Question : In $\mathrm{\Delta ABC, \angle BAC = 90^{\circ}}$ and $\mathrm{AD}$ is drawn perpendicular to $\mathrm{BC}$. If $\mathrm{BD} = 7\;\mathrm{cm}$ and $\mathrm{CD }= 28\;\mathrm{cm}$, what is the length of $\mathrm{AD}$?
Option 1: $3.5 \text{ cm}$
Option 2: $7 \text{ cm}$
Option 3: $10.5 \text{ cm}$
Option 4: $14 \text{ cm}$
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Correct Answer: $14 \text{ cm}$
Solution :
In the right-angled $\triangle \mathrm{ABC}$, $\mathrm{AD}$ is the altitude on the hypotenuse $\mathrm{BC}$.
According to the property of triangles, the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and each other.
$\therefore\mathrm{AD^2 = BD × DC}$
$⇒\mathrm{AD = \sqrt{BD × DC}}$
$⇒\mathrm{AD = \sqrt{BD×DC} = \sqrt{7 \text{ cm}× 28 \text{ cm}} = \sqrt{196 \text{ cm}^2} = 14 \text{ cm}}$
Hence, the correct answer is $14 \text{ cm}$.
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