Question

In fig two circular flower bds have been shown on two sides of a lawn $ABCD$ of side $56$m . If the centre of each circular flower bed is the point of intersection $O$ of the diagonals of the square lawn, find the sum of the areas of the lawn and flower beds

• A. $4032m^2$
• B. $4428m^2$
• C. $4628m^2$
• D. $Noneofthese$

Answer 1

Answer: (A)

$4032m^2$

We have,

AC = BD = $\sqrt{{56}^2+{56}^2}$ =$56\sqrt{2}$m

Therefore,OA = OB = $\frac{1}{2}$AC = $28\sqrt{2}$m

So, radius of the circle having centre at the point of intersection of diagonals is $28\sqrt{2}$m

Now,

Area of one circular end = Area of a segment of angle ${90}^o$ in a circle of radius $28\sqrt{2}$m.

=($\frac{22}{7}\times \frac{90}{360}-sin{45}^{o}cos{45}^o$)$\times (28\sqrt{2}{)}^2{m}^2$

$=(\frac{11}{14}-\frac{1}{2})\times 28\times 28\times 2m^2$

$=448m^2$

Therefore, area of flower beds = $2\times 448=896$ m${}^2$

Area of square lown = $56\times 56=3136$ m${}^2$

Hence, total area = $3136+896=4032$ m${}^2$