The right-angled triangle ABC is such that ∠B = 90°. Point D is picked on BC such that triangles ABC and DBA are similar. If AB : BC = m : n, what is ΔABC : ΔABD, where Δ denotes the area of a triangle?

Correct — B, n² : m². Both triangles stand on the same vertex B and share the right angle there, so ΔABC = ½ × AB × BC and ΔABD = ½ × AB × BD. The common factor ½ × AB cancels and the answer is simply BC : BD. The similarity fixes BD: reading ABC ~ DBA letter by letter pairs AB with DB and BC with BA, so AB/DB = BC/BA, which gives AB² = BD × BC. Putting AB = m and BC = n makes BD = m²/n, and BC : BD = n : m²/n = n² : m². Check it with numbers — AB = 3, BC = 4 gives BD = 9/4, areas 6 and 27/8, and 6 ÷ 27/8 = 16/9.

• (a)n : m — That is the ratio of corresponding sides. Areas of similar figures go as the square of the side ratio, so this answer is one squaring short.
• (c)(m + n) : n — Nothing in the configuration is m + n. D lies on BC, so the lengths that matter are BD and BC, and BD is fixed by AB² = BD × BC rather than by any sum of the two given sides.
• (d)(m + n)² : n² — The squaring is right but the ratio being squared is invented; there is no pair of corresponding sides in these triangles standing in the ratio (m + n) : n.

Two similar triangles have their areas in the ratio of the squares of any pair of corresponding sides. When one triangle sits inside the other on a shared vertex, there is a shortcut: express both areas with the same base or the same height, cancel it, and the area ratio reduces to a single length ratio. Here the shared leg AB is the height for both, so the areas stand in the ratio of the bases BC and BD.

The dangerous step is the correspondence of vertices. The stem says triangles ABC and DBA are similar, and the order of the letters is not decoration — A matches D, B matches B, C matches A. Get that wrong and you pair AB with BA, which gives a ratio of 1 and no answer at all. This configuration, a right triangle with the altitude-like segment drawn to the leg, is also where AB² = BD × BC comes from: AB is the geometric mean of BD and BC. Note that D is on BC and not the foot of the perpendicular from B, so do not import the standard altitude-on-hypotenuse relations by reflex.

• Areas of similar triangles are in the ratio of the squares of corresponding sides.
• The order of letters in a similarity statement fixes which sides correspond — ABC ~ DBA pairs AB with DB and BC with BA.
• That correspondence gives AB² = BD × BC, so AB is the geometric mean of BD and BC.
• Sharing the leg AB lets both areas be written as ½ × AB × (base), so the ratio collapses to BC : BD.
• With AB : BC = m : n, BD = m²/n and the area ratio is n² : m².

• Answering with the side ratio n : m instead of squaring it.
• Matching the vertices in the order they are printed rather than as the similarity statement dictates.
• Assuming D is the foot of a perpendicular and importing altitude formulas that do not apply.

Asked in symbols rather than numbers, so the examiner is checking whether you can read a similarity statement's vertex order and turn it into a length relation.

No directly related past PYQ was found.

• practice — not a real PYQ

  Two similar triangles have corresponding sides in the ratio 3 : 5. What is the ratio of their areas?

  • (a)3 : 5
  • (b)5 : 3
  • (c)9 : 25
  • (d)27 : 125

  Answer(c) 9 : 25 — areas of similar figures go as the square of the ratio of corresponding sides.

• practice — not a real PYQ

  In a right-angled triangle the altitude to the hypotenuse divides it into segments of 4 cm and 9 cm. What is the length of that altitude?

  • (a)5 cm
  • (b)6 cm
  • (c)6.5 cm
  • (d)13 cm

  Answer(b) 6 cm — the altitude is the geometric mean of the two segments, √(4 × 9) = 6.