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?
+2666 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 \(18 - (2 \times {1 \over 2}) = \color{brown}\boxed{17}\)
+2666 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 \(18 - (2 \times {1 \over 2}) = \color{brown}\boxed{17}\)
