Question

Three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with centre at O. Horses are tied up at P, R and S such that PO and RO are the radii of semi-circles with centres at P and R respectively, and S is the centre 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-circle and the horse tied at S can graze within the circle centred at S. The percentage of the area of the semicircles with diameter AB that cannot be grazed by the horses is nearest to:

• A. $20$
• B. $28$
• C. $36$
• D. $40$

Answer 1

Answer: (B)

$28$

Using Pythagoras theorem:

$P{S}^2=P{O}^2+S{O}^2\Rightarrow (R+r{)}^2=R^2+(2R-r{)}^2$

$\Rightarrow R^2+r^2+2Rr$

$=R^2+4R+r^2-4Rr$

$\Rightarrow 4R^2-6Rr=0$

$\Rightarrow 2R(2R-3r)=0$

$\Rightarrow R=\frac{3r}{2}$

Total grazed area = $\frac{\pi R^2}{2}+\frac{\pi R^2}{2}+\pi {\left(\frac{2R}{3}\right)}^2$

$=\pi R^2+\frac{4}{9}\pi R^2$

Ungazed area = $\pi \frac{(2R{)}^2}{2}-\pi R^2-\frac{4}{9}\pi R^2$

$=\frac{5\pi R^2}{9}$

Percentage of gazed area = $\frac{\frac{5}{9}\pi R^2}{2\pi R^2}\times 100$

$=\frac{500}{18}=28$%