Question

The drawing below shows a square with side a. A straight line intersects the square and encloses

an area A. The heights x and y on the left and right side (in a distance d from the square) of

the intersecting line can be varied. Assuming that x  y and x; y  a, nd an expression for

the enclosed area A(x; y) with respect to x and y.

Answer 1

Area of the trapezoid $BCDE$

$$A=\dfrac{BE+CD}{2}\cdot ED$$ $$\dfrac{BM}{FM}=\dfrac{CN}{FN}=\dfrac{GK}{FK}$$

Substitute

$$\dfrac{BE-x}{d}=\dfrac{CD-x}{a+d}=\dfrac{y-x}{a+2d}$$ $$d(BE-x)+a(BE-x)=d(CD-x)$$ $$d=\dfrac{a(BE-x)}{CD-BE}$$ $$a+d=\dfrac{a(CD-BE+BE-x)}{CD-BE}=\dfrac{a(CD-x)}{CD-BE}$$ $$a+2d=\dfrac{a(CD-BE+2BE-2x)}{CD-BE}$$ $$=\dfrac{a(CD+BE-2x)}{CD-BE}$$ $$\dfrac{BE-x}{d}=\dfrac{CD-BE}{a}$$ $$\dfrac{y-x}{a+2d}=\dfrac{(y-x)(CD-BE)}{a(CD+BE-2x)}$$

Then

$$\dfrac{CD-BE}{a}=\dfrac{(y-x)(CD-BE)}{a(CD+BE-2x)}$$ $$CD+BE-2x=y-x$$ $$CD+BE-2x=y+x$$

Area of the trapezoid $BCDE$

$$A(x, y)=\dfrac{x+y}{2}\cdot a \ \ ({units}^2)$$