20)
diagram of two rectangles: left rectangle with sides 40, 56, 40, x; right rectangle with sides 7, 5, 7, 5
22)
diagram of two rectangles: left rectangle with top 14, bottom x, sides 1, 1; right rectangle with sides 112, 8, 112, 8
24)
partial diagram of two rectangles: left rectangle with top 21; right rectangle with top 12, sides 3, 4

Question

Accepted Answer

### Problem 20:
# Explanation:
## Step1: Identify similar rectangles
The two rectangles are similar, so their corresponding sides are proportional. The small rectangle has sides 7 and 5, the large one has sides 40 and 56 (and $ x $). Wait, actually, let's check the sides. Wait, the small rectangle: length 7, width 5. The large rectangle: one side is 40, another is 56, and the other side is $ x $. Wait, maybe the ratio of length to width should be equal. Wait, maybe the small rectangle's length is 7, width 5, and the large rectangle's length is 40, width $ x $, or maybe 56 and $ x $. Wait, no, let's see: the small rectangle has length 7 and width 5. The large rectangle has two sides 40 and one side 56, and the other side $ x $. So the ratio of length to width in the small rectangle is $ \frac{7}{5} $, and in the large rectangle, it should be $ \frac{40}{x} $ or $ \frac{56}{40} $? Wait, maybe I got it wrong. Wait, the large rectangle: sides 40, 56, 40, and $ x $. So it's a rectangle, so opposite sides are equal. So the small rectangle is 7 (length) and 5 (width). The large rectangle: let's say the length is 40 and the width is $ x $, or length 56 and width 40? Wait, no, maybe the ratio of corresponding sides. Let's assume that the small rectangle's length is 7 and width is 5, and the large rectangle's length is 40 and width is $ x $, but that doesn't make sense. Wait, maybe the large rectangle has length 56 and width 40, and the small one has length 7 and width 5. Let's check the ratio: $ \frac{56}{7} = 8 $, $ \frac{40}{5} = 8 $. Oh, so the scale factor is 8. Wait, but the large rectangle has a side $ x $? Wait, no, the large rectangle: sides 40, 56, 40, and $ x $. Wait, maybe the small rectangle is 7 (length) and 5 (width), and the large rectangle is 40 (length) and $ x $ (width), but that would mean $ \frac{40}{7} = \frac{x}{5} $, but that doesn't match the 56. Wait, maybe the large rectangle has length 56 and width $ x $, and the small one has length 7 and width 5. Then $ \frac{56}{7} = \frac{x}{5} $. Let's calculate that. $ \frac{56}{7} = 8 $, so $ x = 5 \times 8 = 40 $? But that's already a side. Wait, maybe I misread the diagram. Wait, the large rectangle: top side 40, left side 56, bottom side 40, right side $ x $. The small rectangle: top 7, left 5, bottom 7, right 5. So the small rectangle is 7 (length) and 5 (width). The large rectangle: length 40, width $ x $? No, the left side is 56, so maybe the height is 56 and the width is 40, and the small rectangle's height is 5 and width is 7. Wait, that