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IPMAT Indore 2024 - QA (MCQ) Q22 Explanation
November 24, 2024
Explanation:
(b) Given: BA = ‘c’
BD = ‘k’
∠BDA = θ (acute angle)
Now in △BDP, BP = k ∙ sinθ
Now in △ABD,
cos θ = x^2 +k^2 −c^2/2.x.k
2xk ⋅ cos θ = x^2 + k^2 - c^2
= x^2 −2x⋅k⋅cos θ+k^2 −c2 = 0
Solving the quadratic equation for x, we get
x = kcosθ ±√c^2−k^2sin2 θ
Now, the area of the ∆ABC = 1/2 × 2x × ksinθ
= x ∙ ksinθ
= (kcosθ±√c^2 −k^2sin^2θ ) ⋅ ksinθ
= k^2sinθcosθ ± ksinθ ⋅ √c^2 −k^2sin^2θ
= 1/2 k^22sinθcosθ ± ksinθ.√c^2 −k^2sin^2θ
=1/2 k^2 sin^2θ ±ksinθ ⋅ √c^2 −k^2sin^2θ
=1/2 k^2sin^2θ± ksinθ ⋅ √c^2 − k^2sin2θ .
Hence option (b). Ans.