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(D) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.By the property of geometric mean in a right-angled triangle,$BM^2

Vedclass · Accepted Answer

$25$

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(D) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.By the property of geometric mean in a right-angled triangle,$BM^2 = AM \cdot MC$.Given $BM = 12$ and $AM = 9$,we have $12^2 = 9 \cdot MC$.$144 = 9 \cdot MC \implies MC = \frac{144}{9} = 16$.In $\Delta ABM$,by Pythagoras theorem,$AB^2 = AM^2 + BM^2 = 9^2 + 12^2 = 81 + 144 = 225$.So,$AB = \sqrt{225} = 15$.In $\Delta ABC$,by Pythagoras theorem,$AC^2 = AB^2 + BC^2$.First,find $BC^2$ in $\Delta BMC$: $BC^2 = BM^2 + MC^2 = 12^2 + 16^2 = 144 + 256 = 400$.Thus,$AC^2 = 225 + 400 = 625$.$AC = \sqrt{625} = 25$.Alternatively,$AC = AM + MC = 9 + 16 = 25$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $BM = 12$ and $AM = 9$,then $AC = \ldots$

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