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Sum
A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.
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Solution
Let BC = 15 m be the tower and its shadow AB is 24 m.
At that time ∠CAB = 8, Again, let EF = h be a telephone pole and its shadow DE = 16 m.
At the same time ∠EDF = 8
Here, ΔASC and ΔDEF both are right-angled triangles.
In ΔABC and ΔDEF,
∠CAB = ∠EDF = θ
∠B = ∠E ......[Each 90°]
∴ ΔABC ∼ ΔDEF ......[By AAA similarity criterion]
Then, `(AB)/(DE) = (BC)/(EF)`
⇒ `24/16 = 15/h`
∴ h = `(15 xx 16)/24` = 10
Hence, the height of the telephone pole is 10 m.
Concept: Similarity of Triangles
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