Question

streak and counting! fantastic consist... 1 streak level 2 find the area of each part. a =, b =, c =, d = 14. what is the total area?

Explanation:

Step1: Find the height of D (and C, B, A)

We know the area of D is 14 and its width is 7. The formula for the area of a rectangle is \( \text{Area} = \text{width} \times \text{height} \), so we can find the height \( h \) by rearranging the formula: \( h=\frac{\text{Area of D}}{\text{width of D}} \).
\( h = \frac{14}{7}=2 \) (this also tells us the height of C, B, and A since they are in the same rows).

Step2: Calculate area of A

The width of A is 50 and the height is 6 (wait, no, looking at the diagram, the height for A and B row is 6? Wait, no, the left side has 60? Wait, maybe I misread. Wait, the row with A and B has a left number 6 (maybe 6 units height?), and the row with C and D has 2? Wait, no, the area of D is 14, width 7, so height is 14/7 = 2. So the row with C and D has height 2. Then the row with A and B: let's check the left number, maybe 6? Wait, no, maybe the height of A's row is, let's see, if D has height 2, and the left side for A's row is 6 (maybe 6 units). Wait, no, maybe the height of A is 6? Wait, no, let's re - examine.

Wait, the area of a rectangle is length × width. For D: area = 14, width = 7, so height (let's call it \( h \)) is \( h=\frac{14}{7} = 2 \). So the row with C and D has height 2. Then the row with A and B: looking at the left - hand side, maybe the height is 6? Wait, the left number for A's row is 6 (maybe 6 units). So for A: width = 50, height = 6? Wait, no, that can't be. Wait, maybe the height of A's row is the same as the height we found for D's row? No, D's row has height 2, A's row has a different height. Wait, maybe the left number for A's row is 6 (vertical side), so height is 6, width is 50. Then area of A is \( 50\times6 = 300 \)? Wait, no, maybe the height of A is 6? Wait, let's check B. B has width 7, height same as A's row. If A's height is 6, then area of B is \( 7\times6=42 \). For C: width 50, height 2, so area is \( 50\times2 = 100 \). Then total area: 300 + 42+100 + 14=456? Wait, but let's do it step by step correctly.

Wait, first, find the height of each row. The row with D: area of D is 14, width is 7. So height \( h_D=\frac{14}{7}=2 \). So the row with C and D has height \( h = 2 \). Now, the row with A and B: let's assume the height of this row is \( H \). Wait, maybe the left - hand side numbers are the heights. The row with A and B has a left number of 6 (maybe 6 units), and the row with C and D has a left number of 2 (2 units). So:

• Area of A: width = 50, height = 6. So \( \text{Area of A}=50\times6 = 300 \)
• Area of B: width = 7, height = 6. So \( \text{Area of B}=7\times6 = 42 \)
• Area of C: width = 50, height = 2. So \( \text{Area of C}=50\times2 = 100 \)
• Area of D: 14 (given)

Now, total area: \( 300 + 42+100 + 14=456 \)

Answer:

• \( A = 300 \)
• \( B = 42 \)
• \( C = 100 \)
• \( D = 14 \)
• Total Area = 456