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Areas of Combination of Plane Figures

Earlier, we have learnt how to find areas of different figures separately.

Now, in this topic, we will learn how to calculate the areas of combined plane figures.

These figures can be seen in our daily life as variety of designs. For example, flower beds, foot path, table cloths, window and door designs, drain covers etc.

We will learn in detail  how to calculate these areas. Following are some of the examples of the
combined plain figures related to circles, the areas of which will be calculated:

 

Example

In the adjoining figure, OACB represents a quadrant of circle of radius 3.5 cm with centre O.

(i) Calculate the area of quadrant OACB.

 (ii)  Given OD = 2 cm, calculate the area of shaded portion (Take

  π = ).
Solution

(i) θ = 90° , r = 3.5 cm

 Area of the quadrant = 

                                 =

π (3.5)× 

                                 = 

× × =

                                 = 9.625 cm2


(ii) Area of Δ AOD = × AO × OD = × 3.5 × 2 = 3.5 cm2.

Therefore area of the shaded portion = Area of the quadrant - Area of the triangle AOD
=9.625 - 3.5
= 6.125 sq units
 


 

Example

The area of an equilateral triangle is 17300 cm2. With each vertex as centre, a circle is described with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles (Use

π = 3.14 and  = 1.73).

Solution

Let a be the side of the equilateral Δ.

Then area of the equilateral triangle = a= 17300 cm2

                 

 a=   =  = 40000

                 

 a =  = 200 cm.

side of equilateral triangle = 200 cm.

Radius of one sector (r) =  = 100 cm

Angle measure of the sector (θ) = 60°

Area of the sector  = 

                          = 3.14

× (100)2 × 

                                = 

                                = 

cm Area of three sectors = × 3 = 15700 cm Area of the triangle not included in the circle = 17300 - 15700 = 1600 cm2.
 



 

Example

 ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm and at  the centre a circle of radius 1 cm are drawn, as shown. Find the area of the green colour shaded  region (Use π =).

Solution

Area of the square ABCD = (4)= 16 cm2

Area of the circle at the centre =

π r= 3.14 × (1)2 = 3.14 cm2

Area of one quarter = 

                             =

π × (1)2 × 

                                   = cm2

 Area of four quarter parts = 4 × cmπ cm= 3.14 cm2 ∴ Area of the shaded region = 16 - (3.14 +3.14) = 9.72 cm2
 


 

Example

The shaded portion of figure represents the area swept by a wiper of a car. Calculate the area swept by the wiper, if OA = 7 cm and OC = 21 cm (Use

π ).

Solution

Radius of the sector OBC = r = 21 cm, θ = 30°.

Area of the sector OBC = 

                                 = × (21)2 × 

                                        =

                                 =cm2

Radius of the sector OAD = r = 7 cm, θ = 30°

 Area of the sector OAD = 

                                   = 

× (7)2 × 

                                          = 

                                         = cm2

 The area swept by the wiper= -

                                          =  = 102.67 cm2.

 



 

Example

It is proposed to lay to a square lawn measuring 58 m a side, two circular ends, the centre of each circle being the point of intersection of the diagonals of the square. Find the area of the whole lawn.

 
Solution

Area of a square lawn = 58m2 = 3364 m2 
In a right triangle ABC, using Pythagorean theorem we have:
AC2 = AB2 + BC2
AC2 = 582 + 582
AC2 = 2 x 582 
AC = ±
AC = ± 58

2 m
Since the diagonal of a square cannot be negative, we reject AC = - 58
2
We know that the diagonals of a square bisect each other

  r =  =  = 292 m 

Area of a sector BOC = 

                                = 

                                = 

                                = 
                                =1321.571 m2

 The area of the whole lawn  =  (the area of a square) + 2(Area of the sector)

                                            = 

                                            = 

                                            = 1682 + 2643.14
                                            = 4325.14 m2.
 


 

Example

The square water tank has its sides equal to 40 m. There are four semi-circular grassy plots all round it. Find the cost of surfing the plots at Rs.1.25 per sq. m. (Use π = 3.14).

Solution

ABCD is the square plot of side = 40 m.

(Here 4 semicircles combine to form two circle of diameter 40 m)

The area of the four semi circular plots    = area of two circles with radius 20 m.

Hence area of the four semi circular plots = 2

π (20 × 20).

                                                           = 800

π sq. m.

The cost of turfing the plot at Rs. 1.25 per sq. m.  = 800

π × 1.25
                                                                         = 800 
× 3.14 × 1.25
                                                                         = Rs. 3140. 





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