CAT 2019 Set-2 | Question: 93

Question

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

• A. $10$
• B. $6\sqrt{2}$
• C. $8\sqrt{2}$
• D. $5$

Answer 1

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