Isosceles right triangle $${ABC}$$ is shown. In a right triangle $$D E F, \angle D$$ is the right angle, the length of $$\overline{D E}$$ is $$2$$ times the length of $$\overline{A B}$$, and the length of $$\overline{{DF}}$$ is 4 times the length of $$\overline{{AC}}$$. ![](http://admin.getprephub.com/media/editor/Screenshot%202023-04-27%20at%201.46.33%20PM_20230427114647038065.png) How many times greater is the length of $$\overline{{EF}}$$ than the length of $$\overline{{BC}}$$ ?

Answer Given that, In $$\triangle A B C, A B=A C=3$$ unit,$$B C=\sqrt{3^{2}+3^{2}}=\sqrt{18}$$ unit In $$\triangle D E F, D E=2 A B=6$$ unit, $${DF}=4 {AC}=12$$ unit Since, In $$\triangle D E F, \angle D$$ is the right angle. $$D E^{2}+D F^{2}=E F^{2}$$ $$E F=\sqrt{6^{2}+12^{2}}=\sqrt{180}$$ unit $$\frac{E F}{B C}=\frac{\sqrt{180}}{\sqrt{18}}=\frac{\sqrt{10} \cdot \sqrt{18}}{\sqrt{18}}=\sqrt{10}$$

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Isosceles right triangle ${ABC}$ is shown.

In a right triangle $D E F, \angle D$ is the right angle, the length of $\overline{D E}$ is $2$ times the length of $\overline{A B}$ , and the length of $\overline{{DF}}$ is 4 times the length of $\overline{{AC}}$ .

How many times greater is the length of $\overline{{EF}}$ than the length of $\overline{{BC}}$ ?

$\sqrt{10}$

$\sqrt{17}$

$10$

$17$

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