A person has a photo of a rabbit and a rectangular piece of card. The card is cut a...

1. Identify the dimensions of the original card and the photo:
The original rectangular piece of card has dimensions $30$ cm by $42$ cm. The photo of the rabbit has dimensions $20$ cm by $21$ cm.

2. Determine the ratio of the sides of the photo:
The ratio of the width to the height of the photo is $\frac{20}{21}$.

3. Understand the cutting process and the remaining piece:
Lydia cuts the card along a dotted line. She throws away the top piece of card that is $x$ cm by $30$ cm. The remaining piece of card has dimensions $30$ cm by $(42-x)$ cm.

4. Apply the condition of mathematical similarity:
The problem states that the piece of card she has left is mathematically similar to the photo. This means the ratio of the sides of the remaining card must be equal to the ratio of the sides of the photo.

5. Set up the proportion:
There are two possible orientations for similarity. Either the ratio of width to height is the same, or the ratio of width to height of one is the inverse of the other.

Case 1: The ratio of the width to the height of the remaining card is the same as the photo's width to height.
$$\frac{30}{42-x} = \frac{20}{21}$$

Case 2: The ratio of the width to the height of the remaining card is the inverse of the photo's width to height.
$$\frac{30}{42-x} = \frac{21}{20}$$

6. Solve for x in Case 1:
$$30 \times 21 = 20 \times (42-x)$$
$$630 = 840 - 20x$$
$$20x = 840 - 630$$
$$20x = 210$$
$$x = \frac{210}{20}$$
$$x = 10.5$$

7. Solve for x in Case 2:
$$30 \times 20 = 21 \times (42-x)$$
$$600 = 882 - 21x$$
$$21x = 882 - 600$$
$$21x = 282$$
$$x = \frac{282}{21}$$
$$x = \frac{94}{7} \approx 13.43$$

8. Re-evaluate the problem statement and diagram:
The problem states that Lydia cuts the card along the dotted line and throws away the piece that is $30$ cm by $x$. This implies that the cut is made horizontally, dividing the $42$ cm side. The remaining piece is $30$ cm wide and $(42-x)$ cm tall. The photo is $20$ cm wide and $21$ cm tall. For the remaining piece to be similar to the photo, the ratio of its sides must match the ratio of the photo's sides. The photo's ratio of width to height is $\frac{20}{21}$. The remaining card's ratio of width to height is $\frac{30}{42-x}$.

However, the problem also states that the piece of card she has left is mathematically similar to the photo. The photo itself is $20$ cm by $21$ cm. The original card is $30$ cm by $42$ cm. The ratio of the original card's sides is $\frac{30}{42} = \frac{5}{7}$. The ratio of the photo's sides is $\frac{20}{21}$. These are not the same, so the original card is not similar to the photo. This means the similarity condition applies to the remaining piece of card and the photo.

Let's re-examine the dimensions and similarity. The photo has dimensions $20$ cm (width) and $21$ cm (height). The ratio of width to height is $\frac{20}{21}$.

The original card has dimensions $30$ cm (width) and $42$ cm (height). The ratio of width to height is $\frac{30}{42} = \frac{5}{7}$.

Lydia cuts off a piece of $30$ cm by $x$ cm. This means the cut is horizontal, and the height $x$ is removed from the $42$ cm side. The remaining piece has dimensions $30$ cm (width) and $(42-x)$ cm (height).

For the remaining piece to be similar to the photo, the ratio of their corresponding sides must be equal. There are two possibilities for similarity:

Possibility A: The ratio of width to height of the remaining card is equal to the ratio of width to height of the photo.
$$\frac{30}{42-x} = \frac{20}{21}$$
Solving this: $30 \times 21 = 20 \times (42-x) \implies 630 = 840 - 20x \implies 20x = 210 \implies x = 10.5$.

Possibility B: The ratio of width to height of the remaining card is equal to the ratio of height to width of the photo (i.e., the orientation is flipped).
$$\frac{30}{42-x} = \frac{21}{20}$$
Solving this: $30 \times 20 = 21 \times (42-x) \implies 600 = 882 - 21x \implies 21x = 282 \implies x = \frac{282}{21} = \frac{94}{7} \approx 13.43$.

Let's consider the context. The photo is $20$ cm by $21$ cm. The remaining card is $30$ cm by $(42-x)$ cm. If the remaining card is similar to the photo, the ratio of sides should be consistent. The ratio of the photo is $\frac{20}{21}$. The ratio of the original card is $\frac{30}{42} = \frac{5}{7}$.

If the remaining card is similar to the photo, then the ratio of its sides must be $\frac{20}{21}$ or $\frac{21}{20}$.

If $\frac{30}{42-x} = \frac{20}{21}$, then $x = 10.5$. The remaining card is $30$ cm by $31.5$ cm. The ratio is $\frac{30}{31.5} = \frac{300}{315} = \frac{60}{63} = \frac{20}{21}$. This matches the photo's ratio. This is a valid solution.

If $\frac{30}{42-x} = \frac{21}{20}$, then $x = \frac{94}{7}$. The remaining card is $30$ cm by $(42 - \frac{94}{7}) = 30$ cm by $(\frac{294-94}{7}) = 30$ cm by $\frac{200}{7}$ cm. The ratio is $\frac{30}{200/7} = \frac{30 \times 7}{200} = \frac{210}{200} = \frac{21}{20}$. This also matches the photo's ratio (i...