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Question 1: If the circumference of a circle and the perimeter of a square are equal, then:
Solution: (A) Area of the circle > Area of the square
Step 1: Let radius be $r$ and side of square be $a$. $2\pi r = 4a \Rightarrow a = \frac{\pi r}{2}$.
Step 2: Area of circle $= \pi r^2$. Area of square $= a^2 = \frac{\pi^2 r^2}{4}$.
Step 3: Ratio $\frac{\text{Circle}}{\text{Square}} = \frac{\pi r^2}{\pi^2 r^2 / 4} = \frac{4}{\pi} = \frac{28}{22} > 1$. -
Question 2: Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is:
Solution: (D) $\frac{p}{720} \times 2\pi R^2$
Step 1: Area of sector $= \frac{\theta}{360} \times \pi R^2$. Here $\theta = p$.
Step 2: Option (D) simplifies to $\frac{p}{360} \times \pi R^2$. -
Question 3: If the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle of radius $R$, then:
Solution: (A) $R_1 + R_2 = R$
Step 1: $2\pi R_1 + 2\pi R_2 = 2\pi R$.
Step 2: Dividing by $2\pi$, we get $R_1 + R_2 = R$. -
Question 4: If the perimeter of a circle is equal to that of a square, then the ratio of their areas is:
Solution: (B) 14 : 11
Step 1: $2\pi r = 4a \Rightarrow a = \frac{\pi r}{2}$.
Step 2: Ratio $= \frac{\pi r^2}{a^2} = \frac{\pi r^2}{(\pi r/2)^2} = \frac{4}{\pi} = \frac{4}{22/7} = \frac{28}{22} = \frac{14}{11}$. -
Question 5: The area of a circle that can be inscribed in a square of side 6 cm is:
Solution: (D) $9\pi$ cm$^2$
Step 1: Diameter of inscribed circle = Side of square = 6 cm.
Step 2: Radius $r = 3$ cm. Area $= \pi r^2 = 9\pi$ cm$^2$. -
Question 6: The area of the square that can be inscribed in a circle of radius 8 cm is:
Solution: (B) 128 cm$^2$
Step 1: Diagonal of square = Diameter of circle = 16 cm.
Step 2: Area of square $= \frac{1}{2} d^2 = \frac{1}{2} (16)^2 = 128$ cm$^2$. -
Question 7: The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is:
Solution: (C) 28 cm
Step 1: $R = r_1 + r_2$. Radii are 18 cm and 10 cm.
Step 2: $R = 18 + 10 = 28$ cm. -
Question 8: The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is:
Solution: (D) 50 cm
Step 1: $R^2 = r_1^2 + r_2^2 = 24^2 + 7^2 = 576 + 49 = 625$.
Step 2: $R = 25$ cm. Diameter $= 50$ cm. -
Question 9: If the perimeter of a semicircular protractor is 36 cm, then its diameter is:
Solution: (C) 14 cm
Step 1: Perimeter $= \pi r + 2r = r(\frac{22}{7} + 2) = r(\frac{36}{7})$.
Step 2: $r(\frac{36}{7}) = 36 \Rightarrow r = 7$ cm. Diameter $= 14$ cm. -
Question 10: The area of a quadrant of a circle with circumference of 22 cm is:
Solution: (B) 77/8 cm$^2$
Step 1: $2\pi r = 22 \Rightarrow 2(\frac{22}{7})r = 22 \Rightarrow r = 3.5 = 7/2$.
Step 2: Area $= \frac{1}{4} \pi r^2 = \frac{1}{4} (\frac{22}{7}) (\frac{49}{4}) = \frac{77}{8}$ cm$^2$.