Answer :

Here,

In the above Figure,

DE is the distance of student and mirror.

EG is the distance of mirror and flagpole.

CD is the height of student till its eyes.

FG is the height of flagpole which we need to find.

Now,

In ΔCDE and ΔFGE

∠FEG = ∠CED (by mirror property)

∠CDE = ∠FGE (both perpendicular given)

∴ ΔCDE ∼ ΔFGE

(∵ ΔCDE ∼ ΔFGE)

⇒ FG = 9 m

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