FidgetBoss 4000's 2019 Mock AMC 12B Problems/Problem 2


(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 $\triangle ABC$.

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


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 $\boxed{\textbf{(C) }28}$.

See also

FidgetBoss 4000's 2019 Mock AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 12 Problems and Solutions

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