+0

# help geometry

0
32
2

The vertex of the right isosceles triangle is the center of the square.   Each leg of the right isoscles triangle is 6. What is the area of the overlapping region? Jul 14, 2022

#2
+1

Note that the area of the overlapping region is basically the area of the triangle without the 2 smaller triangles in the corner.

Because the distance from the center vertice to the edge of the square is $$10 \div 2 = 5$$, the smaller triangle has a height of 1.

This means that the 2 smaller triangles are congruent and similar to the larger triangle by a ratio of $$1:6$$

So, the area of each triangle is $${1 \over 6} ^2 \times 18= {{1 \over 2}}$$, meaning the area of the overlapping region is $$18 - (2 \times {1 \over 2}) = \color{brown}\boxed{17}$$

Jul 14, 2022

#1
-1

The area of the overlapping region is 12 - 2*sqrt(2).

Jul 14, 2022
#2
+1

Note that the area of the overlapping region is basically the area of the triangle without the 2 smaller triangles in the corner.

Because the distance from the center vertice to the edge of the square is $$10 \div 2 = 5$$, the smaller triangle has a height of 1.

This means that the 2 smaller triangles are congruent and similar to the larger triangle by a ratio of $$1:6$$

So, the area of each triangle is $${1 \over 6} ^2 \times 18= {{1 \over 2}}$$, meaning the area of the overlapping region is $$18 - (2 \times {1 \over 2}) = \color{brown}\boxed{17}$$

BuilderBoi Jul 14, 2022