Express the area of the shaded region below as a function of x. The dimensions in the figure are centimeters.

Question

Answer 1

I am making this shape into a triangle; a=?, b=x, c=(a^2+b^2)^.5;

and a square; 3x.

In this case "b"=x; "a"=(c^2-x^2)^.5

So ab/2+3x=(((c^2-x^2)^.5)x)/2+3x.

Answer 2

I'm sure you are aware that the area of a trapezoid is (B₁ + B₂)/2(h) in this case, B₁ = 3 and h = x but we don't know what B₂ is.

But let's put a coordinate axis in at the lower right hand corner and put points on the on vertices like this:

P₁ is at (0,3) and P₂ is at (6,5) so we can create the equation of a line between them in the form of y = mx + b and that "y" will be the B₂ length. The y-intercept is 3 so b = 3, the slope is rise/run = (5-3)/(6-0) = 1/3 so the equation of the line is y = 1/3x + 3.

Using B₂ = 1/3x + 3, we get the area of the trapezoid = (B₁ + B₂)/2(h) = (3 + 1/3x + 3)/2(x) = (x)(1/3x + 6)/2 = (x)(1/6x + 3) = 1/6x² + 3x

A(x) = 1/6x² + 3x where x is in cm and A(x) is in square cm