Question

A stadium is in circular shape. Within the stadium some areas have been allotted for a hockey court and a javelin range, as given in the figure. Assume the shape of the hockey court and the javelin range to be square and triangle, resp. The curators would like to accommodate a few more sports in the stadium. Help them by measuring the unallocated region within the stadium (the radius of the stadium is 200 mts.).

• A. 22853.14 m²
• B. 22857.14 m²
• C. 20000 m²
• D. 62857.14 m²

Answer 1

The correct option is B

22857.14 m²

We need to find the unallocated area within the stadium.

The unallocated area should be = The total circular area of the stadium – Area of the hockey court - Area of the Javelin Range.

The area of circular stadium = $\pi \times r^2$

= $\frac{22}{7}\times {200}^2$

= $\frac{440000}{7}$

= 62857.14 $m^2$

The area of hockey court (square), we know that the radius of the stadium forms the diagonal of the hockey court.

Therefore the sides of the hockey court will be a =$\frac{r}{\sqrt{2}}$, (applying Pythagoras Theorem in a square)

Then, the area of the square = $a^2$

=$(\frac{r}{\sqrt{2}}{)}^2$

=$\frac{r^2}{2}$

= $\frac{{200}^2}{2}$

= 20000 $m^2$

Coming to the Javelin Range, since the angle formed by triangle at the centre of the stadium is ${90}^{\circ }$consider the triangle to be a right triangle. And the sides are equal to the radius of the stadium.

Therefore the area of the right triangle = $\frac{1}{2}\times r\times r$

= $\frac{1}{2}\times 200\times 200$

= 20000 $m^2$

Therefore the unallocated area in the stadium =Total area of stadium – Area of the hockey court – Area of the Javelin Range

= 62857.14 - 20000 - 20000

= 22857.14 $m^2$

The unallocated area within the stadium is 22857.14 $m^2$.