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.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. |
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Answer» 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 |
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