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Question 1: Find the area of a sector of a circle with radius 6 cm if angle of the sector is $60^\circ$.
Solution:
Step 1: Area of sector $= \frac{\theta}{360} \times \pi r^2$.
Step 2: Substitute $\theta = 60^\circ, r = 6$. Area $= \frac{60}{360} \times \frac{22}{7} \times 6 \times 6$.
Step 3: Area $= \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{132}{7}$ cm$^2$.
Answer: $132/7$ cm$^2$ or $18.86$ cm$^2$. -
Question 2: Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Step 1: Circumference $2\pi r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = 3.5$ cm.
Step 2: Area of quadrant $= \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5$.
Step 3: Area $= \frac{1}{4} \times 22 \times 0.5 \times 3.5 = \frac{77}{8}$ cm$^2$.
Answer: $9.625$ cm$^2$. -
Question 3: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Step 1: Angle swept in 60 mins $= 360^\circ$. In 5 mins, angle $= \frac{360}{60} \times 5 = 30^\circ$.
Step 2: Area $= \frac{30}{360} \times \frac{22}{7} \times 14 \times 14$.
Step 3: Area $= \frac{1}{12} \times 22 \times 2 \times 14 = \frac{154}{3}$ cm$^2$.
Answer: $51.33$ cm$^2$. -
Question 4: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use $\pi = 3.14$)
Solution:
Step 1: Area of sector $= \frac{90}{360} \times 3.14 \times 10 \times 10 = \frac{1}{4} \times 314 = 78.5$ cm$^2$.
Step 2: Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 10 = 50$ cm$^2$.
Step 3: Area of segment $= \text{Area of sector} - \text{Area of triangle} = 78.5 - 50 = 28.5$ cm$^2$.
Answer: $28.5$ cm$^2$. -
Question 5: In a circle of radius 21 cm, an arc subtends an angle of $60^\circ$ at the centre. Find the length of the arc.
Solution:
Step 1: Length of arc $= \frac{\theta}{360} \times 2\pi r$.
Step 2: Length $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 21$.
Step 3: Length $= \frac{1}{6} \times 2 \times 22 \times 3 = 22$ cm.
Answer: 22 cm. -
Question 6: Find the area of the major sector of a circle of radius 14 cm and central angle $60^\circ$.
Solution:
Step 1: Angle of major sector $= 360^\circ - 60^\circ = 300^\circ$.
Step 2: Area $= \frac{300}{360} \times \frac{22}{7} \times 14 \times 14$.
Step 3: Area $= \frac{5}{6} \times 22 \times 2 \times 14 = \frac{5}{6} \times 616 = \frac{1540}{3}$ cm$^2$.
Answer: $513.33$ cm$^2$. -
Question 7: A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find the area of that part of the field in which the horse can graze.
Solution:
Step 1: The horse can graze in a quadrant of a circle with radius 5 m (since corner angle is $90^\circ$).
Step 2: Area $= \frac{90}{360} \times 3.14 \times 5 \times 5$.
Step 3: Area $= \frac{1}{4} \times 3.14 \times 25 = 19.625$ m$^2$.
Answer: $19.625$ m$^2$. -
Question 8: A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find the total length of the silver wire required.
Solution:
Step 1: Length of wire = Circumference + 5 diameters.
Step 2: Circumference $= \pi d = \frac{22}{7} \times 35 = 110$ mm.
Step 3: Length of 5 diameters $= 5 \times 35 = 175$ mm.
Step 4: Total length $= 110 + 175 = 285$ mm.
Answer: 285 mm. -
Question 9: Find the area of the shaded region in a figure where ABCD is a square of side 14 cm and two semicircles are drawn with AD and BC as diameters inside the square.
Solution:
Step 1: Area of square $= 14 \times 14 = 196$ cm$^2$.
Step 2: Two semicircles make one full circle with diameter 14 cm (radius 7 cm).
Step 3: Area of circle $= \frac{22}{7} \times 7 \times 7 = 154$ cm$^2$.
Step 4: Shaded area $= 196 - 154 = 42$ cm$^2$.
Answer: 42 cm$^2$. -
Question 10: The area of an equilateral triangle ABC is $17320.5$ cm$^2$. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region (area of triangle excluding sectors). (Use $\pi = 3.14$ and $\sqrt{3} = 1.73205$)
Solution:
Step 1: Area of equilateral $\Delta = \frac{\sqrt{3}}{4} a^2 = 17320.5$.
Step 2: $a^2 = \frac{17320.5 \times 4}{1.73205} = 10000 \times 4 = 40000 \Rightarrow a = 200$ cm.
Step 3: Radius $r = a/2 = 100$ cm. Area of 3 sectors $= 3 \times \frac{60}{360} \times 3.14 \times 100 \times 100 = \frac{1}{2} \times 31400 = 15700$ cm$^2$.
Step 4: Shaded area $= 17320.5 - 15700 = 1620.5$ cm$^2$.
Answer: $1620.5$ cm$^2$.