Areas Of Combinations Of Figures

 

We see different kind of shapes in our daily life. Most of the time they are combination of two shapes.

In the picture below, the lawn has two circular flower beds on two sides of a square area ABCD of side measuring 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.

Solution:

(1) Area of the square lawn ABCD = 56 × 56 m^2 ---------------------------------------------

Let OA = OB = x metres

So,  

x^2 + x^2 = 562

or,

2x^2 = 56 × 56

or,                               

x^2 = 28 × 56 ------------------------------------------------------------

(2) Now, area of sector OAB = 90/ 360 × π x^2) = 1/4 × π x^2 ---------------------------

= ¼  X  22/7 X 28 X 56 m^2--------------------- [From (2)] 

(3) Also, area of ∆ OAB = ¼  ×56 ×56 m^2 (∠ AOB = 90°)-------------------------------------------

So, area of flower bed AB = (1/4  * 22/7 * 28 *56 -1/4 *56*56) m^3

= ¼  * 28 * 56 ( 22/7 – 2)m^2

= ¼ * 28 * 56 * 8/7 m^2

(4) Area of the other flower bed = ¼ * 28*56 *8/7 m^2

Therefore, total area = (56 * 56 + ¼ *28 *56 * 8/7 + ¼ *28 * 56 *8/7) m^2

= 28*56( 2+2/7+2/7 )m^2

= 28 * 56 * 18/7 m2

= 4032 m^2