In the figure (i) given below, two circular flower beds have been shown on the two sides of a square 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 the flower beds.

Area of lawn ABCD = (side)²

= (56)² = 3136 m².

In square,

Length of diagonal (AC) = $\sqrt{2}$ side = $56\sqrt{2}$ m.

Since, diagonals of square are equal and bisect each other,

∴ AO = OC = OD = OB.

Radius of quadrant ODC = Radius of quadrant OAB = $\dfrac{AC}{2} = \dfrac{56\sqrt{2}}{2} = 28\sqrt{2}$ m.

Area of quadrant ODC = Area of quadrant OAB = $\dfrac{1}{4}πr^2$

$= \dfrac{1}{4} \times \dfrac{22}{7} \times 28\sqrt{2} \times 28\sqrt{2} \\[1em] = 22 \times 28 \times 2 \\[1em] = 1232 \text{ m}^2.$

Since, diagonals of square divide it into four equal triangles.

Area of △ODC = Area of △OAB = $\dfrac{\text{Area of square}}{4} = \dfrac{3136}{4}$ = 784 m².

From figure,

Area of flower bed = Area of quadrant ODC - Area of △ODC

= 1232 - 784 = 448 m².

Since there are two flower beds so area = 2 × 448 = 896 m².

Area of lawn and flower beds = 3136 + 896 = 4032 m².

Hence, sum of the areas of the lawn and the flower beds = 4032 m².