Difference between revisions of "1954 AHSME Problems/Problem 13"

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== Solution ==
 
== Solution ==
I think the answer should be B
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The arcs created by all sides of the quadrilateral will sum to 360 because they cover the entire circle. The inscribed angles created by each side of the trapezoid will all be half the measure of each arc created by their respective side. Thus, the sum of the all the inscribed angles is half the sum of all the arcs. <math>360/2 = 180\
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\fbox{(A)}</math>
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--bozz

Revision as of 17:17, 27 February 2020

Problem 13

A quadrilateral is inscribed in a circle. If angles are inscribed in the four arcs cut off by the sides of the quadrilateral, their sum will be:

$\textbf{(A)}\ 180^\circ \qquad \textbf{(B)}\ 540^\circ \qquad \textbf{(C)}\ 360^\circ \qquad \textbf{(D)}\ 450^\circ\qquad\textbf{(E)}\ 1080^\circ$

Solution

The arcs created by all sides of the quadrilateral will sum to 360 because they cover the entire circle. The inscribed angles created by each side of the trapezoid will all be half the measure of each arc created by their respective side. Thus, the sum of the all the inscribed angles is half the sum of all the arcs. $360/2 = 180\  \fbox{(A)}$

--bozz