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Chapter 11: Areas Related to Circles

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This page provides comprehensive Chapter 11: Areas Related to Circles - Assertion Reason Worksheet - SJMaths. Assertion and Reason type questions for Class 10 Areas Related to Circles. Practice for CBSE Board Exams.

Assertion-Reason Worksheet

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Directions:

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

  • (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.
  1. Question 1:
    Assertion (A): The area of a sector of a circle with radius 6 cm and angle of sector $60^\circ$ is $132/7$ cm$^2$.
    Reason (R): Area of sector with angle $\theta$ and radius $r$ is given by $\frac{\theta}{360} \times \pi r^2$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Using R, Area $= \frac{60}{360} \times \frac{22}{7} \times 6 \times 6 = \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{132}{7}$ cm$^2$.
    Step 2: A is true and R is the correct formula.
  2. Question 2:
    Assertion (A): If the circumference of a circle is 176 cm, then its radius is 28 cm.
    Reason (R): Circumference of a circle $= 2\pi r$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: $2 \times \frac{22}{7} \times r = 176 \Rightarrow \frac{44}{7} r = 176 \Rightarrow r = \frac{176 \times 7}{44} = 4 \times 7 = 28$ cm.
    Step 2: Both A and R are true, and R explains A.
  3. Question 3:
    Assertion (A): The length of the minute hand of a clock is 7 cm. The area swept by the minute hand in 30 minutes is 77 cm$^2$.
    Reason (R): The minute hand sweeps $6^\circ$ in one minute.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Angle in 30 mins $= 30 \times 6^\circ = 180^\circ$.
    Step 2: Area $= \frac{180}{360} \times \frac{22}{7} \times 7 \times 7 = \frac{1}{2} \times 154 = 77$ cm$^2$.
    Step 3: R is used to find the angle, so it explains A.
  4. Question 4:
    Assertion (A): If the perimeter and area of a circle are numerically equal, then the radius of the circle is 2 units.
    Reason (R): $2\pi r = \pi r^2 \Rightarrow r = 2$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Equating perimeter and area formulas gives $r=2$. A is true.
    Step 2: R shows the calculation.
  5. Question 5:
    Assertion (A): The area of the largest triangle that can be inscribed in a semicircle of radius $r$ is $r^2$.
    Reason (R): The height of the largest triangle inscribed in a semicircle is $r$ and base is $2r$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2r \times r = r^2$.
    Step 2: R correctly identifies the dimensions for maximum area.
  6. Question 6:
    Assertion (A): Distance moved by a wheel of diameter 14 cm in 1 revolution is 44 cm.
    Reason (R): Distance moved in 1 revolution = Circumference of the wheel.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Circumference $= \pi d = \frac{22}{7} \times 14 = 44$ cm.
    Step 2: A is true, R is the principle used.
  7. Question 7:
    Assertion (A): The area of a circle is $220$ cm$^2$. The area of a square inscribed in it is $140$ cm$^2$.
    Reason (R): Area of square inscribed in circle of radius $r$ is $2r^2$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: $\pi r^2 = 220 \Rightarrow \frac{22}{7} r^2 = 220 \Rightarrow r^2 = 70$.
    Step 2: Area of square $= 2r^2 = 2(70) = 140$ cm$^2$. A is true.
    Step 3: R is the correct formula ($d=2r$, Area $= d^2/2 = 2r^2$).
  8. Question 8:
    Assertion (A): If the outer and inner radii of a circular ring are $R$ and $r$, then the area of the ring is $\pi(R^2 - r^2)$.
    Reason (R): Area of ring = Area of outer circle - Area of inner circle.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$. Both A and R are true and R explains A.
  9. Question 9:
    Assertion (A): A wire is bent in the form of a circle of radius 7 cm. It is bent into a square. The area of the square is 121 cm$^2$.
    Reason (R): Length of wire = Circumference of circle = Perimeter of square.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Circumference $= 2 \times \frac{22}{7} \times 7 = 44$ cm.
    Step 2: Perimeter of square $= 44 \Rightarrow$ Side $= 11$. Area $= 11^2 = 121$.
  10. Question 10:
    Assertion (A): In a circle of radius 21 cm, an arc subtends an angle of $60^\circ$ at the centre. The length of the arc is 22 cm.
    Reason (R): Length of arc $= \frac{\theta}{360} \times \pi r^2$.
    (A)
    (B)
    (C)
    (D)
    Solution: (C)
    Step 1: Length $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22$ cm. A is true.
    Step 2: R gives the formula for Area, not Length. R is false.
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