No need of any jugglery - 4

Geometry Level pending

Given is $\triangle ABC$ of side length $10$ . $\triangle ABD$ is a right angle triangle with $\overline { AD }$ as its hypotenuse. The value of $\overline { BD }$ such that $\overline { BC } \bot \overline { AD }$ is $\ell$ . Find $\left\lceil \ell \right\rceil$ .

Note: $\left\lceil x \right\rceil$ represents the smallest integer greater than or equal to $x$ .

7 8 5 6

1 solution

Chew-Seong Cheong
Jul 4, 2015

When $\overline{BC} \bot \overline{AD}$ , $\angle DAB = 30^\circ$ . Then we have:

$\overline{BD} = \ell = 10 \tan{30^\circ} = \dfrac{10}{\sqrt{3}}$

$\Rightarrow \lceil \ell \rceil = \left \lceil \dfrac{10}{\sqrt{3}} \right \rceil = \lceil 5.773502692 \rceil = \boxed{6}$

Nice solution sir. :)

Rohit Ner - 5 years, 11 months ago

No need of any jugglery - 4

1 solution

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