To solve the problem of finding the length of the fold when folding corner (A) to corner (C) of a rectangular paper (12 cm × 10 cm), follow these steps:

Step 1: Coordinate Setup

Place the rectangle (ABCD) in a coordinate system:

• (A(0,0)), (B(12,0)), (C(12,10)), (D(0,10)).

The fold line is the perpendicular bisector of diagonal (AC) (since it maps (A) to (C)).

Step 2: Midpoint and Slope of (AC)

• Midpoint (M) of (AC): ((6,5)).
• Slope of (AC): (\frac{10-0}{12-0}=\frac{5}{6}).
• Slope of perpendicular bisector: (-\frac{6}{5}) (negative reciprocal).

Step 3: Equation of Fold Line

Using midpoint (M(6,5)) and slope (-\frac{6}{5}):
[y - 5 = -\frac{6}{5}(x - 6)]
Simplify:
[y = -\frac{6}{5}x + \frac{61}{5}]

Step 4: Intersection Points of Fold Line with Rectangle

• Intersection with top side (y=10):
  (10 = -\frac{6}{5}x + \frac{61}{5} \implies x=\frac{11}{6}) → Point (P\left(\frac{11}{6},10\right)).

• Intersection with bottom side (y=0):
  (0 = -\frac{6}{5}x + \frac{61}{5} \implies x=\frac{61}{6}) → Point (Q\left(\frac{61}{6},0\right)).

Step 5: Calculate Length of Fold

Distance between (P) and (Q):
[PQ = \sqrt{\left(\frac{61}{6}-\frac{11}{6}\right)^2 + (0-10)^2}]
[= \sqrt{\left(\frac{50}{6}\right)^2 + (-10)^2}]
[= \sqrt{\left(\frac{25}{3}\right)^2 + 100}]
[= \sqrt{\frac{625}{9} + \frac{900}{9}} = \sqrt{\frac{1525}{9}} = \frac{5\sqrt{61}}{3} \approx 13\ \text{cm}]

Answer: (\boxed{13}) (rounded to nearest integer).

(\boxed{13})

作者声明:本文包含人工智能生成内容。