would be a different ratio. Wait, maybe the rectangles are similar, so the ratio of length to height is the same. So small rectangle: length 7, height 5. Large rectangle: length 40, height $ x $, or length 56, height 40. Wait, $ \frac{56}{7} = 8 $, $ \frac{40}{5} = 8 $. So the scale factor is 8. So if the small rectangle's height is 5, the large one's height is $ 5 \times 8 = 40 $, and the small one's length is 7, the large one's length is $ 7 \times 8 = 56 $. Oh! So the large rectangle has length 56 and height 40, which matches the diagram (sides 40, 56, 40, and $ x $? Wait, no, the diagram shows the large rectangle with sides 40, 56, 40, and $ x $. Wait, maybe $ x $ is 56? No, that can't be. Wait, maybe I made a mistake. Wait, the small rectangle: 7 (top and bottom), 5 (left and right). The large rectangle: 40 (top and bottom), $ x $ (left and right), and 56 (left and right)? Wait, no, the diagram shows the large rectangle with left side 56, top 40, bottom 40, and right side $ x $. So it's a rectangle, so left and right sides should be equal, top and bottom equal. So left side 56, right side $ x $, so $ x = 56 $? But that doesn't involve the small rectangle. Wait, maybe the two rectangles are similar, so the ratio of their sides is equal. So small rectangle: length 7, width 5. Large rectangle: length 40, width $ x $. Then $ \frac{7}{5} = \frac{40}{x} $, so $ 7x = 200 $, $ x = \frac{200}{7} \approx 28.57 $, but that doesn't match 56. Wait, maybe the large rectangle's length is 56 and width is 40, and the small one's length is 7 and width is 5. Then $ \frac{56}{7} = 8 $, $ \frac{40}{5} = 8 $, so the scale factor is 8. So if the small rectangle has length 7 and width 5, the large one has length $ 7 \times 8 = 56 $ and width $ 5 \times 8 = 40 $. So the large rectangle's sides are 56 (length) and 40 (width), so the right side $ x $ should be 56? But that's the left side. Wait, maybe the diagram is labeled wrong, or I misread it. Alternatively, maybe the large rectangle has a side $ x $ and the small one has a side 5, and the other sides are 7 and 40, 56. Wait, let's try again. Let's set up the proportion: $ \frac{40}{7} = \frac{x}{5} $. No, that gives $ x = \frac{200}{7} $. Or $ \frac{56}{5} = \frac{x}{7} $, $ x = \frac{392}{5} = 78.4 $. No, that doesn't make sense. Wait, maybe the two rectangles are similar, so the ratio of length to width is the same. Small rectangle: length 7, width 5. Large rectangle: length 56, width $ x $. Then $ \frac{7}{5} = \frac{56}{x} $, so $ 7x = 280 $, $ x = 40 $. Ah! That works. So $ \frac{7}{5} = \frac{56}{x} $, cross-multiplying: $ 7x = 56 \times 5 $, $ 7x = 280 $, $ x = 40 $. Wait, but the large rectangle already has a side 40. Maybe the large rectangle's width is $ x $ and length is 56, and the small one's width is 5 and length is 7. So the ratio of length to width is $ \frac{7}{5} = \frac{56}{x} $, so $ x = 40 $. So that's the solution.

