Question

In circle K with m/JKL = 74° and JK = 4, find the area of sector

JKL. Round to the nearest hundredth.

Answer 1

10.33 square units

Explanation:

Area of the sector:

∠JKL = Ф= 74°

JK = r = 4

`\boxed{\text{\bf Area of sector = $ (\theta)/(360)\pi r^2$}}`

Ф is the central angle of the sector.

r is the radius

`\sf Area \ of \ the \ sector = (74)/(360)*3.14*4*4`

= 10.33 square units