Project 1 [PERIMETERS AND AREAS OF RECTANGLES]
BACKGROUND
The most common problems that people come across in daily life are to fence their fields and to determine the size of the field to grow a particular type of crop. These two problems are closely related with the mathematical concepts - perimeter and area, respectively. There is an impression that if the perimeter of a figure is increased, then its area will also increase and vice-versa. However, this impression is not always correct. This project tries to explain this idea by taking rectangle as a basic figure.
OBJECTIVE
To explore the changes in behaviours of perimeters and areas of rectangles with respect to each other.
DESCRIPTION
(A) Rectangles with equal perimeters
Let us consider the rectangles of the following lengths and breadths:
(1) Length = 10 cm : Breadth = 6 cm
Length = 11 cm: Breadth = 5 cm
Length = 8 cm: Breadth = 8 cm
(IV) Length = 9.5 cm: Breadth = 6.5 cm
(V) Length = 8.5 cm : Breadth = 7.5 cm
We may note that perimeter of each of the above rectangles is the same.
namely. 32 cm. Let us calculate the areas of these rectangles:
[1] Area of the rectangle of length 10 cm and breadth 6 em
= 10 cm x 6 cm = 60 cm³.
(ii) Area of the rectangle of length 11 cm and breadth 5 cm
= 11 cm x 5 cm = 55 cm
(iii) Area of the rectangle of length 8 cm and breadth 8 cm
=8 cm x 8 cm = 64 cm² (Maximum)
(iv) Area of the rectangle of length 9.5 cm and breadth 6.5 cm
9.5 cm x 6.5 cm = 61.75 cm²
(v) Area of the rectangle of length 8.5 cm and breadth 7.5 cm
8.5 cm x 7.5 cm = 63.75 cm
From the above, it can be observed that in case (u), the rectangle has t maximum area. Also, in this case, the rectangle is a square.
(1)
(B) Rectangles with equal areas
Let us consider the rectangles with following lengths and breadths
Length = 16 cm: Breadth = 4 cm
Length = 8 cm: Breadth = 8 cm
Length 10 cm: Breadth=6,4 cm
length-22 cm Breadth - 2 cm
length=64 cm: Breadth-1cm
We may note that area of each of the above rectangles is the same namely
64 cm². Let us calculate the perimeter of each of these rectangles:
(1) Perimeter of rectangle of length 16 cm and breadth 4 cm
=2 (16+4) cm 40 cm
(1) Perimeter of rectangle of length 8 cm and breadth 8 cm
= 2 (8+8) cm = 32 cm (Minimum) (iii) Perimeter of rectangle of length 10 cm and breadth 6.4 cm
= 2 (10 +6.4) cm = 32.8 cm
(iv) Perimeter of rectangle of length 32 cm and breadth 2 cm = 2 (32 + 2) cm = 68 cm
(v) Perimeter of rectangle of length 64 cm and breadth 1 cm
= 2 (64+ 1) cm = 130 cm
From the above, it can be observed that in case (ii), the rectangle has the minimum perimeter. Further, in this case, the rectangle is a square.
CONCLUSION
The impression that when the perimeter of a rectangle is increased, its area is also increased and vice-versa is not correct. In fact,
(i) of all the rectangles with equal perimeters, the square has the
maximum area. (ii) of all the rectangles with equal areas, the square has t inimum perimeter.
APPLICATION
1. This project is useful in estimating a field of maximum area within a given fencing (perimeter) and also a fencing (or boundary) of minimum length enclosed in a given area. T
2. A square is a regular quadrilateral. Increasing the number of sides of a regular polygon, we can arrive at the result in Description (A) above, that with a given perimeter, a circle has the maximum area.
3. If the measures of length and breadth are considered only in terms of natural numbers, then in Description (B) above, it can be seen that with a given area, the perimeter is maximum, when one of the sides is of unit length [case (v) of BJ. However, if this condition is not there, then there is no limit for the perimeter to be maximum. That is, for a given area, a rectangle can have infinitely large perimeter. nom Is it not surprising that you can make a belt of infinite length having a given area?
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