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Dissection of shapes · 4 marks

Since ΔABD and ΔACD have equal area, you may wonder — Can we divide ΔABD using straight cuts into two or more pieces that we can then rearrange to exactly cover ΔACD? What do you think? Is it possible?

ABCD
Answer: Yes, with just one cut (2 pieces). Let E be the midpoint of AB and cut ΔABD along DE. Slide piece EBD along BC by the length BD: it becomes NDC (N = midpoint of AC). Turn piece AED half round the midpoint M of AD: it becomes DNA. Together NDC and DNA make ΔACD exactly.

Step-by-step solution

Idea: A slide by BD (= DC) carries B to D and D to C; a half-turn about M swaps A and D. The midpoint E is chosen so that both pieces land on the same point N.

ABCDENM1212
  1. Mark E, the midpoint of AB, and cut ΔABD along the straight line DE. This gives piece 1 = ΔEBD and piece 2 = ΔAED. Let N be the midpoint of AC and M the midpoint of AD.1 mark
  2. Piece 1: slide it to the right by the distance BD (BD = DC). B goes to D, D goes to C, and E goes to N, because EN is the line joining midpoints of AB and AC, so EN ∥ BC and EN = 12BC = BD. Piece 1 now covers ΔNDC.1 mark
  3. Piece 2: turn it through 180° about M. Since M is the midpoint of AD, A and D swap places. Where does E go? EM joins the midpoints of AB and AD, so EM ∥ BD and EM = 12BD; MN joins the midpoints of AD and AC, so MN ∥ DC and MN = 12DC. So E, M, N lie on one line and EM = MN. The half-turn about M therefore sends E to N, and piece 2 now covers ΔDNA.1 mark
  4. ΔDNA and ΔNDC together are ΔACD (DN splits ΔACD into these two). So yes, it is possible, with only 2 pieces. One piece (no cut) is impossible unless the two triangles happen to be congruent, so 2 is the least number of pieces in general.1 mark
Yes. Cut ΔABD along DE (E the midpoint of AB). Slide ΔEBD along BC by BD to ΔNDC, and turn ΔAED half round the midpoint of AD to ΔDNA. The two pieces exactly cover ΔACD, so 2 pieces (one cut) are enough.

Check: In coordinates A(1, 3), B(0, 0), D(3, 0), C(6, 0): E = (0.5, 1.5), N = (3.5, 1.5), M = (2, 1.5). Sliding E by BD = 3 gives (3.5, 1.5) = N ✓. Turning E half round M gives (2 × 2 − 0.5, 2 × 1.5 − 1.5) = (3.5, 1.5) = N ✓.

Answer to write in the exam

E = midpoint of AB; cut along DE ⇒ pieces △EBD, △AED

Slide △EBD by BD (= DC): B → D, D → C, E → N (EN ∥ BC, EN = 12BC)

Rotate △AED by 180° about M (midpoint of AD): A → D, D → A, E → N

△NDC ∪ △DNA = △ACD

∴ Yes; 2 pieces (one cut) are enough.

Common mistakes that cost marks

  • Thinking equal area means the triangles must be congruent. They usually are not; they are only equal in area.
  • Cutting from D to a point of AB other than its midpoint: after the slide and the half-turn the two pieces no longer meet at the same point N, leaving a gap and an overlap.
  • Forgetting that a piece may be turned over or rotated, not just slid.

How this can come in the exam

MCQ (1 mark)

AD is a median of ΔABC and area(ΔABC) = 48 cm2. Then area(ΔABD) is

  1. 12 cm2
  2. 16 cm2
  3. 24 cm2
  4. 48 cm2
Show answer

(C) 24 cm2
A median divides a triangle into two triangles of equal area: 48 ÷ 2 = 24 cm2.

Try one yourself

In ΔPQR, S is the midpoint of QR and T is the midpoint of PQ. Which single cut of ΔPQS gives two pieces that can be rearranged into ΔPRS?

Show answer

Cut ΔPQS along ST. Slide ΔTQS along QR by QS, and turn ΔPTS half round the midpoint of PS.

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