Difference between revisions of "1960 AHSME Problems/Problem 6"
Rockmanex3 (talk | contribs) m (→See Also) |
Rockmanex3 (talk | contribs) m (→See Also) |
||
| Line 28: | Line 28: | ||
==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1960|num-b=5|num-a=7}} | {{AHSME 40p box|year=1960|num-b=5|num-a=7}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 17:57, 17 May 2018
Problem
The circumference of a circle is 
inches. The side of a square inscribed in this circle, expressed in inches, is:
Solution
First, find the diameter of the circle. Plug in the circumference for the formula 
to solve for 
.
![\[100 = \pi d\]](http://latex.artofproblemsolving.com/5/c/e/5ce01d3633e9564527e94d7492209688814e931d.png)
![\[d = \frac{100}{\pi}\]](http://latex.artofproblemsolving.com/6/a/0/6a0bc39d7182fcc2f04a1181879f48bd492fc5ec.png)
Since all of an inscribed square's vertices touch the circle, and one of the angles of the circle is 
, the square's diagonal is the diameter of the circle.
![[asy] draw(circle((0,0),50)); draw((-50,0)--(0,50)--(50,0)--(0,-50)--cycle); draw((-50,0)--(50,0)); draw((-3,47)--(0,44)--(3,47)); label("$\frac{100}{\pi}$",(0,-10)); [/asy]](http://latex.artofproblemsolving.com/2/4/7/24707203747249e65ba033f4b1c2e36e3a5e43ba.png)
From the Pythagorean theorem (or 45-45-90 triangles), the side length is 
inches, which is answer choice 
.
See Also
| 1960 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
