Area of a Sector- Why squared?

[opus]

Homework Statement

Find the area, A, of a sector of a circle with a radius of 9 inches and a central angle of 30°.

Homework Equations

$$Area~of~a~Sector:$$
$$A=\left( \frac 1 2 \right)r^2θ$$

The Attempt at a Solution

[/B]
$$θ=30°$$
$$θ=30°\left( \frac π {180} \right)$$
$$θ=\left( \frac π 6 \right)$$

$$A=\left( \frac 1 2 \right)\left(9\right)^2\left(\frac π 6 \right)$$
$$A=\left( \frac {81π} {12} \right)$$
$$A≈21.2 in^2$$

My question:
I know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result. However in this case, I don't see how the units for a sector of a circle are squared, as it doesn't seem like we're multiplying two things of equal value to each other.
So why is this result squared?

 

[Mark44]

Homework Statement

Find the area, A, of a sector of a circle with a radius of 9 inches and a central angle of 30°.

Homework Equations

$$Area~of~a~Sector:$$
$$A=\left( \frac 1 2 \right)r^2θ$$

The Attempt at a Solution

[/B]
$$θ=30°$$
$$θ=30°\left( \frac π {180} \right)$$
$$θ=\left( \frac π 6 \right)$$

$$A=\left( \frac 1 2 \right)\left(9\right)^2\left(\frac π 6 \right)$$
$$A=\left( \frac {81π} {12} \right)$$
$$A≈21.2 in^2$$

My question:
I know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result. However in this case, I don't see how the units for a sector of a circle are squared, as it doesn't seem like we're multiplying two things of equal value to each other.
So why is this result squared?

Because it's an area. The shape doesn't matter.
The standard units of area are always squared, $\text{length} \times \text{length}$, except for some special cases such as acres or hectares (which involve implicitly squared units such as ft² for acres and m² for hectares.

 

[opus]

So the length x length in this particular case, would be length(radius) x length(arc). However the length of the radius is in inches, and the length of the arc is in radians. So how can this results in inches squared?

 

[Mark44]

So the length x length in this particular case, would be length(radius) x length(arc). However the length of the radius is in inches, and the length of the arc is in radians. So how can this results in inches squared?

The angle in radians is just an angle, with no length. Think about it this way, as a, say, peach pie. If an 8" diameter pie is cut into 6 pieces, each slice (a sector) will subtend an angle of $\pi/3$, and the radius will be 4". The arc length of the curved edge of the slice has to take into account the radius, otherwise the arc length of an 8" pie would be the same as for a 16" pie. So in fact, the curved dimension of the pie sector is radius * angle (in radians), or $4 \times \pi/3$. So you have one radius for the radius of the sector and another radius for the arc length, making the sector area equal to $\frac 1 2 r^2 \theta$.

 

[opus]

Ahhh ok. That makes complete sense. Great explanation, thank you Mark.

 

[DrClaude]

I know that when you find the area of a space, it will be in $units^2$. But I've always thought of it as a square- that is, one equal side multiplied by the other equal side obviously results in a squared result.

Let me add that one way to visualize the "units square" is to think that a wedge with an area of 22.2 in² has the same area as a square with sides of √(22.2) ≈ 4.6 in.

 

[opus]

Interesting! Thanks DrClaude

 

[alijan kk]

imagine the arc as a triangle , becuase area would be same even if you make the arc straight line.
now the base of this triangle is "r=radius" and the perpendicular side is the arc which is equal to "theta*r"

area of triangle = 0.5base*height

0.5(r)(r)(theta)=formula of area of sector