Find the areas of the following figures by counting squares:
(a) By observing the figure, we see that it contains 9 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 9 square units.
(b) By observing the figure, we see that it contains 5 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 5 square units.
(c) By observing the figure, we see that it contains 2 fully filled squares and 4 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 2 + (4× 0.5) = 2+ 2 = 4 square units.
(d) By observing the figure, we see that it contains 8 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 8 square units.
(e) By observing the figure, we see that it contains 10 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 10 square units.
(f) By observing the figure, we see that it contains 2 fully filled squares and 4 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 2 + (4× 0.5) = 2+ 2 = 4 square units.
(g) By observing the figure, we see that it contains 4 fully filled squares and 4 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 4 + (4× 0.5) = 4 + 2 = 6 square units.
(h) By observing the figure, we see that it contains 5 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 5 square units.
(i) By observing the figure, we see that it contains 9 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 9 square units.
(j) By observing the figure, we see that it contains 2 fully filled squares and 4 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 2 + (4× 0.5) = 2+ 2 = 4 square units.
(k) By observing the figure, we see that it contains 4 fully filled squares and 2 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 4 + (2× 0.5) = 4 + 1 = 5 square units.
(l) In the figure given, we find that there are variations in the filling of the squares.
Thus, we form a table to organise the data by observing the above figure.
Covered Area | Total Numbers | Area |
Fully filled squarea | 2 | 2 |
Half filled squares | 0 | 0 |
More than half filled squares | 6 | 6 |
Less than half filled squares | 6 | 0 |
Therefore,
Total Area = 2 + 6 = 8 square units
(m) In the figure given, we find that there are variations in the filling of the squares.
Thus, we form a table to organise the data by observing the above figure.
Covered Area | Total Numbers | Area |
Fully filled squarea | 5 | 5 |
Half filled squares | 0 | 0 |
More than half filled squares | 9 | 9 |
Less than half filled squares | 12 | 0 |
Therefore,
Total Area = 5 + 9 = 14 square units
(n) By observing above figure, we get
Covered Area | Total Numbers | Area |
Fully filled squarea | 8 | 8 |
Half filled squares | 0 | 0 |
More than half filled squares | 10 | 10 |
Less than half filled squares | 9 | 0 |
Therefore,
Total Area = 8 + 10 = 18 square units