A lotus is 20 cm above the water surface in a pond and its stem is partly below the water surface. As the wind blew, the stem is pushed aside so that the lotus touched the water at 40 cm away from the original position of the stem. Originally how much of the stem was below the water surface? (From ‘Leelavathi’ of Bhaskaracharya).

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Solution

Let AD be the height of the lotus above the water surface, BD be the part of the stem below the water surface and CD be the distance of the lotus from the original position of the stem.

It is given that AD = 20 cm and CD = 40 cm.

Let the stem be initially x cm below the water surface.

∴BD = x cm

Now, AB = AD + BD = (20 + x) cm

Also, AB = BC = (20 + x) cm

In right Δ BCD:

Thus, the stem was initially 30 cm below the water surface.

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