Two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 cm - Maths - Areas Related to Circles

Question

Two circular flower beds have been shown on two sides of a
square lawn ABCD of side 56 cm If the centre of each circular
flower bed is the point of intersection O of the diagonals of of
the square lawn, find the sum of the areas of the lawn and flower
beds

Koka Sri Lakshmi Divya Sai · Accepted Answer

$Given Side of the square=56m From the fig in Δ BCD {BD}^{2} = {BC}^{2} + {DC}^{2} \Rightarrow {BD}^{2} = (56)^{2} + (56)^{2} \Rightarrow {BD}^{2} = 2 {(56)}^{2} \Rightarrow BD=56 \sqrt{2} m we know that diagnols of a square are equal and perpendicular bisector of each other \Rightarrow AC=BD= 56 \sqrt{2} m and AO=BO=CO=DO= \frac{56 \sqrt{2}}{2} = 28 \sqrt{2} m Also area (Δ AOD) = area (Δ DOC)=area(Δ BOC)=area(Δ AOB) = \frac{1}{4} area(ABCD)= \frac{1}{4} \times {(56)}^{2} = \frac{3136}{4} = 784 m^{2} and area sector OAB = area sector OCD= \frac{1}{4} π \times (28 \sqrt{2})^{2} = \frac{1}{4} \times \frac{22}{7} \times 1568 = 1232 m^{2} T hus required area=Area of Δ AOD+Area of sector ODC+Area of Δ BOC + Area of sector OAB = 784 + 1232 + 784 + 1232 = 4032 m^{2}$

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