A circular sector with radius R and central angle \theta (measured in radians) has the sam… — ACT

A circular sector with radius $R$ and central angle $\theta$ (measured in radians) has the same area and the same perimeter as a square with side length $s$ . What is the value of $\theta$ ?

2Answer
B
4
C
8
D
$\frac{1}{\pi}$
E
$2\pi$

Answer

The value of theta must be 2

The correct value is 2. By equating the sector's area and perimeter to the square's area and perimeter, we set up a system of equations:

\frac{1}{2} R^2 \theta = s^2

and

2R + R\theta = 4s

. Solving for

s

from the area equation gives

s = R\sqrt{\theta/2}

. Substituting this into the perimeter equation and dividing by

R

yields

2 + \theta = 2\sqrt{2\theta}

. Squaring both sides and simplifying leads to the quadratic equation

\theta^2 - 4\theta + 4 = 0

, which has a single solution of

\theta = 2

radians.

Step-by-Step Solution

Write down the equations for the area and perimeter of both shapes.

For the circular sector: Area =

\frac{1}{2}R^2\theta

, Perimeter =

2R + R\theta

(where

\theta

is in radians). For the square: Area =

s^2

, Perimeter =

4s

These are the standard geometric formulas for a sector in radians and a square.

Set the corresponding areas and perimeters equal to each other to form a system of equations.

Equation 1:

\frac{1}{2} R^2 \theta = s^2

Equation 2:

2R + R\theta = 4s

The problem statement specifies that the two shapes have equal areas and equal perimeters.

Solve Equation 1 for the side length

s

of the square.

s = R \sqrt{\frac{\theta}{2}}

This allows us to substitute

s

in Equation 2 and solve for

\theta

in terms of

R

Substitute the expression for

s

into Equation 2 and simplify.

2R + R\theta = 4\left(R \sqrt{\frac{\theta}{2}}\right) \implies 2 + \theta = 4\sqrt{\frac{\theta}{2}}

Since the radius

R

is a positive length, we can divide both sides of the equation by

R

Solve the simplified equation for

\theta

by squaring both sides.

2 + \theta = 2\sqrt{2\theta} \implies (2+\theta)^2 = (2\sqrt{2\theta})^2 \implies 4 + 4\theta + \theta^2 = 8\theta \implies \theta^2 - 4\theta + 4 = 0 \implies (\theta - 2)^2 = 0 \implies \theta = 2

Squaring both sides and setting the quadratic equation to zero yields a perfect square trinomial with a single real solution.

Key Concept

Relating the area and perimeter of a circular sector using radian measures to those of a square.

Estimated Time:3m 0s

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