Please Wait... While Loading Full Video#### class-10 chapter-12

### 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.

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

`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

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

Class 10 Chapter 12 Example 5

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

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

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`