Difference between revisions of "1954 AHSME Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | <math> | + | 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\ |
| + | \fbox{(A)}</math> | ||
| + | |||
| + | --bozz | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME 50p box|year=1954|num-b=12|num-a=14}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 00:26, 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:
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. 
--bozz
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
