Question

fill in the missing numbers along the sides of the triangle so that it contains each of the numbers from 4 through 12 exactly once. furthermore each side of the triangle should contain four numbers whose sum is 32the pair of numbers that can be used for A and B is ___. the pair of numbers that can be used for C and D is ____. and the pair of numbers that can be used for E and F is___.

Answer 1

The missing numbers along the sides of the triangle

A and B: (10, 6)
C and D: (12, 8)
E and F: (12, 8)

Calculate the Missing Numbers
For A and B:
1. Add the two numbers at the vertex: 10 + 6 = 16
2. Subtract it from 32: 32 - 16 = 16
3. Divide by 2: 16 / 2 = 8
4. Add 2 to it to get the first number: 8 + 2 = 10
5. Subtract 2 to get the second number: 8 - 2 = 6
So, the missing numbers for A and B are 10 and 6.

For C and D:
1. Add the two numbers at the vertex: 12 + 8 = 20
2. Subtract it from 32: 32 - 20 = 12
3. Divide by 2: 12 / 2 = 6
4. Add 2 to it to get the first number: 6 + 2 = 8
5. Subtract 2 to get the second number: 6 - 2 = 4
So, the missing numbers for C and D are 8 and 4.

For E and F:
1. Add the two numbers at the vertex: 12 + 8 = 20
2. Subtract it from 32: 32 - 20 = 12
3. Divide by 2: 12 / 2 = 6
4. Add 1 to it to get the first number: 6 + 1 = 7
5. Subtract 1 to get the second number: 6 - 1 = 5
So, the missing numbers for E and F are 7 and 5.

Therefore, the missing numbers along the sides of the triangle
A and B: (10, 6)
C and D: (8, 4)
E and F: (7, 5)

Answer 2

For A and B, we have (10, 6)

For C and D, we have (12, 8)

For E and F, we have (12, 8)

Step-by-step explanation:

To determine the missing numbers along each side of the triangle, we have to

*Add the two numbers at the vertex

*Subtract it from 32

*Divide it by 2

*Add 2 to it to have the 1st number

*Subtract 2 from it to have the 2nd number

So to find the pair of numbers that can be used for A and B, we'll have;

`\begin{gathered} 12+4=16 \\ 32-16=16 \\ (16)/(2)=8 \\ 8+2=10 \\ 8-2=6 \end{gathered}`

Therefore the missing numbers for A and B are 10 and 6.

So to find the pair of numbers that can be used for C and D, we'll have;

`\begin{gathered} 4+8=12 \\ 32-12=20 \\ (20)/(2)=10 \\ 10+2=12 \\ 10-2=8 \end{gathered}`

Therefore the missing numbers for C and D are 12 and 8.

So to find the pair of numbers that can be used for E and F, we'll have;

`\begin{gathered} 12+8=20 \\ 32-20=12 \\ (12)/(2)=6 \\ 6+1=7 \\ 6-1=5 \end{gathered}`

So to avoid repetition of any of the numbers between 4 and 12, we have to add and subtract 1 instead of 2.