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The vertex of the right isosceles triangle is the center of the square. What is the area of the overlapping region?

 May 4, 2019

Best Answer 

 #1
avatar+8103 
+4

Note that area in question  =  area of △UYZ  +  area of trapezoid UYXW

 

m∠VUW + m∠WUY  =  90°

m∠ZUY + m∠WUY  =  90°

So

m∠VUW  =  m∠ZUY

 

By ASA congruence, △UVW is congruent to △UYZ.

 

area of △UVW  +  area of trapezoid UYXW  =  area of square UVXY  =  5 * 5  =  25 sq units

 

Since △UVW is congruent to △UYZ,

 

area of △UYZ  +  area of trapezoid UYXW  =  area of △UVW  +  area of trapezoid UYXW  =  25 sq units

 

(Also, I made this to help me figure it out because I was very lost on this problem for awhile! You can see that as the value of  a  changes, the area in question stays the same, at 25 sq units.)

 May 5, 2019
 #1
avatar+8103 
+4
Best Answer

Note that area in question  =  area of △UYZ  +  area of trapezoid UYXW

 

m∠VUW + m∠WUY  =  90°

m∠ZUY + m∠WUY  =  90°

So

m∠VUW  =  m∠ZUY

 

By ASA congruence, △UVW is congruent to △UYZ.

 

area of △UVW  +  area of trapezoid UYXW  =  area of square UVXY  =  5 * 5  =  25 sq units

 

Since △UVW is congruent to △UYZ,

 

area of △UYZ  +  area of trapezoid UYXW  =  area of △UVW  +  area of trapezoid UYXW  =  25 sq units

 

(Also, I made this to help me figure it out because I was very lost on this problem for awhile! You can see that as the value of  a  changes, the area in question stays the same, at 25 sq units.)

hectictar May 5, 2019
 #2
avatar+101099 
+2

HECTICTAR!!!!..good to see you!!!!..where have you been keeping yourself????????

 

Very nice answer.....this one had me stumped   !!!!!

 

 

 

 

cool cool cool

CPhill  May 5, 2019
 #3
avatar+101734 
+2

Yes it is excellent to see you back again Hectictar!

Melody  May 5, 2019
 #5
avatar+122 
0

Welcome back, Hectictar!!

 

~~Hypotenuisance

Hypotenuisance  May 10, 2019
 #4
avatar+4249 
+1

We have to place an isosceles triangle, 45-45-90 triangle such that the center of the isosceles triangle is the center of the square. The only way to do this is by rotating the isosceles triangles, such that the side is opposite the 90-degree angle. This is 1/4 of the total square or 1/4*100-25 square units.

 

 

Welcome back, hectictar!

 May 5, 2019

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