Difference between revisions of "FidgetBoss 4000's 2019 Mock AMC 12B Problems/Problem 2"

Revision as of 15:18, 19 November 2020

Problem

(fidgetboss_4000) In the diagram below, $ABC$ is an isosceles right triangle with a right angle at $B$ and with a hypotenuse of $40\sqrt2$ units. Find the greatest integer less than or equal to the value of the radius of the quarter circle inscribed inside $\triangle ABC$ .

$\textbf{(A)} 26\qquad\textbf{(B)} 27\qquad\textbf{(C)} 28\qquad\textbf{(D)} 29\qquad\textbf{(E)} 30\qquad$

Solution

The quarter circle is centered at $B$ and just touches the hypotenuse at the midpoint of the hypotenuse due to symmetry in an isosceles right triangle. The value of its radius is equal to the distance between $B$ and the midpoint of the hypotenuse, and it is well known that this is half the length of the hypotenuse, or $\frac{1}{2}\cdot 40\sqrt2=20\sqrt2$ . The greatest integer less than or equal to $20\sqrt2$ is $28$ , thus we pick answer $\boxed{\textbf{(C) }28}$ .

2019 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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Difference between revisions of "FidgetBoss 4000's 2019 Mock AMC 12B Problems/Problem 2"

Revision as of 15:18, 19 November 2020

Problem

Solution

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