Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 5"
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| − | == Solution == | + | == Solution == |
| + | We see that the area of triangle A and B is 450 from the area of square A doubled. Noticing that all triangles made are 45-45-90,we write the area of the figure in terms of s, a side of square b. Writing this as (s^2)/4 + 2s^2 = 450, solving by algebra is trivial and we see s is <math>\sqrt{200}</math>, and the area of square b is therefore <math>\boxed{200}</math>. | ||
== See also == | == See also == | ||
| − | {{ | + | {{UNCO Math Contest box|year=2009|n=II|num-b=4|num-a=6}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 01:35, 29 January 2019
Problem
The two large isosceles right triangles are congruent.
If the area of the inscribed square 
is 
square
units, what is the area of the inscribed square 
?
![[asy] draw((0,0)--(1,0)--(0,1)--cycle,black); draw((0,0)--(1/2,0)--(1/2,1/2)--(0,1/2)--cycle,black); MP("A",(1/4,1/8),N); draw((2,0)--(3,0)--(2,1)--cycle,black); draw((2+1/3,0)--(2+2/3,1/3)--(2+1/3,2/3)--(2,1/3)--cycle,black); MP("B",(2+1/3,1/8),N); [/asy]](http://latex.artofproblemsolving.com/2/5/3/2531f32f5e9af71cd7c0f6b81eb97a0c18905b18.png)
Solution
We see that the area of triangle A and B is 450 from the area of square A doubled. Noticing that all triangles made are 45-45-90,we write the area of the figure in terms of s, a side of square b. Writing this as (s^2)/4 + 2s^2 = 450, solving by algebra is trivial and we see s is 
, and the area of square b is therefore 
.
See also
| 2009 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
