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

in $\Delta AKF:$

$tan\theta=\frac{FK}{AK}=\frac{EF-KE}{BC+CD+DE}=\frac{y-x}{\alpha+a+\alpha}$

$tan\theta=\frac{y-x}{a+2\alpha}$

$in\ \Delta AJG:$

$tan\theta=\frac{GJ}{AJ}=\frac{GJ}{\alpha+a}$

$GJ=(a+\alpha)tan\theta=\frac{(y-x)(a+\alpha)}{a+2\alpha}$

$DG=GJ+JD=(a+\alpha)tan\theta+x$

$in\ AID$

$tan\theta=\frac{HI}{AI}=\frac{HI}{\alpha}$

$HI=\alpha tan\theta$

$CH=IC+HI=x+\alpha tan\theta$

enclosed area is a shape of trapezoidal AC//CD

$Area=\frac{1}{2}(CH+DG)\times CD$

$=\frac{1}{2}(x+\alpha tan\theta+(a+\alpha)tan\theta+x)\times a$

$=\frac{1}{2}(2x+(a+\alpha)tan\theta)\times a$

$=\frac{1}{2}(2x+(a+2\alpha)\frac{y-x}{(a+2\alpha)})\times a$

$=\frac{1}{2}(2x+y-x)a$

$=\frac{1}{2}(x+y)a$

$A(x,y)=\frac{a}{2}(x+y)$