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Sum

A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.

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#### Solution

Let MN = 18 m be the flag pole and its shadow be LM = 9.6 m.

The distance of the top of the pole, N from the far end, L of the shadow is LN.

In right-angled ∆LMN,

LN^{2} = LM^{2} + MN^{2} ......[By Pythagoras theorem]

⇒ LN^{2} = (9.6)^{2} + (18)^{2}

⇒ LN^{2} = 9.216 + 324

⇒ LN^{2} = 416.16

∴ LN = `sqrt(416.16)` = 20.4 m

Hence, the required distance is 20.4 m

Concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle

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