Question

what is the area of a sector of a circle with a radius of 6 inches and formed by a central angle that measures 90°?
a 3π in²
b 6π in²
c 9π in²
d 12π in²

Explanation:

Step1: Recall the formula for the area of a sector

The formula for the area of a sector of a circle is \(A=\frac{\theta}{360}\times\pi r^{2}\), where \(\theta\) is the central angle in degrees and \(r\) is the radius of the circle.

Step2: Substitute the given values into the formula

Given that \(r = 6\) inches and \(\theta=90^{\circ}\).
Substitute into the formula: \(A=\frac{90}{360}\times\pi\times(6)^{2}\).
First, simplify \(\frac{90}{360}=\frac{1}{4}\).
Then, \((6)^{2}=36\).
So, \(A=\frac{1}{4}\times\pi\times36\).

Step3: Calculate the value of the area

\(\frac{1}{4}\times36 = 9\).

Answer:

B. \(6\pi\ in^{2}\) is incorrect. The correct calculation: \(A=\frac{90}{360}\times\pi\times6^{2}=\frac{1}{4}\times36\pi = 9\pi\ in^{2}\). But if there is a mistake in the problem - assuming the formula was mis - remembered as \(A=\frac{\theta}{180}\times\pi r\) (a wrong formula, but if we follow the options):
If we use \(A=\frac{\theta}{180}\times\pi r\) (wrong formula for area of a sector, the correct formula has \(r^{2}\)), with \(\theta = 90\), \(r = 6\), \(A=\frac{90}{180}\times\pi\times6=3\pi\) (incorrect).
Using the correct formula \(A=\frac{\theta}{360}\times\pi r^{2}\), \(\theta = 90\), \(r = 6\), \(A=\frac{90}{360}\times\pi\times6^{2}=\frac{1}{4}\times36\pi=9\pi\). But if we assume the problem had a typo in the formula presentation (maybe a mis - write of the formula in the source), and if we consider \(A=\frac{\theta}{60}\times\pi\) (another wrong approach), with \(\theta = 90\), \(A = 6\pi\) (incorrect).
If we use \(A=\frac{\theta}{30}\times\pi\) (wrong), \(\theta=90\), \(A = 3\pi\) (incorrect).
Assuming the formula was \(A=\frac{\theta}{60}\times\pi r\) (wrong), \(\theta = 90\), \(r = 6\), \(A=\frac{90}{60}\times\pi\times6=9\pi\).
If we consider the options again, maybe the intended formula was \(A=\frac{\theta}{60}\times\pi\) (a wrong sector - area - like formula), \(\theta = 90\), then \(A = 6\pi\) (but this is based on an incorrect formula derivation).
If we use the standard formula \(A=\frac{\theta}{360}\times\pi r^{2}=\frac{90}{360}\times\pi\times6^{2}=9\pi\). But if we assume that the problem had a mis - print and the radius was \(2\) (then \(A=\frac{90}{360}\times\pi\times2^{2}=\pi\)), no. If radius \(r = \sqrt{6}\) (then \(A=\frac{90}{360}\times\pi\times6 = \frac{3}{2}\pi\)), no.
If we use \(A=\frac{1}{2}r^{2}\theta\) (where \(\theta\) is in radians, \(\theta=\frac{\pi}{2}\) radians for \(90^{\circ}\)), \(A=\frac{1}{2}\times6^{2}\times\frac{\pi}{2}=9\pi\).
Since the options have \(9\pi\) as option C.
C. \(9\pi\ in^{2}\)