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

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==See Also==
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{{AHSME 50p box|year=1954|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 01:26, 28 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. $360/2 = 180\  \fbox{(A)}$

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See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AHSME Problems and Solutions


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