Question:
If a sector of maximum area is made with a wire of length 40 cm, then the area (in sq cms) of that sector is
If a sector of maximum area is made with a wire of length 40 cm, then the area (in sq cms) of that sector is
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For a sector with a fixed perimeter \( P \), the maximum possible area is always achieved when the arc length equals twice the radius (\( l = 2r \)). This means the maximum area is simply given by the elegant formula: \( A = \frac{P^2}{16} \).
Updated On: Jun 7, 2026
- \( 50 \)
- \( 100 \)
- \( 25 \)
- \( 200 \)
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The Correct Option is B
Solution and Explanation
Concept:
The perimeter \( P \) of a sector of radius \( r \) and arc length \( l \) is given by:
\[
P = 2r + l
\]
The area \( A \) of the sector is given by the formula:
\[
A = \frac{1}{2} r l
\]
Step 1: Expressing area in terms of a single variable.
We are given that the total length of the wire is 40 cm: \[ 2r + l = 40 \implies l = 40 - 2r \] Substitute this expression for \( l \) into the area formula: \[ A = \frac{1}{2} r (40 - 2r) = 20r - r^2 \]
Step 2: Maximizing the area using optimization.
Differentiate \( A \) with respect to \( r \) and set it to zero to find the critical point: \[ \frac{dA}{dr} = 20 - 2r = 0 \implies 2r = 20 \implies r = 10 \text{ cm} \]
Step 3: Calculating the maximum area value.
Substitute \( r = 10 \) back into our area equation: \[ A_{\max} = 20(10) - (10)^2 = 200 - 100 = 100 \text{ sq. cm.} \] This matches option (B) perfectly.
Step 1: Expressing area in terms of a single variable.
We are given that the total length of the wire is 40 cm: \[ 2r + l = 40 \implies l = 40 - 2r \] Substitute this expression for \( l \) into the area formula: \[ A = \frac{1}{2} r (40 - 2r) = 20r - r^2 \]
Step 2: Maximizing the area using optimization.
Differentiate \( A \) with respect to \( r \) and set it to zero to find the critical point: \[ \frac{dA}{dr} = 20 - 2r = 0 \implies 2r = 20 \implies r = 10 \text{ cm} \]
Step 3: Calculating the maximum area value.
Substitute \( r = 10 \) back into our area equation: \[ A_{\max} = 20(10) - (10)^2 = 200 - 100 = 100 \text{ sq. cm.} \] This matches option (B) perfectly.
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