A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the altitude x of the rectangle is half the base of the rectangle, then:
Explanation:
Let ABC be the triangle with h as the altitude and BC as base. Let PQRS be the rectangle inscribed in triangle ABC.
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
∴ 2xb = h-xh ∴ 2xb = 1 - xh ∴ 2xb + xh = 1 ∴ x = bhb+2h
Hence, option (c).
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