Difference between revisions of "2007 iTest Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | An equilateral triangle with side length 1 has an area of <math>\frac{(1)^2\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</math>, and a square with that area has a side length of <math>\sqrt{\frac{\sqrt{3}}{4}}=\frac{\sqrt[4]{3}}{2} \Rightarrow \mathrm{(A)}</math> | + | An equilateral triangle with side length 1 has an area of <math>\frac{(1)^2\sqrt{3}}{4}=\frac{\sqrt{3}}{4}</math>, and a square with that area has a side length of <math>\sqrt{\frac{\sqrt{3}}{4}}=\frac{\sqrt[4]{3}}{2} \Rightarrow \mathrm{(A)}.</math> |
==See Also== | ==See Also== | ||
Latest revision as of 16:13, 19 December 2018
Problem
An equilateral triangle with side length 
has the same area as a square with side length 
.
Find .
![$\mathrm{(A)}\,\frac{\sqrt[4]{3}}{2}\quad\mathrm{(B)}\,\frac{\sqrt[4]{3}}{\sqrt{2}}\quad\mathrm{(C)}\,1\quad\mathrm{(D)}\,\frac{3}{4}\quad\mathrm{(E)}\,\frac{4}{3}\quad\mathrm{(F)}\,\sqrt{3}\quad\mathrm{(G)}\,\frac{\sqrt6}{2}$](http://latex.artofproblemsolving.com/0/f/d/0fd9825fdc1605c6a9bd15b665c79728f942f87c.png)
Solution
An equilateral triangle with side length 1 has an area of 
, and a square with that area has a side length of ![$\sqrt{\frac{\sqrt{3}}{4}}=\frac{\sqrt[4]{3}}{2} \Rightarrow \mathrm{(A)}.$](http://latex.artofproblemsolving.com/5/4/f/54f7bc510cc8fccf00dccdf5655c81e4ac795d16.png)
See Also
| 2007 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 6 |
Followed by: Problem 8 | |
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