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ABCD is a rectangle. Δ ADE and Δ ABF are two triangles such that ∠E=∠F as shown in the figure. Prove that AD x AF=AE x AB.
Ans: Consider Δ ADE and Δ ABF
∠D = ∠B = 90o
∠E = ∠F (given)
∴Δ ADE ≅ Δ ABF
AD/AB = AE/AF
⇒ AD x AF = AB x AE Proved
f Y is a discrete random variable with expected value E[Y ] = µ and if X = a + bY , prove that Var (X) = b2Var (Y ) .
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