Solutions: Areas Related to Circles - Test 1

Q1. (d) \(\frac{p}{720} \times 2\pi R^2\)

Area of sector = \(\frac{p}{360} \times \pi R^2\).
Option (d) simplifies to: \(\frac{p}{720} \times 2\pi R^2 = \frac{p}{360} \times \pi R^2\).

Q2. (a) \(150^\circ\)

Angle swept in 60 minutes = \(360^\circ\).
Angle swept in 25 minutes = \(\frac{25}{60} \times 360^\circ = 25 \times 6^\circ = 150^\circ\).

Q3. (b) \(126^\circ\)

Area of sector = \(\frac{7}{20} \times\) Area of circle.
\(\frac{\theta}{360} \pi r^2 = \frac{7}{20} \pi r^2 \Rightarrow \frac{\theta}{360} = \frac{7}{20}\).
\(\theta = \frac{7}{20} \times 360 = 7 \times 18 = 126^\circ\).

Q4. (a) Both A and R are true and R is the correct explanation of A.

Area = \(\frac{60}{360} \times 3.14 \times 6^2 = \frac{1}{6} \times 3.14 \times 36 = 18.84\) cm\(^2\).
The calculation follows the formula given in Reason.

Q5. Find area of sector.

Area = \(\frac{\theta}{360} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 6 \times 6\)
\(= \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{132}{7}\) cm\(^2\) (or approx 18.86 cm\(^2\)).

Q6. Find area of quadrant.

Circumference \(2\pi r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = 3.5\) cm.
Area of quadrant = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8}\) cm\(^2\) (or 9.625 cm\(^2\)).

Q7. Find area swept by minute hand.

Angle in 5 mins = \(30^\circ\). Radius \(r = 14\) cm.
Area = \(\frac{30}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{12} \times 22 \times 2 \times 14 = \frac{154}{3}\) cm\(^2\) (or 51.33 cm\(^2\)).

Q8. Find arc length and sector area.

(i) Length of arc = \(\frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 44 \times 3 = 22\) cm.

(ii) Area of sector = \(\frac{1}{2} \times l \times r = \frac{1}{2} \times 22 \times 21 = 231\) cm\(^2\).

Q9. Find total area cleaned by wipers.

Total Area = \(2 \times \frac{115}{360} \times \frac{22}{7} \times 25 \times 25\)
\(= \frac{230}{360} \times \frac{22}{7} \times 625 = \frac{23}{36} \times \frac{22}{7} \times 625\)
\(= \frac{23 \times 11 \times 625}{18 \times 7} = \frac{158125}{126} \approx 1254.96\) cm\(^2\).

Q10. Grazing area problem.

(i) Area with 5m rope = Area of quadrant with \(r=5\).
\(= \frac{1}{4} \times 3.14 \times 5^2 = \frac{78.5}{4} = 19.625\) m\(^2\).

(ii) Area with 10m rope = Area of quadrant with \(r=10\).
\(= \frac{1}{4} \times 3.14 \times 10^2 = \frac{314}{4} = 78.5\) m\(^2\).
Increase in area = \(78.5 - 19.625 = 58.875\) m\(^2\).

(i) Total wire length = Circumference + \(5 \times\) Diameter.
\(= \pi d + 5d = \frac{22}{7} \times 35 + 5 \times 35 = 110 + 175 = 285\) mm.

(ii) Area of each sector = \(\frac{1}{10} \times\) Area of circle.
\(= \frac{1}{10} \times \frac{22}{7} \times (\frac{35}{2})^2 = \frac{1}{10} \times \frac{22}{7} \times \frac{1225}{4} = \frac{385}{4} = 96.25\) mm\(^2\).

(iii) If diameter is doubled, radius is doubled (\(r \to 2r\)).
New Area \(\propto (2r)^2 = 4r^2\). So area becomes 4 times.
New Area = \(4 \times 96.25 = 385\) mm\(^2\).