, 2024

To determine the coordinates of $F$, we find the point of intersection of the line through $A$ and $C$ and the line through $B$ and $E$. The line through $A(0,0)$ and $C(12,12)$ has slope $\frac{12-0}{12-0}=1$. Since it passes through $(0,0)$, this line has equation $y=x$. The line through $B(12,0)$ and $E(0,6)$ has slope $\frac{6-0}{0-12}=-\frac12$. Since it passes through $(0,6)$, this line has equation $y=-\frac12x+6$. To determine the $x$-coordinate of the point of intersection, $F$, we solve $x=-\frac12x+6$, which gives $\frac32x=6$ or $3x=12$, and so $x=4$. Since $F$ lies on the line with equation $y=x$, then the coordinates of $F$ are $(4,4)$. Solution 1: Consider $\triangle AEF$ as having base $AE=6$. Then $\triangle AEF$ has height equal to the perpendicular distance from $F$ to $AE$, which is 4, the $x$-coordinate of $F$. The area of $\triangle AEF$ is thus $\frac12\cdot6\cdot4=12$. Solution 2: We can determine the area of $\$\triangle$
AEF$by subtracting the area of$$\triangle$ AFB$from the area of$$\triangle$ AEB$. Consider$$\triangle$ AFB$as having base$AB=12$. Then$$\triangle$ AFB has height equal to the perpendicular distance from

F$to$AB$, which is 4, the$y$-coordinate of$F$. The area of$$\triangle$ AFB$is thus$$\frac12\cdot12\cdot4=24$$. The area of$$\triangle$ AEB$is$$\frac12\cdot12\cdot6=36$$, and so the area of$$\triangle$ AEF$is$36-24=12$. To determine the area of quadrilateral$GDEF, our strategy will be to subtract the area of

$\triangle$ AEF$and the area of$$\triangle$ CDG$from the area of$$\triangle$ ACD$. We need to find the area of$$\triangle$
CDG still, which means finding the coordinates of

G. We can find the coordinates of

G by determining the intersection of the line through

A$and$C with the given circle. Thus, we proceed by finding the equation of the circle. Since the circle has diameter

EB, then its centre is the midpoint of

EB$, which is$$\left(\frac{0+12}{2},\frac{6+0}{2}\right)$$or$(6,3). The diameter has length

$EB=\sqrt{(12-0)^2+(0-6)^2}$$or$$EB=\sqrt{180}$$, which simplifies to$$EB=6\sqrt{5}$. Thus the radius of the circle is

$r=\frac12\cdot6\sqrt{5}=3\sqrt{5}$, and so the circle has equation

$(x-6)^2+(y-3)^2=(3\sqrt{5})^2$$or$(x-6)^2+(y-3)^2=45$. Suppose the$x$-coordinate of$G$is$g$. Since$G lies on the line with equation

y=x$, then the coordinates of$G$are$(g,g)$. The point$G also lies on the circle, and thus the coordinates of

G satisfy the equation of the circle. That is,

(g-6)^2+(g-3)^2=45$, and solving for$g$, we get$$\text{g 2-12g+36+g 2-6g+9 =45 2g 2-18g+45 =45 2g 2-18g =0 2g(g-9) =0}$$and so$g=0$or$g=9$. Since$G$is distinct from$A$, then$g=9$and$G$has coordinates$(9,9). We may now determine the area of

$\triangle$
CDG$. Consider$$\triangle$ CDG$as having base$CD=12$. Then$$\triangle$ CDG has height equal to the perpendicular distance from

G$to$CD$, which is$12-9=3$, since$CD lies along the line

y=12$and the$y$-coordinate of$G$is 9. The area of$$\triangle$ CDG$is thus$$\frac12\cdot12\cdot3=18$$. The area of$$\triangle$ ACD$is half the area of square$ABCD$or$$\frac12\cdot12^2=72$. From part (b), the area of

$\triangle$
AEF$is$12$, and so the area of$GDEF$is$72-18-12=42$.