Question
The height of a conical tent is 9m. A vertical pole of
6m height is placed 4 m away from its centre such that it touches the surface. Find the slant height of the tent from the base to the point where the pole touches it?Solution
Since, β’s are similar, AD/AB = DE/BC⇒ 3/9 = 4/BC ∴ BC = 12m In β ABC, AC = √(γABγ^2+BC² ) = √(9^2+12² ) = √(81+144) = = √(225 ) = 15 m In β ADE, AE = √(γADγ^2+DE² ) = √(3^2+4² ) = √(9+16) = = √(25 ) = 5 m ∴ CE = AC – AE = 15 – 5 = 10 m
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