Correction: 7/15 – Found the point of intersection of the two lines and then calculated the area of the small triangle. Trying to figure out a non-algebraic way of solving this.

Taking the side of the square as 6 units and the axes along the two sides of the square, the intersection is (1.2, 3.6) The shaded region is area of trapezoid minus area of triangle, which is 18 – 1.2 = 16.8 sq units. The fraction is therefore 16.8/36 = 7/15.

7/15 – The slope of the bottom right boundary of the shaded area is -1/3 and the equation of the line is y = -1/3*x -1/3. The slope of the bottom left boundary line is -2 and the equation of this boundary line is y = -2*x. Solving for the intersection of these two line gives the point x = 1/5, y = -2/5. The shaded area = (1/2)*(1/5)*(2/5) + (2/5)*(1-1/5) + (1/2)*(1-1/5)*(2/3-2/5) = 7/15.

260/9? Or 28.8 repeating percent?

13/30

Correction: 7/15 – Found the point of intersection of the two lines and then calculated the area of the small triangle. Trying to figure out a non-algebraic way of solving this.

Taking the side of the square as 6 units and the axes along the two sides of the square, the intersection is (1.2, 3.6) The shaded region is area of trapezoid minus area of triangle, which is 18 – 1.2 = 16.8 sq units. The fraction is therefore 16.8/36 = 7/15.

7/15 – The slope of the bottom right boundary of the shaded area is -1/3 and the equation of the line is y = -1/3*x -1/3. The slope of the bottom left boundary line is -2 and the equation of this boundary line is y = -2*x. Solving for the intersection of these two line gives the point x = 1/5, y = -2/5. The shaded area = (1/2)*(1/5)*(2/5) + (2/5)*(1-1/5) + (1/2)*(1-1/5)*(2/3-2/5) = 7/15.