Let's look at the pink rectangle (on the right-hand side). Its area is $20,$ and its width is $4.$ Therefore, the length of this rectangle must be

\[ 20 \div 4 = 5. \]

Therefore, the missing number is $\color{blue}5.$

Let's look at the blue rectangle (the one on the left-hand side). Its area is $560,$ and its width is $7.$ Therefore, the length of the rectangle must be \[ 560 \div 7 \]

To carry out this division, we write $560$ as a product with $10$ as one of the factors and then swap the order of multiplication and division: \begin{align*} 560 \div 7 & =\\[5pt] 56 \times 10 \div 7 &=\\[5pt] 56 \div 7 \times 10 &=\\[5pt] ({\color{black}{56 \div 7}}) \times 10 &=\\[5pt] {\color{black}{8}} \times 10 & =\\[5pt] {\color{blue}80}& \end{align*}

Let's add this to our model:

We need to look at the blue and pink rectangles:

• Let's first look at the blue rectangle (the one on the left-hand side). Its area is $360,$ and its width is $6.$ Therefore, the length of the rectangle must be \[ 360 \div 6. \] To carry out this division, we write $360$ as a product with $10$ as one of the factors and then swap the order of multiplication and division: \begin{align*} 360 \div 6 & = \\[5pt] 36 \times 10 \div 6 &=\\[5pt] 36 \div 6 \times 10 &=\\[5pt] ({\color{black}{36 \div 6}}) \times 10 &=\\[5pt] {\color{black}{6}} \times 10 &=\\[5pt] {\color{blue}60}& \end{align*}

  Let's add this to our model:

• Now, let's look at the pink rectangle (the one on the right-hand side). Its area is $42,$ and its width is $6.$ Therefore, the length of the rectangle must be \[ 42 \div 6 = {\color{blue}7}. \] Let's add this to our model:
  

Let's look at the big rectangle:

• its length is $200+3=203$

• its width is $5$

• its area is $1,000+15=1,015$

Therefore, this area model can be used to represent the following multiplication problem:

\[ 1,015 = 203 \times 5 \]

Writing this as a division problem, we get

\[ 1,015 \div 5 = 203. \]

Therefore, the missing number is $5.$

Let's look at the big rectangle:

• its length is $10+2=12$

• its width is $9$

• its area is $90+18=108$

Therefore, this area model can be used to represent the following multiplication problem:

\[ 108 = 12 \times 9 \]

Writing this as a division problem, we get

\[ 108 \div 9 = 12. \]

Therefore, the missing number is $108.$

Let's look at the big rectangle:

• its length is $70 + 7 = 77$

• its width is $6$

• its area is $420 + 42 = 462$

Therefore, this area model can be used to represent the following multiplication problem: \[ 462 = 77 \times 6 \]

Writing this as a division problem, we get \[ 462 \div 6 = \bbox[3pt,Gainsboro]{\color{blue}77}. \]

Therefore, the missing number is $\bbox[3pt,Gainsboro]{\color{blue}77}.$

Let's look at the pink rectangle (on the right-hand side). Its area is $49,$ and its width is $7.$ Hence, the length of this rectangle must be \[ 49 \div 7 = {\color{blue}7}. \]

Let's add this to our model:

Now, notice the following regarding the big rectangle:

• its length is $120 + 7 = 127$

• its width is $7$

• its area is $840 + 49 = 889$

Therefore, this area model can be used to represent the following multiplication problem: \[ 889 = 127 \times 7 \]

Writing this as a division problem, we get \[ 889 \div 7 = {\color{blue}127}. \]

To determine the amount of money each child will receive, we need to calculate $172 \div 4.$

Let's look at the blue rectangle (the one on the left-hand side). Its area is $160,$ and its width is $4.$ Hence, the length of this rectangle must be \[ 160 \div 4. \]

To carry out this division, we write $160$ as a product with $10$ as one of the factors and then swap the order of multiplication and division: \begin{align*} 160 \div 4 &= \\[5pt] 16 \times 10 \div 4 &=\\[5pt] 16 \div 4 \times 10 &=\\[5pt] ({\color{black}{16 \div 4}}) \times 10 &=\\[5pt] {\color{black}{4}} \times 10 &=\\[5pt] \bbox[3pt, Gainsboro]{\color{blue}40}& \end{align*}

Let's add this to our model:

Now, notice the following regarding the big rectangle:

• its length is $40 + 3 = 43$

• its width is $4$

• its area is $160 + 12 = 172$

Therefore, this area model can be used to represent the following multiplication problem: \[ 172 = 43 \times 4 \]

Writing this as a division problem, we get \[ 172 \div 4 = {\color{blue}43}. \]

Therefore, each child will receive $\$\bbox[3pt, Gainsboro]{\color{blue}43}.$

We need to look at the blue and pink rectangles:

• Let's first look at the blue rectangle (the one on the left-hand side). Its area is $5,600,$ and its width is $8.$ Hence, the length of this rectangle must be \[ 5,600 \div 8 \] To carry out this division, we write $5,600$ as a product with $100$ as one of the factors and then swap the order of multiplication and division: \begin{align*} 5,600 \div 8 &= \\[5pt] 56 \times 100 \div 8 &=\\[5pt] 56 \div 8 \times 100 &=\\[5pt] ({\color{black}{56\div 8}}) \times 100 &=\\[5pt] {\color{black}{7}} \times 100 &=\\[5pt] \bbox[3pt, Gainsboro]{\color{blue}700}& \end{align*} Let's add this to our model:

• Now, let's look at the pink rectangle (the one on the right-hand side). Its area is $32,$ and its width is $8.$ Hence, the length of this rectangle must be \[ 32 \div 8 =\bbox[3pt, Gainsboro]{\color{blue}4}. \] Let's add this to our model:

Now, notice the following regarding the big rectangle:

• its length is $700 + 4 = 704$

• its width is $8$

• its area is $5,600 + 32 = 5,632$

Therefore, this area model can be used to represent the following multiplication problem: \[ 5,632 = 704 \times 8 \]

Writing this as a division problem, we get \[ 5,632 \div 8 =\bbox[3pt, Gainsboro]{\color{blue}704}. \]