The surface area, S, of a sphere of radius r feet is S = S(r) = 4πr^2. Find the instantaneous rate of change of the surface area
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Answer:
The length of the track is 247 m.
The area enclosed by the track is 5391 . 76 sq m.
Step-by-step explanation:
The length of the rectangle = 88 m
The width of the rectangle = 44 m
Now, as we know the semicircles are at the end of rectangular track.
So, the diameter of semicircles = Width of the rectangle
⇒ The diameter of the semi circle = 44 m
⇒Radius of the semi circle = 44/2 = 22 m
TOTAL LENGTH OF TRACK
= 2 ( Length of rectangle) + 2 (Circumference of 1 semicircle)
= 2 x (88 m) + 2 ()
= 176 m + 2 (3.14)(22 m) = 176 + 71.08 = 247 m
Hence, the length of the track is 247 m
TOTAL AREA OF TRACK
= (Area of rectangle) + 2 (Area of 1 semicircle)
= (Length x Width) + 2 ()
= (44 m x 88 m) + (3.14)(22 m)(22 m) = (3872 + 1519.76) sq m
= 5391 . 76 sq m
Hence, the area enclosed by the track is 5391 . 76 sq m.