Now in the above square piece
side of square = 1 unit
area of square = 1 × 1 = 1 sq. unit.
and perimeter of square = 1 + 1 + 1 + 1 = 4 units.
Now after folding the above square piece in half becomes 2 rectangles
Perimeter of rectangle R
1
= 1 + \(\frac{1}{2}\) + 1 + \(\frac{1}{2}\) = 3 units.
Area of rectangle R
1
= \(\frac{1}{2}\) × 1 = \(\frac{1}{2}\) sq. unit.
Perimeter of rectangle R
2
= 1 + \(\frac{1}{2}\) + 1 + \(\frac{1}{2}\) = 3 units.
Area of rectangle R
2
= \(\frac{1}{2}\) × 1 = \(\frac{1}{2}\) sq. unit.
(a) Now, area of rectangle R
1
= area of rectangle R
2
= \(\frac{1}{2}\) < 1.
Hence, option (a) is not true.
(b) Here perimeter of square = 4 units
and perimeters of both the rectangles = 3 + 3 = 6 units.
which is greater than 4 units.
Hence option (b) is not true.
(c) Here perimeters of both the rectangles = 6 units
and perimeter of square = 4 units × 1\(\frac{1}{2}\) = 4 × \(\frac{3}{2}\) = 6 units.
The perimeters of both the rectangles added together are 1\(\frac{1}{2}\) times the perimeter of the square.
Hence, option (c) is true.
(d) Here, the area of the square = 4 units
and areas of both the rectangles = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1 unit.
The area of the square is four times the area of both rectangles.
Hence, option (d) is not true.
Intext Questions
Matha Pachchi! (Page No. 133)
Akshi and Toshi start running along the rectangular tracks as shown in the figure. Akshi runs along the outer track and completes 5 rounds. Toshi runs along the inner track and completes 7 rounds. Now, they are wondering who ran more. Find out who ran the longer distance.
Solution:
Here, perimeter of rectangular track PQRS = 2 × (l + b)
= 2 × (70 + 40)
= 2 × 110
= 220 m
and perimeter of rectangular track ABCD = 2 × (60 + 30)
= 2 × 90
= 180 m
Deep Dive: (Page No. 134)
In races, usually, there is a common finish line for all the runners. Here are two square running tracks with an inner track of 100 m on each side and an outer track of 150 m on each side. The common finishing line for both runners is shown by the flags in the figure which are in the center of one of the sides of the tracks. If the total race is 350 m, then we have to find out where the starting positions of the two runners should be on these two tracks so that they both have a common finishing line after they run for 350 m. Mark the starting points of the runner on the inner track as ‘A ’ and the runner on the outer track as ‘B’.
Solution:
Inner Track (100 m per side)
Perimeter Calculation: The perimeter of the inner track is (4 times 100 = 400) meters.
Distance to Run: The runner on the inner track needs to run 350 meters.
Starting Position (A): Since the perimeter is 400 meters, the runner will start 50 meters before the common finish line (400 – 350 = 50 meters).
Outer Track (150 m per side)
Perimeter Calculation: The perimeter of the outer track is (4 times 150 = 600) meters.
Distance to Run: The runner on the outer track also needs to run 350 meters.
Starting Position (B): Since the perimeter is 600 meters, the runner will start 250 meters before the common finish line (600 – 350 = 250 meters).
Perimeter of a Regular Polygon (Page No. 135)
Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalize your understanding of the perimeter of other regular polygons.
Solution:
Some common objects with regular shapes and calculating their perimeters:
Here are a few examples:
1. Square Table:
Shape: Square
Side Length = 1 meter
Perimeter = 4 × 1 = 4 meters
2. Equilateral Triangle Clock:
Shape: Equilateral Triangle
Side Length = 30 cm
Perimeter = 3 × 30 = 90 cm
3. Hexagonal Tile:
Shape: Regular Hexagon
Side Length = 10 cm
Perimeter = 6 × 10 = 60 cm
In general, the Perimeter of a Regular Polygon = (Number of sides) × (Side length of a polygon) units.
Split and Rejoin (Page No. 136)
A rectangular paper chit of dimension 6 cm × 4 cm is cut as shown into two equal pieces. These two pieces are joined in different ways.
For example, the arrangement a. has a perimeter of 28 cm.
Find out the length of the boundary (i.e., the perimeter) of each of the other arrangements below.
Solution:
(b)
∴ Length of boundary = AB + BC + CD + DE + EF + FG + GA
= 6 + 2 + 6 + 2 + 4 + 6 + 2
= 28 cm
(c)
Total length of boundary = AB + BC + CD + DE +EF + FG + GH + HA
= 2 + 6 + 2 + 2 + 6 + 2 + 6 + 2
= 28 cm
(d)
Total length of boundary = AB + BC + CD + DE + EF + FG + GH + HA
= 6 + 2 + 3 + 2 + 6 + 2 + 3 + 2
= 26 cm
Arrange the two pieces to form a figure with a perimeter of 22 cm.
Solution:
Arranging the two pieces in such a way that they form a new shape with the desired perimeter:
Total length of boundary = AB + BC + CD + DE + EF + FG + GH + HA
= 2 + 1 + 2 + 6 + 2 + 1 + 2 + 6
= 22 cm
Find the area of the following figures. (Page No. 140)
Solution:
(i)
∴ Total area of the figure = 3 + 1 = 4 sq. units
(ii)
∴ Total area of the figure = 6 + 3 = 9 sq. units
(iii)
∴ Total area of the figure = 7 + 3 = 10 sq. units
(iv)
∴ Total area of the figure = 8 + 3 = 11 sq. unit
Let’s Explore! (Page No. 141)
On a squared grid paper (1 square = 1 square unit), make as many rectangles as you can whose lengths and widths are a whole number of units such that the area of the rectangle is 12 square units.
(a) Which rectangle has the greatest perimeter?
(b) Which rectangle has the least perimeter?
(c) If you take a rectangle of area 32 sq cm, what will your answers be? Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter? Give examples and reasons for your answer.
Solution:
Perimeter of (a) = 2(1 + 24) = 2 × 25 = 50 units
Perimeter of (b) = 2(2 + 12) = 2 × 14 = 28 units
Perimeter of (c) = 2(4 + 6) = 2 × 10 = 20 units
Perimeter of (d) = 2(3 + 8) = 2 × 11 = 22 units
(a) Clearly rectangle (a) has the greatest perimeter.
(b) Obviously rectangle (c) has the least perimeter.
(c) Yes, it is possible to predict the shape of a rectangle with the greatest and least perimeter for a given area. Here’s how:
Greatest Perimeter: For a given area, the rectangle with the greatest perimeter will have one side as small as possible. This essentially means that the rectangle becomes very elongated.
For example, if the area is 24 square units, a rectangle with dimensions 1 unit by 24 units will have the greatest perimeter.
Example: Area = 24 square units
Dimensions = 1 unit by 24 units
Perimeter = 2(1 + 24) = 50 units
Least Perimeter: The rectangle with the least perimeter for a given area will be as close to a square as possible. This is because a square has the smallest perimeter for a given area among all rectangles.
Example: Area = 24 square units
Dimensions = 4 units by 6 units (since 4 × 6 = 24)
Perimeter = 2(4 + 6) = 20 units
Reasoning
Greatest Perimeter: When one side is minimized, the other side must be maximized to maintain the same area. This increases the sum of the sides, thus increasing the perimeter.
Least Perimeter: A square or a shape close to a square minimizes the sum of the sides for a given area, thus minimizing the perimeter.