Question

if these two shapes are similar, what is the measure of the missing length n? 2 yd n 2 yd 1 yd n = \boxed{} yards submit work it out

Explanation:

Step1: Recall similarity of rectangles

For similar rectangles, the ratios of corresponding sides are equal. Let the sides of the small rectangle be \(2\) yd (length) and \(1\) yd (width), and the large rectangle has length \(2\) yd (corresponding to small's length) and height \(n\) (corresponding to small's width). So the ratio of length to width for small is \(\frac{2}{1}\), and for large is \(\frac{2}{n}\)? Wait, no, wait. Wait, the small rectangle: length \(2\) yd, width \(1\) yd. The large rectangle: base \(2\) yd (same as small's length), so the height \(n\) should correspond to small's width? Wait, no, maybe I mixed up. Wait, similar figures: corresponding sides are proportional. So if the small rectangle has length \(2\) and width \(1\), and the large rectangle has length \(2\) (same as small's length) and height \(n\) (corresponding to small's width). Wait, no, maybe the base of large is \(2\), same as small's length, so the height \(n\) should be proportional to small's width. Wait, the ratio of length to width in small is \(2:1\), so in large, length is \(2\), so width (height) should be \(n\) such that \(2/n = 2/1\)? No, that would be \(n=1\), which is wrong. Wait, maybe the large rectangle's base is \(2\) (same as small's length), and the small's width is \(1\), large's height is \(n\). Wait, no, maybe the small rectangle: length \(2\), width \(1\); large rectangle: width \(2\) (same as small's length), so length \(n\) (corresponding to small's width? No, rectangles have length and width. Let's denote: small rectangle: length \(l_1 = 2\) yd, width \(w_1 = 1\) yd. Large rectangle: length \(l_2 = n\) yd, width \(w_2 = 2\) yd. Since they are similar, \(l_1/w_1 = l_2/w_2\). So \(2/1 = n/2\). Then solve for \(n\): \(n = (2\times2)/1 = 4\)? Wait, no, wait. Wait, maybe the small rectangle's length is \(2\), width \(1\); large rectangle's width is \(2\) (same as small's length), so the length of large should be proportional. So ratio of length to width in small is \(2:1\), so in large, width is \(2\), so length is \(2\times2 = 4\)? Wait, no, let's do it properly. Similar figures: corresponding sides are in proportion. So if the small rectangle has dimensions \(2\) (length) and \(1\) (width), and the large rectangle has length \(n\) and width \(2\) (since the base is \(2\) yd, same as small's length). Wait, no, the large rectangle's base is \(2\) yd (horizontal side), and the small's horizontal side is \(2\) yd. So the vertical side of small is \(1\) yd, vertical side of large is \(n\) yd. So the ratio of horizontal to vertical in small is \(2/1\), so in large, horizontal is \(2\), vertical is \(n\), so \(2/n = 2/1\)? No, that would be \(n=1\), which is wrong. Wait, maybe I got the correspondence wrong. Maybe the small rectangle: length \(2\), width \(1\); large rectangle: width \(2\) (same as small's length), length \(n\) (same as small's width? No, that can't be. Wait, maybe the two rectangles: the small one is \(2\) (length) by \(1\) (width), the large one is \(2\) (width) by \(n\) (length). So the ratio of length to width in small is \(2/1\), so in large, length is \(n\), width is \(2\), so \(n/2 = 2/1\). Then \(n = 4\). Yes, that makes sense. So the proportion is \(\frac{\text{length of large}}{\text{width of large}} = \frac{\text{length of small}}{\text{width of small}}\). So \(n/2 = 2/1\), so \(n = 2\times2 = 4\).

Step2: Set up the proportion

Let’s define the sides:
- Small rectangle: length \(= 2\) yd, width \(= 1\) yd.
- Large rectangle: length \(= n\) yd, width \(= 2\) yd.

Since the rectangles are similar, the ratios of corresponding sides are equal:
\[ \frac{\text{length of large}}{\text{width of large}} = \frac{\text{length of small}}{\text{width of small}} \]
Substitute the values:
\[ \frac{n}{2} = \frac{2}{1} \]

Step3: Solve for \(n\)

Cross - multiply:
\[ n\times1 = 2\times2 \]
\[ n = 4 \]

Answer:

\(4\)