Solutions: Areas Related to Circles - Test 2

Q1. (b) Minor segment only

The area of a minor segment is always less than the area of the corresponding minor sector. For a major segment, the area is greater than the major sector if we consider the triangle area addition, but strictly speaking, the definition usually applies to the minor segment subtraction.

Q2. (b) \(\frac{\theta}{360}\pi r^2 - \frac{1}{2}r^2 \sin \theta\)

Area of Segment = Area of Sector - Area of Triangle.
Area of Triangle = \(\frac{1}{2}r^2 \sin \theta\).

Q3. (a) \(28.5\) cm\(^2\)

Area of Sector = \(\frac{90}{360} \times 3.14 \times 100 = 78.5\) cm\(^2\).
Area of Triangle = \(\frac{1}{2} \times 10 \times 10 = 50\) cm\(^2\).
Area of Segment = \(78.5 - 50 = 28.5\) cm\(^2\).

Q4. (b) Both A and R are true but R is not the correct explanation of A.

Assertion is the formula/method for finding area. Reason is the definition of a segment. Both are true, but the definition doesn't explain the subtraction formula directly without geometric derivation.

Q5. Find area of minor segment (\(r=14, \theta=90^\circ\)).

Area of Sector = \(\frac{90}{360} \times \frac{22}{7} \times 14 \times 14 = 154\) cm\(^2\).
Area of Triangle = \(\frac{1}{2} \times 14 \times 14 = 98\) cm\(^2\).
Area of Segment = \(154 - 98 = 56\) cm\(^2\).

Q6. Find area of minor segment (\(r=10, \theta=60^\circ\)).

Area of Sector = \(\frac{60}{360} \times 3.14 \times 100 = \frac{314}{6} = 52.33\) cm\(^2\).
Area of Triangle (Equilateral) = \(\frac{\sqrt{3}}{4} \times 10^2 = \frac{1.73}{4} \times 100 = 43.25\) cm\(^2\).
Area of Segment = \(52.33 - 43.25 = 9.08\) cm\(^2\).

Q7. Find area of major segment (\(r=35, \theta=90^\circ\)).

Area of Minor Segment = Area of Sector - Area of Triangle.
Sector = \(\frac{1}{4} \times \frac{22}{7} \times 35 \times 35 = 962.5\) cm\(^2\).
Triangle = \(\frac{1}{2} \times 35 \times 35 = 612.5\) cm\(^2\).
Minor Segment = \(962.5 - 612.5 = 350\) cm\(^2\).
Area of Circle = \(\frac{22}{7} \times 35 \times 35 = 3850\) cm\(^2\).
Major Segment = Area of Circle - Minor Segment = \(3850 - 350 = 3500\) cm\(^2\).

Q8. Find area of segment (\(r=12, \theta=120^\circ\)).

Area of Sector = \(\frac{120}{360} \times 3.14 \times 12 \times 12 = \frac{1}{3} \times 3.14 \times 144 = 150.72\) cm\(^2\).
Area of Triangle: Draw \(OM \perp AB\). \(\angle AOM = 60^\circ\).
\(OM = 12 \cos 60^\circ = 6\) cm. \(AM = 12 \sin 60^\circ = 6\sqrt{3}\) cm. Base \(AB = 12\sqrt{3}\) cm.
Area \(\triangle OAB = \frac{1}{2} \times 12\sqrt{3} \times 6 = 36\sqrt{3} = 36 \times 1.73 = 62.28\) cm\(^2\).
Area of Segment = \(150.72 - 62.28 = 88.44\) cm\(^2\).

Q9. Find area of shaded region.

Area of Square = \(14 \times 14 = 196\) cm\(^2\).
Area of two semicircles = Area of one circle with diameter 14 cm (\(r=7\)).
Area = \(\frac{22}{7} \times 7 \times 7 = 154\) cm\(^2\).
Shaded Area = Area of Square - Area of Semicircles = \(196 - 154 = 42\) cm\(^2\).

Q10. Round table cover designs.

Radius \(r = 28\) cm. 6 equal designs \(\Rightarrow\) 6 segments with \(\theta = 60^\circ\).
Area of 1 Sector = \(\frac{60}{360} \times \frac{22}{7} \times 28 \times 28 = \frac{1}{6} \times 22 \times 4 \times 28 = \frac{1232}{3} \approx 410.67\) cm\(^2\).
Area of 1 Equilateral Triangle = \(\frac{\sqrt{3}}{4} \times 28^2 = \frac{1.7}{4} \times 784 = 1.7 \times 196 = 333.2\) cm\(^2\).
Area of 1 Design (Segment) = \(410.67 - 333.2 = 77.47\) cm\(^2\).
Total Area of 6 Designs = \(6 \times 77.47 = 464.82\) cm\(^2\).
Cost = \(464.82 \times 0.35 = \text{Rs } 162.68\).

(i) Area of circle = \(\pi r^2 = \frac{22}{7} \times 32 \times 32 = \frac{22528}{7} \approx 3218.28\) cm\(^2\).

(ii) In \(\triangle ABC\) inscribed in circle, side \(a = r\sqrt{3}\).
Side = \(32\sqrt{3}\) cm.

(iii) Area of Triangle = \(\frac{\sqrt{3}}{4} (\text{side})^2 = \frac{\sqrt{3}}{4} (32\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 1024 \times 3 = 768\sqrt{3} \approx 768 \times 1.732 = 1330.18\) cm\(^2\).
Area of Design = Area of Circle - Area of Triangle = \(3218.28 - 1330.18 = 1888.1\) cm\(^2\).