**Problem:** A cow is grazing with a rope around its neck and the other end of the rope is tied to a pol. The length of the rope is 10 metres. There are two boundary walls perpendicular to each other, one at a distance of 5 metres to the east of the pole and another at a distance of \(5 \sqrt{2}\) metres to the north of the pole. Find the area the cow can graze on.

**Solution:** First we need to analyse the question and draw a diagram. The figure for the problem would look something like this

We have a circle with centre P and radius 10m. Let AC and AB be the boundary walls present. The striped region shows the area available for grazing. Now if we see carefully we can see that area of the shaded region is nothing but the sum of the areas of the rectangle AEPC, the two right angled triangles EPB and DPC and the sector BPC.

PB = PC = radius of the circle = 10m

PE = \(5\sqrt{2}\)m

AS \(PE^2 + EB^2 = PB^2\) (Pythagorean property of a right angled triangle)

From here, we get \(EB = 5\sqrt{2}\)m

Similarly, as PD = 5m, PC = 10m

\(CD = 5\sqrt{3}\)m (pythagorean property)

Read More…

*Related*

## No comments, be the first one to comment !