SECTION F — Assertion and Reasoning Type Questions [1 Marks Each]
Directions: Each of the following questions consists of two statements, namely, Assertion (A) and Reason (R). Select the correct option from the choices given below:
- (a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
- (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- (c) Assertion (A) is true, but Reason (R) is false.
- (d) Assertion (A) is false, but Reason (R) is true.
-
Q1. Assertion (A): The area of a quadrant of a circle of radius $7\text{ cm}$ is $38.5\text{ cm}^2$.
Reason (R): A quadrant of a circle represents a sector whose central angle is $90^\circ$, and its area is calculated using the formula $\frac{90^\circ}{360^\circ} \times \pi r^2$. -
Q2. Assertion (A): If an arc of a circle with radius $14\text{ cm}$ subtends an angle of $60^\circ$ at the center, then the length of the arc is $\frac{44}{3}\text{ cm}$.
Reason (R): The length of an arc of a sector with central angle $\theta$ is given by the formula $\frac{\theta}{360^\circ} \times 2\pi r$. -
Q3. Assertion (A): When a wire of fixed length is bent into a circle instead of a square, the enclosed area increases.
Reason (R): For a given fixed perimeter (circumference/boundary length), a circle maximizes the enclosed planar area compared to any other regular polygon. -
Q4. Assertion (A): If the perimeter and the area of a circle are numerically equal, then the radius of the circle is $2\text{ units}$.
Reason (R): Equating the circumference formula ($2\pi r$) and the area formula ($\pi r^2$) directly yields $r = 2$. -
Q5. Assertion (A): The area of a minor segment of a circle is less than the area of its corresponding minor sector.
Reason (R): The area of a minor segment is obtained by subtracting the area of the triangle formed by the two radii and the chord from the area of the corresponding minor sector. -
Q6. Assertion (A): If a car wheel of diameter $80\text{ cm}$ makes complete revolutions, the distance covered in one revolution is equal to its circumference ($\pi \times 80\text{ cm}$).
Reason (R): In one complete rotation, a circular wheel covers a linear distance along the ground exactly equal to its perimeter. -
Q7. Assertion (A): The area of a circular ring (annulus) enclosed between two concentric circles of radii $R$ and $r$ ($R > r$) is given by $\pi(R^2 - r^2)$.
Reason (R): The area of an annulus is found by subtracting the area of the inner circle from the area of the outer circle. -
Q8. Assertion (A): The minute hand of a clock $14\text{ cm}$ long sweeps an area of $154\text{ cm}^2$ in $15\text{ minutes}$.
Reason (R): In $15\text{ minutes}$, the minute hand rotates through an angle of $90^\circ$, covering $\frac{1}{4}\text{th}$ of the total circle area. -
Q9. Assertion (A): If the radius of a circle is doubled, its area gets quadrupled.
Reason (R): The area of a circle is directly proportional to the square of its radius ($A \propto r^2$), so substituting $2r$ for $r$ increases the area by a factor of $2^2 = 4$. -
Q10. Assertion (A): If the radius of a circle is increased by $10\%$, its area increases by $21\%$.
Reason (R): The percentage change in area for a fractional change $x$ in radius is given by the expansion $(1 + x)^2 - 1$, which for $x = 0.1$ gives $1.21 - 1 = 21\%$. -
Q11. Assertion (A): The area of a circle inscribed in a square of side $10\text{ cm}$ is $25\pi\text{ cm}^2$.
Reason (R): The diameter of the largest circle that can be inscribed in a square is equal to the side length of the square ($10\text{ cm}$), making the radius equal to $5\text{ cm}$. -
Q12. Assertion (A): The perimeter of a semicircular protractor of radius $r$ is given by $(\pi + 2)r$.
Reason (R): The perimeter of a semicircle consists of half of the circumference ($\pi r$) plus the length of its diameter ($2r$). -
Q13. Assertion (A): If the radii of two circles are $6\text{ cm}$ and $8\text{ cm}$, the radius of the circle whose area is equal to the sum of the two areas is $10\text{ cm}$.
Reason (R): By Pythagoras' theorem relation for areas, $\pi R^2 = \pi r_1^2 + \pi r_2^2$, which simplifies to $R = \sqrt{6^2 + 8^2} = 10\text{ cm}$. -
Q14. Assertion (A): The area of a square inscribed in a circle of radius $r$ is $2r^2$.
Reason (R): The diagonal of the inscribed square is equal to the diameter of the circle ($2r$), and the area of a square in terms of its diagonal $d$ is $\frac{1}{2}d^2$. -
Q15. Assertion (A): If a sector area is $\frac{5}{18}$ of the total area of the circle, then the central angle of the sector is $100^\circ$.
Reason (R): The central angle is found by multiplying the given fraction by $360^\circ$ ($\frac{5}{18} \times 360^\circ = 100^\circ$). -
Q16. Assertion (A): The hour hand of a clock sweeps an angle of $30^\circ$ in $1\text{ hour}$.
Reason (R): The hour hand completes a full $360^\circ$ rotation in $12\text{ hours}$, meaning it moves at a rate of $\frac{360^\circ}{12} = 30^\circ$ per hour. -
Q17. Assertion (A): If a uniform path of width $w$ runs outside a circular field of radius $r$, the radius of the outer boundary becomes $r + w$.
Reason (R): The outer radius is formed by extending the radius of the inner circular field outward by the exact width of the path in all radial directions. -
Q18. Assertion (A): A horse tied with a $7\text{ m}$ rope at one corner of a square grassy field can graze an area of $38.5\text{ m}^2$.
Reason (R): The accessible grazing area forms a quadrant of a circle with radius equal to the rope length ($7\text{ m}$), evaluated as $\frac{1}{4} \times \frac{22}{7} \times 7^2$. -
Q19. Assertion (A): The difference between the circumferences of two concentric circles depends only on the difference between their radii and is independent of the individual radii values.
Reason (R): The difference is given by $2\pi R - 2\pi r = 2\pi(R - r)$, which relies solely on $(R - r)$. -
Q20. Assertion (A): The area of a regular hexagon circumscribing a circle can be determined using the radius of the inscribed circle acting as the apothem.
Reason (R): A circumscribed regular hexagon can be divided into six congruent equilateral triangles whose height matches the inradius of the circle.
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English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
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