Suppose we are given two polygons P and Q with equal area. Will it always be possible to divide one of them using straight cuts into two or more pieces and then rearrange the pieces to exactly cover the other polygon? Try this out for familiar shapes, e.g.,
- 1. A square and non-square rectangle with equal area,
- 2. Two triangles with different shapes but equal area,
- 3. A triangle and a square with equal area. Formulate a conjecture of your own about this.
Step-by-step solution
Idea: Use a common ‘middle’ shape. Every triangle can be cut into a rectangle, and any rectangle can be cut into any other rectangle of the same area. Going forward from one shape and backward to the other gives the pieces.
1. A square and non-square rectangle with equal area,
- Yes. Example: a 9 × 4 rectangle and a 6 × 6 square (both area 36). Cut the rectangle along a ‘staircase’ line with steps 3 wide and 2 high. Slide the right piece 3 units left and 2 units up: the two pieces make the 6 × 6 square (see the picture). Only 2 pieces are needed.1 mark
2. Two triangles with different shapes but equal area,
- Yes. Cut each triangle along the line joining the midpoints of two sides and turn the top piece half round: each triangle becomes a parallelogram with the same base and half the height. Parallelograms of equal area can be turned into the same rectangle by cutting off a triangle at one end and sliding it to the other. Reversing the steps for the second triangle gives the pieces. (When the triangles have equal bases on one line and a common vertex, like the two halves made by a median, a single cut is enough.)1 mark
3. A triangle and a square with equal area. Formulate a conjecture of your own about this.
- Yes. Triangle → rectangle (base b, height h2) as in 2. Then rectangle → square of the same area, by cutting and sliding (as in 1, or using Baudhāyana’s squaring construction).½ mark
- Conjecture: any two polygons with equal area can be cut into a finite number of pieces that rearrange into each other. (Reason: every polygon can be cut into triangles, each triangle into a rectangle, and all the rectangles into strips of one fixed width stacked into one rectangle; equal areas give the same final rectangle.) This is true; it is called the Wallace–Bolyai–Gerwien theorem.½ mark
Answer to write in the exam
1.
9 × 4 = 36 = 6 × 6
Staircase cut: steps 3 wide (9 − 6) and 2 high (6 − 4)
Slide right piece 3 left, 2 up
∴ Possible with 2 pieces.
2.
Triangle → parallelogram (cut along midline, rotate top 180°)
Parallelogram → rectangle (cut and slide a triangle)
Equal areas ⇒ same rectangle
∴ Possible.
3.
Triangle → rectangle (b × h2) → square
Conjecture: equal-area polygons can always be cut into finitely many pieces and rearranged into each other.
Common mistakes that cost marks
- Thinking two shapes need the same perimeter to be rearranged. Only the area must be equal; perimeters can change.
- Expecting the pieces to be squares or rectangles only. Pieces may be any polygons.
- Forgetting that pieces may be turned (rotated) as well as slid.
How this can come in the exam
A rectangle 8 cm × 2 cm is cut and rearranged without overlap into a square. The side of the square is
- 2 cm
- 4 cm
- 5 cm
- 16 cm
Show answer
(B) 4 cm
Area is unchanged: 8 × 2 = 16 cm2, so the side is √16 = 4 cm.
Try one yourself
A 16 × 9 rectangle is to be cut into a square of the same area with a staircase cut. What is the side of the square, and what are the step width and step height?
Show answer
Area 144, so side 12. Step width 16 − 12 = 4, step height 12 − 9 = 3.
More questions like this
- Think of various rectangles with perimeter 40 units (the sides do not have to be integers).
- Let us test Heron’s formula against some known cases: an equilateral triangle with side a units.
- Let us test Heron’s formula against some known cases: an isosceles triangle with equal sides a units and base 2b units.
- Let us test Heron’s formula against some known cases: a triangle with sides 3 units, 4 units and 5 units.
- In the same way we ask: can we find the area of a 4-gon if we only know the lengths of its sides? The figures below reveal the answer to this question. The problem (see the figure) is about a 4-gon whose sides are known to be 3, 3, 3, 3 (it is a ‘rhombus’). As you can see, the areas of the three figures are different. (We drew the figures using GeoGebra and found the areas using the ‘Area’ tool. Please try this exercise yourself, or by using four rods joined together at their ends.)