In any triangle ABC,AB = 2, BC = 4, CA = 3 an | vedclass

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(D) In $	riangle ABC$,by the Law of Cosines in $	riangle ABC$ at angle $B$:$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 	imes AB 	imes BC} = \frac{2^2 +

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$AD^2 = 2.5$

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(D) In $	riangle ABC$,by the Law of Cosines in $	riangle ABC$ at angle $B$:$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 	imes AB 	imes BC} = \frac{2^2 + 4^2 - 3^2}{2 	imes 2 	imes 4} = \frac{4 + 16 - 9}{16} = \frac{11}{16}$.Since $D$ is the midpoint of $BC$,$BD = 2$.Applying the Law of Cosines in $	riangle ABD$:$AD^2 = AB^2 + BD^2 - 2 	imes AB 	imes BD 	imes \cos B$$AD^2 = 2^2 + 2^2 - 2 	imes 2 	imes 2 	imes \frac{11}{16}$$AD^2 = 4 + 4 - 8 	imes \frac{11}{16}$$AD^2 = 8 - \frac{11}{2} = 8 - 5.5 = 2.5$.Thus,$AD^2 = 2.5$.

In any triangle $ABC$,$AB = 2, BC = 4, CA = 3$ and $D$ is the midpoint of $BC$. Then:

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