Question

in circle k with m∠jkl = 46° and jk = 11, find the area of sector jkl. round to the nearest hundredth.

Explanation:

Step1: Recall the formula for the area of a sector

The formula for the area of a sector of a circle with radius \( r \) and central angle \( \theta \) (in degrees) is \( A=\frac{\theta}{360}\times\pi r^{2} \).
Here, the radius \( r = JK=11 \) (since \( JK \) is a radius of the circle) and the central angle \( \theta=m\angle JKL = 46^{\circ} \).

Step2: Substitute the values into the formula

Substitute \( \theta = 46^{\circ} \), \( r = 11 \) into the formula:
\( A=\frac{46}{360}\times\pi\times(11)^{2} \)
First, calculate \( (11)^{2}=121 \). Then, \( \frac{46}{360}\times\pi\times121=\frac{46\times121\times\pi}{360} \)
Calculate \( 46\times121 = 5566 \). So, \( A=\frac{5566\pi}{360} \)

Step3: Compute the numerical value

Simplify \( \frac{5566\pi}{360}\approx\frac{5566\times3.14159}{360} \)
\( 5566\times3.14159\approx5566\times3.1416 = 5566\times3+5566\times0.1416=16698 + 788.1456 = 17486.1456 \)
Then, \( \frac{17486.1456}{360}\approx48.5726 \)
Rounding to the nearest hundredth, we get \( A\approx48.57 \)

Answer:

\( 48.57 \)