Area Function
The squares of the figures has parallel sides and the same center. The larger square has a side length of 10 and the smaller square has a side length of x . . The graph that better plot the function of the area of the gray region in function of X is:
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1 solution
Let the gray area in question be the sum of the areas of a trapezoid and a right triangle according to:
Trapezoid: Height = (10-x)/2, Base 1 = x, Base 2 = 10 - (10-x)/2 = 5 + x/2.
Triangle: Base = (10-x)/2, Height = x + (10-x)/2 = 5 + x/2.
So now the gray area as a function of x computes to:
A(x) = (1/2) [(10-x)/2] [x + (5 + x/2)] + (1/2) [(10-x)/2] (5 + x/2);
or A(x) = [(10-x)/4] * (5 + 3x/2) + [(10-x)/4] * (5 + x/2);
or A(x) = [(10-x)/4] * (10 + 2x);
or A(x) = (1/2) (10 - x)(5 + x) = (1/2) (50 + 5x - x^2);
or A(x) = -(1/2)*(x^2 - 5x) + 25;
or A(x) = -(1/2)*(x^2 - 5x + 25/4) + 25/8 + 25 (completing the square);
or A(x) = -(1/2)*(x - 5/2)^2 + 225/8
which is a concave-down parabola with vertex at (5/2, 225/8). Graph (B) is therefore the best choice.