A street lamp is fixed on top of a vertical pole 6 m high. A boy 1.5 m tall stands 9 m away from the foot of the pole on level ground. (i) Find the length of his shadow. (ii) How far from the pole must he stand so that his shadow is 4 m long?

Answer: (i) 3 m (ii) 12 m from the pole

[Figure: pole AB (A at top, B on ground) and boy CD (C at head, D at feet) both vertical; the ray from lamp A passes over C and reaches the ground at E, the tip of the shadow DE. B, D, E are on one line.]

AB || CD (both perpendicular to the ground). In \(\Delta EDC\) and \(\Delta EBA\): \(\angle E\) is common and \(\angle EDC \cong \angle EBA\) (each 90°), so \(\Delta EDC \sim \Delta EBA\) (AA test) and \(\dfrac{CD}{AB} = \dfrac{ED}{EB}\).

(i) Let the shadow ED = s. Then EB = s + 9. \(\dfrac{1.5}{6} = \dfrac{s}{s + 9}\) \(\therefore 6s = 1.5s + 13.5\) \(\therefore 4.5s = 13.5\) \(\therefore s = 3\) m.

(ii) Shadow ED = 4, let BD = d. \(\dfrac{1.5}{6} = \dfrac{4}{d + 4}\) \(\therefore 1.5(d + 4) = 24\) \(\therefore d + 4 = 16\) \(\therefore d = 12\) m.