If E the universal set contains all rectangles of length x and breadth y, where x and y are integers and x>y.
A contains rectangles with area of 24m2
B contains rectangles , each with length multiples of 3 and area of 24m2.
Find n(A), n(B), and n(A ^B)
To find the values of n(A), n(B), and n(A ^ B) in this scenario, we need to analyze the sets A and B and understand their properties.
Set A contains rectangles with an area of 24m^2, whereas Set B contains rectangles with lengths that are multiples of 3 and an area of 24m^2.
1. n(A) – Number of rectangles in Set A:
To calculate n(A), we need to determine the number of rectangles in Set A. The given condition is that the area of each rectangle in A is 24m^2. We can start by finding the possible combinations of length and breadth that result in an area of 24m^2.
Since x > y, we can consider the following combinations:
- (x = 24, y = 1)
- (x = 12, y = 2)
- (x = 6, y = 4)
- (x = 4, y = 6)
- (x = 2, y = 12)
- (x = 1, y = 24)
Therefore, n(A) = 6, as there are six different combinations that satisfy the condition of having an area of 24m^2.
2. n(B) – Number of rectangles in Set B:
To calculate n(B), we need to determine the number of rectangles in Set B. The condition is that the length of each rectangle in B is a multiple of 3, and the area is 24m^2.
We can determine the possible values for the length and breadth of the rectangles that satisfy these conditions:
- (length = 6, breadth = 4)
- (length = 12, breadth = 2)
Therefore, n(B) = 2, as there are two combinations that satisfy the conditions.
3. n(A ^ B) – Number of rectangles in the intersection of A and B:
The intersection of sets A and B (A ^ B) consists of rectangles that satisfy both conditions – an area of 24m^2 and a length that is a multiple of 3.
The intersection is given by:
- (length = 6, breadth = 4)
Therefore, n(A ^ B) = 1, as there is only one rectangle that falls within the intersection of A and B.
In summary:
n(A) = 6
n(B) = 2
n(A ^ B) = 1