In $\triangle A B C, \angle A=90^{\circ}, \mathrm{M}$ is the midpoint of $\mathrm{BC}$ and $\mathrm{D}$ is a point on $\mathrm{BC}$ such that $\mathrm{AD} \perp \mathrm{BC}$. If $\mathrm{AB}=7 \mathrm{~cm}$ and $\mathrm{AC}=24 \mathrm{~cm}$, then $\mathrm{AD}: \mathrm{AM}$ is equal to:
Answer & explanation
Correct answer: option 1
In right angled \(\Delta \)ABC, \({90}^\circ\) at A
\( {BC }^{2 } \) = \( {AB }^{2 } \) + \( {AC }^{2 } \) = \( {7 }^{2 } \) + \( {24 }^{2 } \) = 49 + 576 = 625
= BC = \(\sqrt {625 }\) = 25
If, M is the midpoint of BC, then
AM = MC = BM = \(\frac{25}{2}\)
As we know,
AD x BC = AB x AC
= AD x 25 = 7 x 24
= AD = \(\frac{7 \;×\;24 }{25}\)
Therefore, ratio of AD : AM = \(\frac{7 \;×\;24 }{25}\) : \(\frac{25}{2}\) = 336 : 625