Areas of Combinations of plane figure:
Areas of Combinations of plane figure is the process of calculating the areas of different combinations of figures. these types of figures We come across in our daily life and also in the form of various interesting designs.
Flower beds, drain covers, window designs, We come across, designs on th etable covers, are some of such examples.We illustrate the process of calculating areas of these figures through some examples.
The following examples are combined with some plane figures.
Areas of Combinations of Plane Figure Problems:
Example:
Two circular flower beds have been shown on two sides of a square lawn ABCD of side 50m.If the center 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:
Area of the square lawn ABCD = 50 x 50 m2 --------------------- (1)
Let OA = OB = x metres
So x2 + x2 = 502
Or 2x2 = 50 x 50
X2 = 25 x 50 -----------------------(2)
Now ,
Area of sector OAB = `(90)/(360)` *`Pi` *x2
=`(1)/(4)` * `Pi` i*x2
= `(1)/(4)` x `(22)/(7)` x 25 x 50m2 [from (2)] ------(3)
Also, area of `Delta` OAD = `(1)/(4)` * 50 * 50 m2 (<AOB=90)------(4)
So, area of flower bed AB = (`(1)/(4)`* `(22)/(7)`*25*50 – `(1)/(4)`*50*50)m2 [from (3) and (4)]
= `(1)/(4)`*25*50(`(22)/(7)` -2)m2
=`(1)/(4)`*25*50* `(8)/(7)` m2 ----------------(5)
Similarly area of the other flower bed
= `(1)/(4)` * 25 * 50 * `(8)/(7)` m2 -------------------------(6)
Therefore,
Total area =(50*50 + `(1)/(4)` *25*50* `(8)/(7)` +`(1)/(4)`*25*50*`(8)/(7)`)m2 [from (1),(5) and (6)]
=25*50(2+`(2)/(7)`+`(2)/(7)`)m2
=25*50* `(18)/(7)` m2
= 3214. 29 m2
Areas of Combinations of Plane Figure Example 2:
Example 2:
Find the area of the shaded region in the following figure , where ABCD is a square of a side 10 cm.
Solution:
Area of square ABCD
= 10 * 10 cm2
= 100 cm2
Diameter of each circle = `(10)/(2)`cm=5cm
So, radius of each circle =(5)/(2) cm
<br>
=`(22)/(7)` *`(5)/(2)` * `(5)/(2)`cm2
=`(550)/(28)` cm2
Therefore area of the four circles = 4* `(550)/(28)`cm2 = 78.57Cm2
Hence area of the shaded region = (100-78.57)cm2= 21.43Cm2
Areas of Combinations of plane figure is the process of calculating the areas of different combinations of figures. these types of figures We come across in our daily life and also in the form of various interesting designs.
Flower beds, drain covers, window designs, We come across, designs on th etable covers, are some of such examples.We illustrate the process of calculating areas of these figures through some examples.
The following examples are combined with some plane figures.
Areas of Combinations of Plane Figure Problems:
Example:
Two circular flower beds have been shown on two sides of a square lawn ABCD of side 50m.If the center 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:
Area of the square lawn ABCD = 50 x 50 m2 --------------------- (1)
Let OA = OB = x metres
So x2 + x2 = 502
Or 2x2 = 50 x 50
X2 = 25 x 50 -----------------------(2)
Now ,
Area of sector OAB = `(90)/(360)` *`Pi` *x2
=`(1)/(4)` * `Pi` i*x2
= `(1)/(4)` x `(22)/(7)` x 25 x 50m2 [from (2)] ------(3)
Also, area of `Delta` OAD = `(1)/(4)` * 50 * 50 m2 (<AOB=90)------(4)
So, area of flower bed AB = (`(1)/(4)`* `(22)/(7)`*25*50 – `(1)/(4)`*50*50)m2 [from (3) and (4)]
= `(1)/(4)`*25*50(`(22)/(7)` -2)m2
=`(1)/(4)`*25*50* `(8)/(7)` m2 ----------------(5)
Similarly area of the other flower bed
= `(1)/(4)` * 25 * 50 * `(8)/(7)` m2 -------------------------(6)
Therefore,
Total area =(50*50 + `(1)/(4)` *25*50* `(8)/(7)` +`(1)/(4)`*25*50*`(8)/(7)`)m2 [from (1),(5) and (6)]
=25*50(2+`(2)/(7)`+`(2)/(7)`)m2
=25*50* `(18)/(7)` m2
= 3214. 29 m2
Areas of Combinations of Plane Figure Example 2:
Example 2:
Find the area of the shaded region in the following figure , where ABCD is a square of a side 10 cm.
Solution:
Area of square ABCD
= 10 * 10 cm2
= 100 cm2
Diameter of each circle = `(10)/(2)`cm=5cm
So, radius of each circle =(5)/(2) cm
<br>
=`(22)/(7)` *`(5)/(2)` * `(5)/(2)`cm2
=`(550)/(28)` cm2
Therefore area of the four circles = 4* `(550)/(28)`cm2 = 78.57Cm2
Hence area of the shaded region = (100-78.57)cm2= 21.43Cm2
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