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?
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Since, ∆’s are similar, AD/AB = DE/BC⇒ 3/9 = 4/BC ∴ BC = 12m In ∆ ABC, AC = √(〖AB〗^2+BC² ) = √(9^2+12² ) = √(81+144) = = √(225 ) = 15 m In ∆ ADE, AE = √(〖AD〗^2+DE² ) = √(3^2+4² ) = √(9+16) = = √(25 ) = 5 m ∴ CE = AC – AE = 15 – 5 = 10 m