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In the ΔABC, AB = 2 cm, BC = 3 cm and AC = 4 cm. D is the middle-point of AC. If a square is constructed on the side BD, what is the area of the square? 

(a) 4.5 cm2 

(b) 2.5 cm2 

(c) 6.35 cm2 

(d) None of these

1 Answer

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Best answer

(b) 2.5 cm2.

In ΔABC,

cos A = \(\frac{b^2+c^2-a^2}{2bc}\)

\(\frac{AC^2+AB^2-BC^2}{2\times{AC}\times{AB}}\)

\(\frac{4^2+2^2-3^2}{2\times4\times2}\)

\(\frac{16+4-9}{16}=\frac{11}{16}\)

In ΔBAD,

cos A = \(\frac{AD^2+AB^2-BD^2}{2\times{AD}\times{AB}}\)

⇒ \(\frac{11}{16}\) = \(\frac{4+4-BD^2}{2\times2\times2}\) = \(\frac{8-BD^2}{8}\)

⇒ 11 = 16 – 2BD2 ⇒ 2BD2 = 5 ⇒ BD2 = 2.5 

∴ Area of square on BD = (BD)2 = 2.5 cm2

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