Question

.A building 16 m high casts a shadow of 6.4 m length. Find the height of the building which casts a shadow of 1.6 m at the same angle of Sun rays.​

Answer 1

solution:-

Height of pole=AB=6 m

Length of shadow of pole =BC=4 m

Length of shadow of tower=EF=28 m

In △ABC and △DEF

∠B=∠E=90

∘

both 90

∘

as both are vertical to ground

∠C=∠F (same elevation in both the cases as both shadows are cast at the same time)

∴△ABC∼△DEF by AA similarity criterion

We know that if two triangles are similar, ratio of their sides are in proportion

So, AB/DE=BC/EF

⇒ 6/DE=4/28

DE=6×28/4

⇒DE=6×7=42 m

Hence the height of the tower is 42 m