Difference between revisions of "1959 AHSME Problems/Problem 43"
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== Solution == | == Solution == | ||
− | Use the formula : | + | Use the formula : Area = abc/4R |
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+ | == See also == | ||
+ | {{AHSME 50p box|year=1959|num-b=42|num-a=44}} | ||
+ | {{MAA Notice}} |
Revision as of 11:49, 22 July 2024
Problem
The sides of a triangle are , and
. The diameter of the circumscribed circle is:
Solution
Use the formula : Area = abc/4R
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 42 |
Followed by Problem 44 | |
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 |
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