問93
頻出度★★☆
A3判の紙の長辺を半分に折ると, A4判の大きさになり, 短辺：長辺の比率は変わらない。A3判の長辺は A4判の長辺のおよそ何倍か。
令和4年度 ITパスポート試験 問93
テクノロジ系基礎理論 › 応用数学
This question's explanation and your answers so far.
解説
問 93
- ア
正解 A-series paper is designed so that every size has the same ratio of short edge to long edge = 1:√2. Folding the long edge in half gives the next smaller size (A3→A4), and the key point is that this similarity ratio is preserved. Let the short edge of A3 be s and its long edge L; then the A4 obtained by folding has long edge s and short edge L/2. Because the ratio does not change, (L/2):s = s:L holds, giving L²=2s², i.e. L=√2·s. The value asked for, 'A3 long edge ÷ A4 long edge', is therefore L ÷ s = √2 ≒ 1.41. Checking against the actual dimensions (A3 long edge 420mm, A4 long edge 297mm), 420÷297 ≒ 1.41 also matches, so option ア's '1.41' is correct.
Incorrectイ
1.5 matches neither √2≒1.41 nor √3≒1.73, and is not a value that represents the similarity ratio of A-series paper. It is merely a rough round number, different from the √2 derived from the condition that the ratio is constant, so it is wrong.
Incorrectウ
1.73 is an approximation of √3, a ratio that appears in other contexts such as the height of an equilateral triangle. The short edge to long edge of A-series paper is 1:√2, not 1:√3, so as the multiple for the long edge it is wrong.
Incorrectエ
2 comes from confusing this with the 'area ratio'. The area of A3 is exactly twice that of A4, but that concerns area; the ratio of edge lengths is the square root of the area ratio = √2≒1.41. Taking the multiple between the long edges to be 2 is wrong.
ここがポイント
- A-series (and B-series) paper keeps a similarity ratio of short edge to long edge = 1:√2, and folding the long edge in half gives the next smaller size.
- Even when the area is doubled, the ratio of corresponding edge lengths is the square root of the area ratio = √2≒1.41 (be careful not to confuse area ratio with length ratio).
用語