Question

3
in the given figure, two circular flower beds have been shown on 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.

Answer 1

➩Given:

→ Side of square ABCD=56 m

→ AC=BD(diagonals of a square are equal in lengths)

→ Diagonal of square (AC)

➩To Find:

→The sum of the areas of the lawn and the flower beds.

➩Solution:

AC=BD(diagonals of a square are equal in length)

Diagonal of square AC= $\sqrt{2} \times 56 = 56 \sqrt{2} \: m$

(We know that diagonal of square when intersected by the other diagonal bisects itself or divide itself into two equal parts)

$OA=OB= \frac{1}{2} AC = \frac{1}{2} (56\sqrt{2})$

$=28\sqrt{2}m$

→ OA and OB are also the radius for the sector OAB

Area of sector=$(\frac{90}{2})\pi \: {r}^{2}$

$=(\frac{1}{4})\pi \: {r}^{2}$

$=\frac{1}{4}×\frac{22}{7}×\: {28\sqrt{2}}^{2}$

$=\frac{1}{4}×\frac{22}{7}×\:{28}×\:{28}×2$

$={(1232)}m^{2}$

Till now we got the area of the sector OAB.

_______________________________________

Now let us try to find the area of △AOB

Formula will be $\frac{1}{2} × Base × Height$

Note:This formula can only be used if one angle of triangle△ will be 90°

$=\frac{1}{2} × OA× OB$

$=\frac{1}{2} × {28\sqrt{2}} × {28\sqrt{2}}$

$=\frac{1}{2} × {28} × {28}×{2}$

$={28} × {28}$

$={(784)}m^{2}$

Now we also got area of triangle AOB

_______________________________________

→From our knowledge about square, we got to know that area of △AOB=△BOC=△COD=△AOD

Now Sum of the areas of the lawn and the flower beds=Area of the sector OAB+Area of the sector ODC+Area of the △BOC+△AOD

=1232+1232+784+784

$={(4032)}m^{2}$