Three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with center at O. Horses are tied up at P, R and S such that PO and RO are the radii of semi-circles with centers at P and R respectively, and S is the center of the circle touching the two semi-circles with diameters AO and OB. The horses tied at P and R can graze within the respective semi-circles and the horse tied at S can graze within the circle centred at S. The percentage of the area of the semi-circle with diameter AB that cannot be grazed by the horses is nearest to
Show Hint
Take the radius of the big semi-circle as 2 units. Find the radius of the small circle S using the distance between the centres of two touching circles.
Step 1: Choose a convenient unit:
Let the big semi-circle with diameter AB have radius 2 units. Then AO = OB = 2, so the two small semi-circles centred at P and R have radius 1 unit each (their diameters are AO and OB).
We need the radius r of circle S, then we compare areas.
Step 2: Set up coordinates:
Put O at the origin with AB along the x-axis. Then P = (-1, 0), R = (1, 0), and S lies on the y-axis by symmetry, say S = (0, y).
Circle S touches the big arc from inside, so the distance from O to S plus r equals the big radius: \(y + r = 2\), which gives \(y = 2 - r\).
Step 3: Use the tangency with a small semi-circle:
Circle S also touches the small circle centred at P from outside. For two circles touching externally, the distance between centres equals the sum of the radii:
\[ PS = 1 + r \implies \sqrt{1^2 + y^2} = 1 + r \]
Square both sides and put \(y = 2-r\):
\[ 1 + (2-r)^2 = (1+r)^2 \implies 1 + 4 - 4r + r^2 = 1 + 2r + r^2 \]
\[ 5 - 4r = 1 + 2r \implies 6r = 4 \implies r = \frac{2}{3} \]
Step 4: Find the areas:
Area of the big semi-circle: \(\tfrac{1}{2}\pi (2)^2 = 2\pi\).
Grazed by P and R: each is a semi-circle of radius 1, so together \(2 \times \tfrac{1}{2}\pi(1)^2 = \pi\).
Grazed by S: a full circle of radius \(\tfrac{2}{3}\), area \(\pi \cdot \tfrac{4}{9} = \tfrac{4\pi}{9}\).
Total grazed area \(= \pi + \tfrac{4\pi}{9} = \tfrac{13\pi}{9}\).
Step 6: Compare with the options:
The value 27.8 percent is nearest to 28. Option (1) 20 and option (3) 36 are about 8 points away, and option (4) 40 is even further. So only option (2) fits.
Final Answer:
About 27.8 percent of the field is left ungrazed, which is nearest to 28.
\[ \boxed{\text{Option (2): } 28} \]
