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

Guest May 4, 2019

#1**+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.)

hectictar May 5, 2019

#1**+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

#4**+2 **

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!

tertre May 5, 2019