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>
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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 $1$ has the same area as a square with side length $s$. Find $s$.

$\mathrm{(A)}\,\frac{\sqrt[4]{3}}{2}\quad\mathrm{(B)}\,\frac{\sqrt[4]{3}}{\sqrt{2}}\quad\mathrm{(C)}\,1\quad\mathrm{(D)}\,\frac{3}{4}\quad\mathrm{(E)}\,\frac{4}{3}\quad\mathrm{(F)}\,\sqrt{3}\quad\mathrm{(G)}\,\frac{\sqrt6}{2}$

Solution

An equilateral triangle with side length 1 has an area of $\frac{(1)^2\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$, and a square with that area has a side length of $\sqrt{\frac{\sqrt{3}}{4}}=\frac{\sqrt[4]{3}}{2} \Rightarrow \mathrm{(A)}.$

See Also

2007 iTest (Problems)
Preceded by:
Problem 6
Followed by:
Problem 8
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