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-cricles with diameters $AO$ and $OB$. The horses ties 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 semi-circles 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)

$28$

Let R be radius of big circle and r be radius of circle with center S. Radius of 2 semicircles is $\frac{R}{2}$

From Right angled triangle OPS, using Pythagoras theorem we get

$(r+0.5R{)}^2=(0.5R{)}^2+(R-r{)}^2$

We get $R=3r$.

Now the area of big semicircle that cannot be grazed is

$=$ Area of big S.C $-$ area of 2 semicircle $-$ area of small circle.

$=5\times \pi \times \frac{R^2}{36}$ .

This is about 28 % of the area of $\frac{\pi R^2}{2}$.