Question

In circle Q with m/PQR = 82° and PQ = 19, find the area of sector PQR. Round to the nearest hundredth.

Answer 1

`\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360} ~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=19\\ \theta =82 \end{cases}\implies A=\cfrac{(82)\pi (15)^2}{360}\implies A\approx 161.01`

Answer 2

The area of sector PQR with a central angle of 82 degrees and a radius of 19 units is approximately 258.20 square units.

The figure in the image is a circle having a sector PQR.

The area of the sector is expressed as:

Area A = (θ/360º) × πr²

Where θ is the central angle and r is the radius.

From the figure:

Central angle θ = 82 degrees

Radius r = 19 units

Constant pi π = 3.14

Area of the sector =?

Plug the given values into the above formula and solve for the area of the sector:

Area A = (θ/360º) × πr²

Area A = ( 82/360) × 3.14 × 19²

Simplifying, we get:

Area = 258.20 square units

Therefore, the area of the sector measures 258.20 square units.