Question

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

Answer 1

Hint: As flower beds are in circular shape it means they represent the sector of circle hence we will find the area of sector and for this we need radius of circle which is represented by OA or OB or OC or OD, find any one of these using Pythagoras theorem. Then the area of the segment is equal to the difference between the area of the sector and the triangle formed by it. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given that the side length of the square lawn ABCD is 56 meters.
We know that the area of a square whose side length is \[a\] meters is given by \[{a^2}{{\text{m}}^2}\].
So, area of the square lawn ABCD = \[{\left( {56} \right)^2} = 56 \times 56 = 3136{{\text{m}}^2}\]
Now we have to find the area of the flower bed AB
Area of the flower bed AB = area of segment AB
                                               = area of sector OAB – area of \[\Delta AOB\]
First, we will find the sides OA and OB.
We know that, diagonals of a square bisect each other at right angles and are equal in lengths.
So, \[\angle AOB = {90^0}\] and \[OA = OB\]
Let \[OA = OB = x\]
Now, in right angle triangle AOB
Using Pythagoras theorem \[{\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{height}}} \right)^2} + {\left( {{\text{base}}} \right)^2}\], we have
\[
   \Rightarrow {\left( {AB} \right)^2} = {\left( {OA} \right)^2} + {\left( {OB} \right)^2} \\
   \Rightarrow {\left( {56} \right)^2} = {\left( x \right)^2} + {\left( x \right)^2}{\text{ }}\left[ {\because A = 56} \right] \\
   \Rightarrow 56 \times 56 = 2{x^2} \\
  \therefore {x^2} = \dfrac{{56 \times 56}}{2} = 56 \times 28 \\
\]
For the sector OAB, radius = OA = OB = \[x\] and \[\angle AOB = {90^0}\]
We know that the area of a sector with radius \[r\] and by making an angle \[\theta \] at the center is given by \[\dfrac{\theta }{{{{360}^0}}} \times \pi {r^2}\].
So, area of sector OAB = \[\dfrac{{{{90}^0}}}{{{{360}^0}}} \times \pi \times {\left( x \right)^2} = \dfrac{1}{4} \times \dfrac{{22}}{7} \times {x^2} = \dfrac{1}{4} \times \dfrac{{22}}{7} \times 56 \times 28 = 1232{{\text{m}}^2}\]
We know that diagonals of a square divides the square into congruent triangles.
Therefore, \[\Delta AOB \cong \Delta BOC \cong \Delta COD \cong \Delta AOD\]
Congruent triangles in a square have the same area.
So, \[ar\left( {\Delta AOB} \right) = ar\left( {\Delta BOC} \right) = ar\left( {\Delta COD} \right) = ar\left( {\Delta AOD} \right)\]
\[
   \Rightarrow ar\left( {\Delta AOB} \right) = ar\left( {\Delta BOC} \right) = ar\left( {\Delta COD} \right) = ar\left( {\Delta AOD} \right) = \dfrac{1}{4}ar\left( {ABCD} \right) \\
   \Rightarrow ar\left( {\Delta AOB} \right) = ar\left( {\Delta BOC} \right) = ar\left( {\Delta COD} \right) = ar\left( {\Delta AOD} \right) = \dfrac{1}{4}\left( {3136} \right) \\
  \therefore ar\left( {\Delta AOB} \right) = ar\left( {\Delta BOC} \right) = ar\left( {\Delta COD} \right) = ar\left( {\Delta AOD} \right) = 784{{\text{m}}^2} \\
\]
Area of the flower bed AB = area of sector AOB – area of \[\Delta AOB\]
 \[
   = 1232 - 784 \\
   = 448{{\text{m}}^2} \\
\]
Similarly, by symmetry
Area of flower bed CD \[ = 448{{\text{m}}^2}\]
Area of lawn + area of flower beds = area of ABCD + area of flower bed AB + area of flower bed CD
  \[
   = 3136 + 448 + 448 \\
   = 4032{{\text{m}}^2} \\
\]
Hence, total area of the lawn and the flower beds \[ = 4032{{\text{m}}^2}\]

Note: The area of a square whose side length is \[a\] meters is given by \[{a^2}{{\text{m}}^2}\]. Diagonals of a square bisect each other at right angles and are equal in lengths. The area of a sector with radius \[r\] and by making an angle \[\theta \] at the center is given by \[\dfrac{\theta }{{{{360}^0}}} \times \pi {r^2}\]. Diagonals of a square divide the square into congruent triangles. Never forget to write the units after completion of the solution.