A point P on the ground is on the same line as the bases and the tips of a pair of trees A and B such that P is on the left of both these trees. The ratio of heights of A and B is 1 : 3. If the distance between P and A is a metre, then the distance between A and B, in metre, is

Correct — B, 2a. The stem puts P, the two bases and the two tips on a single line, which means the line of sight from P grazes the top of the near tree A and goes on to the top of the far tree B. So both trees subtend the same angle at P, and the two right triangles are similar. Let the height of A be h; then B is 3h. From the near triangle the tangent of the angle is h/a. From the far triangle it is 3h/(a + d), where d is the distance between the trees. Setting them equal: h/a = 3h/(a + d), so a + d = 3a and d = 2a. Note where 3a comes from — it is the distance from P to the far tree B, and reading that as the answer is exactly the slip the option set is built to catch.

• (a)a — Would make the two trees equally far apart as P is from the first, giving a total of 2a from P to B and a height ratio of 1 : 2 rather than 1 : 3.
• (c)3a — This is the distance from P to B, not from A to B. The similar-triangle relation gives a + d = 3a, and the question asks for d, so one subtraction still has to be done.
• (d)4a — Would put B at 5a from P and make the height ratio 1 : 5. The given ratio is 1 : 3, which fixes P-to-B at 3a.

When two objects have their feet and their tops on one straight line seen from a point, the point, the feet and the tops form two similar right triangles sharing the angle at the observer. Heights then stand in the same ratio as horizontal distances from the observer. That single proportion — height to distance is constant along a line of sight — solves most shadow, tree and tower problems without any trigonometric tables.

The trap is a distance measured from the wrong origin. Everything in the algebra is anchored at P, so the natural output is the P-to-B distance of 3a, and the question asks for A-to-B. Sketch the line with P at the left, A at a, and B further right, mark 3a from P to B, and the answer 3a − a = 2a is visible without algebra. It is also worth noticing that the individual heights never appear in the answer; only their ratio matters, which is the signature of a similar-triangles item.

• If the tips of two vertical objects lie on one line of sight from a ground point, the triangles formed are similar.
• Similar right triangles give height to horizontal distance as a constant ratio from the observer.
• Here h/a = 3h/(a + d), so a + d = 3a and d = 2a.
• The distance from P to the far tree is 3a; the distance between the trees is 2a.
• The actual heights cancel out — only the 1 : 3 ratio is needed.

• Answering 3a, which is the distance from P to the far tree.
• Assuming the actual heights are needed when only their ratio is.
• Placing P between the two trees; the stem puts it to the left of both.

A similar-triangles item written as a word problem, with the intended wrong answer being the intermediate quantity.

No directly related past PYQ was found.

• practice — not a real PYQ

  Two vertical poles of heights 4 m and 12 m stand on level ground. A point P on the ground is in line with both poles and lies beyond the shorter one, and the tips of both poles are in line with P. If P is 5 m from the shorter pole, how far apart are the poles?

  • (a)5 m
  • (b)10 m
  • (c)15 m
  • (d)20 m

  Answer(b) 10 m — the ratio of heights is 1 : 3, so P is 15 m from the taller pole and the gap is 15 − 5 = 10 m.

• practice — not a real PYQ

  At a given moment a 6 m pole casts a shadow 4 m long. How long is the shadow of a 15 m tower at the same moment?

  • (a)8 m
  • (b)10 m
  • (c)12 m
  • (d)9 m

  Answer(b) 10 m — the height to shadow ratio is constant, so the shadow is 15 x 4 divided by 6.