Class 10 Maths: Areas related to Circles – Extra Questions with Answers

Ans: D

Solution: The diameter of the largest circle equals the shorter side of the rectangle. Diameter = 6 cm, radius = 3 cm. Area = πr² = π(3²)=9π.
Answer: 9π

Ans: B

Solution:
Area of handkerchief = 42 * 42 = 1764 cm^2
Area of one circle = π * (14/2)^2 = 49π cm^2
Area of nine circles = 9 * 49π = 441π cm^2
Area not covered = 1764 – 441π ≈ 1764 – 441 * 3.14 = 1764 – 1384.74 = 379.26 cm^2
Answer: 379.26

Ans: C

Solution:
Area of shaded region = Area of sector AOB with radius 12 cm – Area of sector AOB with radius 6 cm.
Area of sector = $\frac{\theta}{360} \times \pi r^2$
Area of shaded region = $\frac{60}{360} \times \pi (12^2) – \frac{60}{360} \times \pi (6^2)$
Area of shaded region = $\frac{1}{6} \times \pi (144) – \frac{1}{6} \times \pi (36)$
Area of shaded region = $24\pi – 6\pi = 18\pi$

Answer: $18\pi \text{ cm}^2$

Ans: C

Solution:
Let the side of the equilateral triangle be $a$. Then the area of the triangle is $\frac{\sqrt{3}}{4}a^2 = 16\sqrt{3}$. So, $a^2 = 64$, and $a = 8$. The perimeter of the triangle is $3a = 3(8) = 24$ cm. This is the length of the wire.
When the wire is bent into a circle, the circumference is 24 cm. Let the radius of the circle be $r$. Then $2\pi r = 24$, so $r = \frac{24}{2\pi} = \frac{12}{\pi}$.
The area of the circle is $\pi r^2 = \pi \left(\frac{12}{\pi}\right)^2 = \frac{144}{\pi} = \frac{144}{\frac{22}{7}} = \frac{144 \times 7}{22} = \frac{72 \times 7}{11} = \frac{504}{11} = 45.818…$.
Area of circle = $\frac{144}{\pi} = 144 \times \frac{7}{22} = \frac{1008}{22} = \frac{504}{11}$

Answer: $\frac{504}{11}$

Ans: C

Solution:
Area = (θ/360) * πr² = (90/360) * π * 7² = (1/4) * π * 49 = (49π)/4
Answer: (49π)/4

Ans: D

Solution:
Circumference = 2 * pi * radius = 2 * pi * 0.3 meters ≈ 1.885 meters.
Revolutions = Distance / Circumference = 15 meters / 1.885 meters ≈ 7.96.
Complete revolutions = 7.

Answer: 7

Ans: D

Solution:
Area of square = 6*6 = 36 sq cm
Area of 4 quarter circles = 4 * (1/4) * pi * (1.5)^2 = pi * 2.25 sq cm
Area of central circle = pi * (1.5)^2 = pi * 2.25 sq cm
Total removed area = pi * 2.25 + pi * 2.25 = 4.5 * pi sq cm
Remaining area = 36 – 4.5*pi sq cm

Answer: 36 – 4.5π

Ans: A

Solution:
Area cleaned by one wiper = (100/360) * π * (35)^2
Total area = 2 * (100/360) * π * (35)^2
Total area ≈ 2 * (5/18) * π * 1225
Total area ≈ 2138.01 cm^2

Answer: 2138.01

Ans: C

Solution:
Area of quadrant OACB = (1/4)πr² = (1/4)π(4²)=4π cm²
Area of triangle OBD = (1/2) * OD * OB = (1/2) * 2.5 * 4 = 5 cm²
Area of shaded region = Area of quadrant – Area of triangle = 4π – 5 = 4(3.14) – 5 = 12.56 – 5 = 7.56 cm²

Answer: 7.56 cm²

Ans: C

Solution: The area of the intersection is twice the area of a segment of a circle. The central angle of the segment is 60 degrees (since the triangle formed by the centers and a point of intersection is equilateral). The area of the segment is the area of the sector minus the area of the triangle. Area of sector = (60/360) * pi * 5^2 = (1/6) * 25pi. Area of the triangle = (sqrt(3)/4) * 5^2 = (25sqrt(3))/4. Area of segment = (25pi/6) – (25sqrt(3))/4. Area of intersection = 2 * [(25pi/6) – (25sqrt(3))/4] = (25pi/3) – (25sqrt(3))/2. Approximately 25*3.14/3 – 25*1.73/2 = 26.17 – 21.625 = 4.545, which approximates to 25*(pi/3 – sqrt(3)/2) = 25 * (1.047 – 0.866) = 25 * 0.181 = 4.525.
Answer: (25π/3) – (25√3)/2 cm²