − minor segment = 314 − 28.5 = 285.5 cm².
In simple words: The minor segment is a quarter circle minus a right triangle.

Teacher's Note:
1. The triangle OAB is right-angled and isosceles.
2. Major = circle − minor.

Teacher's Note:
1. Take π/4 common.
2. R² − r² = 441 − 196 = 245.

Exercise-1: Subjective Questions on Applications

Teacher's Note:
1. Diameter 14 cm means radius 7 cm.
2. Add the arc length to the three straight sides.

Teacher's Note:
1. A regular hexagon splits into 6 equilateral triangles of side r.
2. Cost = area × rate.

Teacher's Note:
1. For an equilateral triangle, side = R√3.
2. Subtract the triangle from the circle.

Teacher's Note:
1. The diagonal of the square cuts the lens into two equal segments.
2. Segment = quadrant − triangle.