Difference between revisions of "1950 AHSME Problems/Problem 9"
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| − | == Problem== | + | == Problem == |
The area of the largest triangle that can be inscribed in a semi-circle whose radius is <math>r</math> is: | The area of the largest triangle that can be inscribed in a semi-circle whose radius is <math>r</math> is: | ||
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<math> \textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2} </math> | <math> \textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2} </math> | ||
| − | ==Solution== | + | == Solution == |
| − | The area of a triangle is <math>\frac12 bh.</math> To maximize the base, let it be equal to the diameter of the semi circle, which is equal to <math>2r.</math> To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to <math>r.</math> Therefore the area is <math>\frac12 \cdot 2r \cdot r = \boxed{\mathrm{(A) }r^2 | + | The area of a triangle is <math>\frac12 bh.</math> To maximize the base, let it be equal to the diameter of the semi circle, which is equal to <math>2r.</math> To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to <math>r.</math> Therefore the area is <math>\frac12 \cdot 2r \cdot r = \boxed{\mathrm{(A) }r^2}</math>. |
| − | ==See Also== | + | == See Also == |
{{AHSME 50p box|year=1950|num-b=8|num-a=10}} | {{AHSME 50p box|year=1950|num-b=8|num-a=10}} | ||
Latest revision as of 23:50, 11 October 2020
Problem
The area of the largest triangle that can be inscribed in a semi-circle whose radius is 
is:
Solution
The area of a triangle is 
To maximize the base, let it be equal to the diameter of the semi circle, which is equal to 
To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to 
Therefore the area is 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
