#### Exercise c

Chapter 8, Section 8-1, Page 345

ISBN: 9780328931583

Solution

VerifiedThe length of the hypotenuse of $\triangle ABC$ is $10$ and the length of its sides is $5\sqrt{2}$ so if we divide them, we get:

$\dfrac{10}{5\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\color{#c34632}{\sqrt{2}}$

The length of the hypotenuse of $\triangle ADC$ is $5\sqrt{2}$ and the length of its sides is $5$ so if we divide them, we get:

$\dfrac{5\sqrt{2}}{5}=\color{#c34632}{\sqrt{2}}$

The ratio of the hypotenuse and one of the congruent legs in a $45\text{\textdegree}$-$45\text{\textdegree}$-$90\text{\textdegree}$ triangle is $\sqrt{2}$.

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