Question

the diagram shows the universal set u = {parallelograms}. set a represents parallelograms with four congruent sides, while b represents parallelograms with four congruent angles. how many of the parallelograms fall into the category ( a cup b )? options: ( \bigcirc 6 ), ( \bigcirc 8 ), ( \bigcirc 18 ), ( \bigcirc 26 ). the venn diagram (labeled ( u )) has regions: outside both circles (3), circle b only (6), intersection of a and b (8), circle a only (12).

Explanation:

Step1: Recall the formula for the union of two sets

The formula for \( n(A \cup B) \) is \( n(A) + n(B) - n(A \cap B) \), where \( n(A) \) is the number of elements in set \( A \), \( n(B) \) is the number of elements in set \( B \), and \( n(A \cap B) \) is the number of elements in the intersection of \( A \) and \( B \).

Step2: Identify the values from the Venn diagram

• For set \( A \): The elements in \( A \) are the part only in \( A \) (12) and the intersection (8), so \( n(A)=12 + 8=20 \)? Wait, no, wait. Wait, actually, looking at the Venn diagram: Set \( A \) has the region with 12 (only \( A \)) and 8 (intersection). Set \( B \) has the region with 6 (only \( B \)) and 8 (intersection). Wait, no, maybe I misread. Wait, the Venn diagram: the universal set \( U \) has a region outside both circles with 3. Then circle \( B \) has a part labeled 6 (only \( B \)) and the intersection with \( A \) labeled 8. Circle \( A \) has a part labeled 12 (only \( A \)) and the intersection 8. So to find \( A \cup B \), we need the number of elements in \( A \) only, \( B \) only, and the intersection. So \( n(A \cup B)=n(\text{only } A)+n(\text{only } B)+n(A \cap B) \).

So only \( A \) is 12, only \( B \) is 6, and intersection is 8. So \( 12 + 6+ 8 = 26 \)? Wait, no, wait. Wait, maybe the numbers are: Set \( A \) (parallelograms with four congruent sides) has 12 (only \( A \)) and 8 (intersection with \( B \)). Set \( B \) (parallelograms with four congruent angles) has 6 (only \( B \)) and 8 (intersection with \( A \)). So \( A \cup B \) is the union, so we add the elements in \( A \) (12 + 8) and the elements in \( B \) that are not in \( A \) (6). Wait, no, the formula is \( n(A \cup B)=n(A)+n(B)-n(A \cap B) \). \( n(A)=12 + 8 = 20 \), \( n(B)=6 + 8 = 14 \), \( n(A \cap B)=8 \). So \( 20 + 14 - 8 = 26 \). Alternatively, adding only \( A \) (12), only \( B \) (6), and intersection (8): \( 12 + 6 + 8 = 26 \).

Answer:

26