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Assertion (A): OACB is a quadrant of a circle with centre O and radius $$7\ \text{cm}$$, where ACB is the arc. Then, the perimeter of the quadrant is $$25\ \text{cm}$$.

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Question

Assertion (A): OACB is a quadrant of a circle with centre O and radius $$7\ \text{cm}$$, where ACB is the arc. Then, the perimeter of the quadrant is $$25\ \text{cm}$$.

Reason (R): The perimeter of the sector of a circle of radius $$r$$ cm and central angle $$\theta$$ is $$2r\left(1 + \frac{\pi\theta}{360}\right)\ \text{cm}$$.

Options

  • Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

  • Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (А).

  • Assertion (A) is true and Reason (R) is false.

  • Assertion (A) is false and Reason (R) is true.

MCQ
Assertion and Reasoning
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Solution

Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Explanation:

Step 1 – Assertion: A is true. A quadrant has central angle $$90^{\circ}$$. Substituting $$r = 7\ \text{cm}$$ and $$\theta = 90^{\circ}$$ into the perimeter formula gives $$2\times7\left(1 + \frac{22}{7}\times\frac{90}{360}\right) = 25\ \text{cm}$$.

Step 2 – Reason: R is true. A sector’s perimeter is the sum of its two radii and its arc length, which gives the stated formula.

Step 3 – Link: Applying the sector-perimeter formula to a quadrant gives the asserted perimeter, so R correctly explains A.

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Chapter 20: Additional Questions - Area of Circle, Sector and Segment [Page 1036]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 20 Additional Questions
Area of Circle, Sector and Segment | Q 9. | Page 1036
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