Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

Step-by-step solution

Idea: Equal bases on one line and a common apex give equal areas. For the cutting, use midpoints: a half-turn about the midpoint of a side swaps its two ends, and a slide by one-third of the base carries one division point to the next.

1. Equal areas. Call the top vertex V and the base points B, T₁, T₂, C, with BT₁ = T₁T₂ = T₂C (lines to the trisection points). Blue = △VBT₁, red = △VT₂C. Both have base 13BC and the same height h (the distance from V to BC). So both areas are 12 × 13BC × h: they are equal.1 mark
2. Cutting the blue triangle (3 cuts, 4 pieces). On the side VB mark G, E, H at 13, 12 and 23 of the way from B to V, and let M be the midpoint of VT₁. Cut along T₁G, T₁E and HM. Pieces: 1 = △BT₁G, 2 = △T₁EG, 3 = △VHM, 4 = quadrilateral T₁MHE.1 mark
3. Slides: move piece 1 to the right by BT₂ (two-thirds of the base); it becomes the triangle at C. Move piece 4 to the right by BT₁ (one-third of the base); it lands with T₁ on T₂, M on the midpoint of VC and E on N, the midpoint of VT₂.1 mark
4. Half-turns: turn piece 2 through 180° about M: T₁ goes to V and E goes to N, so it fills the top corner of the red triangle. Turn piece 3 through 180° about N: V goes to T₂, so it fills the gap next to piece 1.1 mark
5. The four pieces now cover △VT₂C exactly (their areas, in units where ar(blue) = 9, are 3 + 1.5 + 1.5 + 3 = 9). Check with coordinates B(0, 0), T₁(3, 0), T₂(6, 0), C(9, 0), V(2, 6): every corner of every moved piece lands on the boundary or a shared cut of the red triangle.1 mark

Check: Areas of the 4 pieces with the coordinates above: 3, 1.5, 1.5 and 3, total 9 = 12 × 3 × 6 ✓; their images in the red triangle have the same areas and fit without gaps.

Answer to write in the exam

Common mistakes that cost marks

• Thinking the triangles cannot have equal area because they look different. Equal base and equal height are enough.
• Cutting along a line and only sliding; at least one piece must be turned.
• Measuring the heights along the slanting sides; the common height is the perpendicular from V to BC.

How this can come in the exam

Try one yourself

The base of a triangle of area 45 cm² is divided into 5 equal parts and joined to the opposite vertex. Find the area of each part and of the two outer parts together.

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