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Explanation:

The figure can be drawn as below.

Sum of areas of three smaller ∆s = Area of ∆ABC

⇒ 12×h1×a12×h2×a + 12×h3×a = 34a2

12×s×a34a2

⇒ a = 2s3

∴ Area of ∆ABC = 34a2 = 34×43×s2 = s23

Hence, option (a).

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