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

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)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by FidgetBoss 4000's Mock American Mathematics Competitions.

Page

Toolbox

Search

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

Problem

Solution

See also