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45%
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The figure above represents a semicircular garden that is enclosed by 20 feet of fencing and a straight garden wall. What is the area, in square feet, of the garden?
A. \(\frac{20}{\pi}\)
B. \(\frac{50}{\pi}\)
C. \(\frac{100}{\pi}\)
D. \(\frac{200}{\pi}\)
E. \(\frac{400}{\pi}\)
Attachment:
1.jpg [ 23.49 KiB | Viewed 7660 times ]
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Bunuel
Explanation:
2πr/2=20
r=20/π
Area of Semi-circle= (πr^2)/2= π{(20/π)^2}/2
= 200/π
IMO-D
Circumference = 2πr
Half of circumference is 20. So full circumference is 40.
Therefore:
2πr=40
=> πr=40/2
=> πr=20
=> r=20/π
Area of the garden
(πr^2)/2
= (π (20/π) (20/π))/2 = (400/π)/2 = 200/π Option D
Half of circumference is 20. So full circumference is 40.
Therefore:
2πr=40
=> πr=40/2
=> πr=20
=> r=20/π
Area of the garden
(πr^2)/2
= (π (20/π) (20/π))/2 = (400/π)/2 = 200/π Option D
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