## Step1: Set up the proportion
Since the rectangles are similar, the ratio of corresponding sides is equal. Let the small rectangle have length $ 7 $ and width $ 5 $, and the large rectangle have length $ 56 $ and width $ x $. Then:
$$
\frac{7}{5} = \frac{56}{x}
$$

## Step2: Cross - multiply to solve for $ x $
Cross - multiplying gives:
$$
7x = 56\times5
$$
$$
7x = 280
$$

## Step3: Divide both sides by 7
$$
x=\frac{280}{7} = 40
$$

# Answer:
$ x = 40 $

### Problem 22:
# Explanation:
## Step1: Identify similar rectangles
The two rectangles are similar, so their corresponding sides are proportional. The small rectangle has top side $ 14 $, bottom side $ x $, and left/right sides $ 1 $. The large rectangle has top/bottom sides $ 112 $ and left/right sides $ 8 $.

## Step2: Set up the proportion
The ratio of the top - bottom sides should be equal to the ratio of the left - right sides. So:
$$
\frac{14}{x}=\frac{1}{8}
$$
Wait, no, maybe the ratio of the small rectangle's side to the large rectangle's side. Wait, the small rectangle: height $ 1 $, length $ 14 $ (top) and $ x $ (bottom). The large rectangle: height $ 8 $, length $ 112 $ (top and bottom). So the ratio of height to length in the small rectangle is $ \frac{1}{14} $, and in the large rectangle is $ \frac{8}{112}=\frac{1}{14} $. Wait, maybe the proportion is $ \frac{14}{112}=\frac{1}{8} $? No, that's not for $ x $. Wait, maybe the small rectangle's bottom side is $ x $, top side $ 14 $, and the large rectangle's bottom side is $ 112 $, top side... Wait, the small rectangle: left side $ 1 $, right side $ 1 $, top $ 14 $, bottom $ x $. The large rectangle: left side $ 8 $, right side $ 8 $, top $ 112 $, bottom $ 112 $. So the ratio of left - right sides (height) is $ \frac{1}{8} $, and the ratio of top - bottom sides (length) should be the same. So:
$$
\frac{14}{x}=\frac{1}{8}
$$
Wait, no, that would be $ x = 112 $, but let's check. Wait, the small rectangle's height is $ 1 $, length is $ 14 $ (top) and $ x $ (bottom). The large rectangle's height is $ 8 $, length is $ 112 $ (top and bottom). So the scale factor from small to large is $ \frac{8}{1}=8 $. So the length of the large rectangle should be $ 14\times8 = 112 $, and the bottom side of the small rectangle $ x $ should correspond to the bottom side of the large rectangle $ 112 $? Wait, no, the small rectangle's bottom side is $ x $, and the large rectangle's bottom side is $ 112 $. So the ratio of small to large is $ \frac{x}{112}=\frac{1}{8} $, so $ x = \frac{112}{8}=14 $? No, that's the top side. Wait, maybe I got the sides wrong. Let's look again: the small rectangle has top $ 14 $, bottom $ x $, left $ 1 $, right $ 1 $. The large rectangle has top $ 112 $, bottom $ 112 $, left $ 8 $, right $ 8 $. So it's a rectangle, so top = bottom, left = right. So the small rectangle is a rectangle with length $ 14 $ (top) and width $ 1 $ (left), and the large rectangle is a rectangle with length $ 112 $ (top) and width $ 8 $ (left). So the ratio of length to width in the small rectangle is $ \frac{14}{1} $, and in the large rectangle is $ \frac{112}{8}=\frac{14}{1} $. So they are similar with a scale factor. But we need to find $ x $, which is the bottom side of the small rectangle. But since it's a rectangle, bottom side should equal top side, so $ x = 14 $? But that seems too easy. Wait, maybe the diagram is different. Wait, the small rectangle: top $ 14 $, bottom $ x $, left $ 1 $, right $ 1 $. So it's a rectangle, so $ x = 14 $? But the large rectangle has top $ 112 $, bottom $ 112 $, left $ 8 $, right $ 8 $. So the ratio of the small rectangle's length (14) to the large rectangle's length (112) is $ \frac{14}{112}=\frac{1}{8} $, and the ratio of the small rectangle's width (1) to the large rectangle's width (8) is $ \frac{1}{8} $, so they are similar with scale factor $ \frac{1}{8} $. But the bottom side of the small rectangle is $ x $, which should be equal to the top side, so $ x = 14 $. Wait, maybe I misread the problem. Wait, the small rectangle: top $ 14 $, bottom $ x $, left $ 1 $, right $ 1 $. So it's a rectangle, so opposite sides are equal, so $ x = 14 $. But let's check the proportion with the large rectangle. The large rectangle: top $ 112 $, bottom $ 112 $, left $ 8 $, right $ 8 $. So the ratio of small length to large length is $ \frac{14}{112}=\frac{1}{8} $, and small width to large width is $ \frac{1}{8} $, so that's consistent. So $ x = 14 $. Wait, but maybe the problem is that the small rectangle has top $ 14 $, bottom $ x $, and the large one has top $ 112 $, bottom $ 112 $, so the proportion is $ \frac{14}{x}=\frac{112}{112} $? No, $ \frac{112}{112}=1 $, so $ 14 = x $. Yes, that makes sense.

## Step1: Recognize the rectangle property
In a rectangle, opposite sides are equal. For the small rectangle, the top side is $ 14 $ and the bottom side is $ x $.

## Step2: Conclude the value of $ x $
Since opposite sides of a rectangle are equal, we have:
$$
x = 14
$$
(We can also verify using similarity. The ratio of the top - bottom sides of the small rectangle to the large rectangle is $ \frac{14}{112}=\frac{1}{8} $, and the ratio of the left - right sides is $ \frac{1}{8} $, confirming similarity, and since the small rectangle's top and bottom sides must be equal, $ x = 14 $.)

# Answer:
$ x = 14 $

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QUESTION IMAGE

Question

Explanation:

Problem 20:

Step1: Identify similar rectangles

Step1: Identify similar rectangles

Step2: Set up the proportion

Answer:

Problem 22: