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Let $ABC$ be a right-angled triangle with hypotenuse $BC$ of length $20$ cm. If $AP$ is perpendicular on $BC$, then the maximum possible length of $AP$, in cm, is

  1. $10$
  2. $6\sqrt{2}$
  3. $8\sqrt{2}$
  4. $5$
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Given that, $\triangle ABC$  is a right-angled triangle. 

Let  $\angle A = 90^{\circ}$

 

 For $AP$ to be maximum triangle $ABC$ has to be an isosceles triangle. Means $AB = AC$

Apply the Pythagorean theorem, in $\triangle ABC.$ 

$\boxed {\text{Hypotenuse(H)}^{2} = \text{Perpendicular (P)}^{2} + \text{Base (B)}^{2}}$

$(20)^{2} = (AB) ^{2} + (AC)^{2}$

$ \Rightarrow 400 = 2 AB^{2} \quad [\because AB = AC]$

$ \Rightarrow 200 = AB^{2}$

$ \Rightarrow \sqrt{200}= AB$

$ \Rightarrow AB =10 \sqrt {2} $



We have, $ \triangle{ APB } = \triangle {APC} $

Then, the area also should be equal.

$ \dfrac{1}{2} AC \times BA = \dfrac {1}{2} BC \times AP$

$ \Rightarrow \boxed {\dfrac{AC\times BA}{BC} = AP}$

$ \Rightarrow \dfrac {10 \sqrt{2} \times 10 \sqrt {2}}{20}=AP$

$ \Rightarrow \dfrac{100 \times 2}{20}=AP$

$\Rightarrow AP=10\;\text{cm}.$

Correct Answer: A

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