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== | ||
{{iTest box|year=2007|num-b=6|num-a=8}} | {{iTest box|year=2007|num-b=6|num-a=8}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
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 | |
| 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 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4 | ||
