Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 5"

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== Solution ==
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== Solution ==  
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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 ==

Latest revision as of 02:35, 29 January 2019

Problem

The two large isosceles right triangles are congruent. If the area of the inscribed square $A$ is $225$ square units, what is the area of the inscribed square $B$?

[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]

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 $\sqrt{200}$, and the area of square b is therefore $\boxed{200}$.

See also

2009 UNCO Math Contest II (ProblemsAnswer KeyResources)
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
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