Discussion

Explanation:

Let ABC be the triangle with h as the altitude and BC as base. Let PQRS be the rectangle inscribed in triangle ABC.

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As height of rectangle = PS = x, and 

base of the rectangle = SR = 2x = PQ

Now, height of ∆APQ = height of triangle – height of rectangle = h – x

Now, ∆APQ and ∆ABC are similar.

∴ PQBC = h-xh

2xbh-xh
2xb = 1 - xh
2xb + xh = 1 
∴ x = bhb+2h

Hence, option (c).

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