Sides other than the hypotenuse of …
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Sia ? 6 years, 3 months ago
Given, sides other than the hypotenuse of a right triangle is 16 cm and 8 cm.
So, Let AC = 16 cm and BC = 8 cm.
Let PQCR be the largest square which can be inscribed in the right triangle ABC. Thus,
Let AC = x cm, So, AQ = 16 - x cm
In {tex}\triangle A P Q{/tex} and {tex}\triangle A BC{/tex}
{tex}\angle A=\angle A{/tex} (common angle)
{tex}\angle A Q P=\angle A C B{/tex} (each has 90o)
{tex}\Rightarrow \triangle A P Q \approx \triangle A B C{/tex} (By AA similarity)
So,
{tex}\frac{A Q}{A C}=\frac{P Q}{B C}{/tex}
{tex}\Rightarrow \frac{16-x}{16}=\frac{x}{8}{/tex}
{tex}\Rightarrow 16-x=2 x{/tex}
{tex}\Rightarrow x=\frac{16}{3} \mathrm{cm}{/tex}
Hence, the length of largest square which can be inscribed in the right triangle ABC is 16/3 cm.
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