Class 10 Areas of Combinations of Plane Figures

### Topic Covered

♦ Areas of Combinations of Plane Figures

### Areas of Combinations of Plane Figures

So far, we have calculated the areas of different figures separately. Let us now try to calculate the areas of some combinations of plane figures.

We come across these types of figures in our daily life and also in the form of various interesting designs. Flower beds, drain covers, window designs, designs on table covers, are some of such examples.

Q 3240501413

In Fig. 12.15, 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.
Class 10 Chapter 12 Example 4
Solution:

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

Let OA = OB = x metres

So, x^2 + x^2 = 56^2

or, 2 x^2 = 56 × 56

or, x^2 = 28 × 56 .......... (2)

Now, area of sector OAB = 90/360 xx pi x^2 = 1/4 xx pi x^2

 = 1/4 xx 22/7 xx 28 xx 56 m^2 [From (2)] ........(3)

Also, area of Delta OAB = 1/4 xx 56 xx 56 m^2 (∠ AOB = 90°) .............(4)

So, area of flower bed AB = (1/4 xx 22/7 xx 28 xx 56 -1/4 xx 56 xx 56) m^2

[From (3) and (4)]

= 1/4 xx 28 xx56 ( 22/7 -2) m^2

 = 1/4 xx 28 xx 56 xx 8/7 m^2 ...............(5)

Similarly, area of the other flower bed

 =1/4 xx 28 xx 56 xx 8/7 m^2  ..............(6)

Therefore, total area = (56 xx 56 +1/4 xx 28 xx 56 xx 8/7 +1/4 xx 28 xx 56 xx 8/7) m^2 [From (1), (5) and (6)]

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

 = 28 xx 56 xx 18/7 m^2 = 4032 m^2

Alternative Solution :

Total area = Area of sector OAB + Area of sector ODC

+ Area of Delta OAD + Area of Delta OBC

 = ( 90/360 xx 22/7 xx 28 xx 56 + 90/360 xx 22/7 xx 28 xx 56 +1/4 xx 56 xx 56 +1/4 xx 56 xx 56) m^2

 = 1/4 xx 28 xx 56 (22/7 + 22/7 +2 +2) m^2

= (7 xx 56)/7 (22 + 22 + 14 +14) m^2

 = 56 xx72 m^2 = 4032 m^2
Q 3250501414

Find the area of the shaded region in Fig. 12.16, where ABCD is a square of side 14 cm.
Class 10 Chapter 12 Example 5
Solution:

Area of square ABCD

= 14 × 14 cm^2 = 196 cm^2

Diameter of each circle = 14/2 cm = 7 cm

So , radius of each circle = 7/2 cm

So, area of one circle = πr^2 = 22/7 xx 7/2 xx 7/2 cm^2

= 154/4 cm = 77/2 cm^2

Therefore, area of the four circles = 4 xx 77/2 cm^2 = 154 cm^2

Hence, area of the shaded region = (196 -154) cm^2 = 42 cm^2 .
Q 3200501418

Find the area of the shaded design in Fig. 12.17, where ABCD is a
square of side 10 cm and semicircles are drawn with each side of the square as
diameter. (Use π = 3.14)
Class 10 Chapter 12 Example 6
Solution:

Let us mark the four unshaded regions as I, II, III and IV (see Fig. 12.18).
Area of I + Area of III

= Area of ABCD – Areas of two semicircles of each of radius 5 cm

 = (10 xx 10 -2 xx 1/2 xx pi xx 5^2) cm^2 = (100 - 3.14 xx 25) cm^2

= (100 - 78.5) cm^2 =21.5 cm^2

Similarly, Area of II + Area of IV = 21.5 cm^2

So, area of the shaded design = Area of ABCD – Area of (I + II + III + IV)

= (100 – 2 × 21.5) cm^2 = (100 – 43) cm^2 = 57 cm^2