Answer:
10.33 square units
Step-by-step explanation:
Area of the sector:
∠JKL = Ф= 74°
JK = r = 4
[tex]\boxed{\text{\bf Area of sector = $ \dfrac{\theta}{360}\pi r^2$}}[/tex]
Ф is the central angle of the sector.
r is the radius
[tex]\sf Area \ of \ the \ sector = \dfrac{74}{360}*3.14*4*4[/tex]
= 10.33 square units
