# Problem of the Week Problem E and Solution Stained Glass

## Problem

A stained glass window hanging is in the shape of a rectangle with a length of $$8$$ cm and a width of $$6$$ cm.

Rectangle $$ABCD$$ represents the window hanging with $$AB=8$$ and $$BC=6$$. The points $$E$$, $$F$$,$$G$$, and $$H$$ are the midpoints of sides $$AB$$, $$BC$$, $$CD$$, and $$AD$$, respectively. The point $$J$$ is the midpoint of line segment $$EH$$. Triangle $$FGJ$$ is coloured blue.

Determine the area of the blue triangle.

## Solution

Solution 1

Since $$ABCD$$ is a rectangle and $$AB=8$$, it follows that $$AE=EB=DG=GC=4$$. Similarly, since $$BC=6$$, it follows that $$BF=FC=AH=HD=3$$.

Consider the four corner triangles, $$\triangle HAE$$, $$\triangle EBF$$, $$\triangle FCG$$, and $$\triangle GDH$$.

Each of these triangles is a right-angled triangle with base $$4$$ and height $$3$$. Therefore, the total area of these four triangles is equal to $$4\times \frac{4 \times 3}{2}=24$$.

The length of the hypotenuse of each of the four corner triangles is equal to $$\sqrt{3^2 + 4^2}=5$$. Thus, $$EF=FG=GH=EH=5$$, so $$EFGH$$ is a rhombus. Thus $$EH \parallel FG$$. The area of rhombus $$EFGH$$ is equal to the area of rectangle $$ABCD$$ minus the area of the four corner triangles. Thus, the area of rhombus $$EFGH$$ is $$8 \times 6 - 24 = 24$$.

Let $$h$$ be the perpendicular distance between $$FG$$ and $$EH$$. Then the area of rhombus $$EFGH$$ is $$h \times FG$$. Thus, $$h \times 5 = 24$$.

Triangle $$FGJ$$ has base $$FG$$ and height $$h$$, so its area is equal to $$\frac{h \times FG}{2} = \frac{h \times 5}{2} = \frac{24}{2}=12 \text{ cm}^2$$.

Solution 2

In this solution we will use analytic geometry and set the coordinates of $$D$$ to $$(0,0)$$. Then $$A(0,6)$$, $$B(8,6)$$, and $$C(8,0)$$ are the other corners of the rectangle. The midpoints $$E$$, $$F$$, $$G$$, and $$H$$ have coordinates $$(4,6)$$, $$(8,3)$$, $$(4,0)$$, and $$(0,3)$$, respectively. Then $$J$$ has coordinates $$(2,4.5)$$. Let $$K$$ have coordinates $$(8,4.5)$$, and $$L$$ have coordinates $$(2,0)$$. Then $$JKCL$$ is a rectangle.

We can then calculate the area of $$\triangle FGJ$$ as follows. \begin{aligned} \text{Area }\triangle FGJ &= \text{Area }JKCL - \text{Area }\triangle JKF - \text{Area }\triangle FCG - \text{Area }\triangle GLJ\\ &= JK \times CK - \frac{JK \times KF}{2} - \frac{FC \times CG}{2} - \frac{GL \times LJ}{2}\\ &= 6 \times 4.5 - \frac{6 \times 1.5}{2} - \frac{3 \times 4}{2} - \frac{2 \times 4.5}{2}\\ &= 27 - 4.5 - 6 - 4.5\\ &= 12 \end{aligned} Therefore, the area of $$\triangle FGJ$$ is equal to $$12~\text{cm}^2$$.

Solution 3

This solution also uses analytic geometry. As in Solution 2, set the coordinates of $$D$$ to $$(0,0)$$. Then $$A(0,6)$$, $$B(8,6)$$, and $$C(8,0)$$ are the other corners of the rectangle. The midpoints $$E$$, $$F$$, $$G$$, and $$H$$ have coordinates $$(4,6)$$, $$(8,3)$$, $$(4,0)$$, and $$(0,3)$$, respectively. Then $$J$$ has coordinates $$(2,4.5)$$.

The base of $$\triangle FGJ$$ is equal to the length of $$FG$$. Since $$\triangle FCG$$ is a right-angled triangle, $$FG=\sqrt{CF^2+CG^2}=\sqrt{3^2+4^2}=5$$. Line segments $$EH$$ and $$FG$$ each have a slope of $$\frac{3}{4}$$, so it follows that they are parallel. Thus, the height of $$\triangle FGJ$$ is equal to the perpendicular distance between $$EH$$ and $$FG$$.

The line passing through $$F$$ and $$G$$ has slope $$\frac{3}{4}$$. The line perpendicular to $$FG$$, passing through $$G$$ has slope $$-\frac{4}{3}$$ and $$y$$-intercept $$\frac{16}{3}$$. Therefore, its equation is $$y=-\frac{4}{3}x+\frac{16}{3}$$.

The line passing through $$EH$$ has slope $$\frac{3}{4}$$ and $$y$$-intercept $$3$$. Therefore, its equation is $$y=\frac{3}{4}x+3$$. We can then determine the point of intersection of $$y=\frac{3}{4}x+3$$ and $$y=-\frac{4}{3}x+\frac{16}{3}$$ by setting $$\frac{3}{4}x+3 = -\frac{4}{3}x+\frac{16}{3}$$.

We multiply both sides of this equation by $$12$$ and solve for $$x$$: \begin{aligned} 9x + 36 &= -16x + 64\\ 25x &= 28\\ x &= \frac{28}{25} \end{aligned} The $$y$$-coordinate for this intersection point is then $$y=\frac{3}{4}\left( \frac{28}{25}\right) + 3 = \frac{21}{25}+3 = \frac{96}{25}$$.

Then, the height of $$\triangle FGJ$$ is equal to the distance between $$\left(\frac{28}{25},\frac{96}{25}\right)$$ and $$G(4,0)$$, which is $\sqrt{\left(\frac{96}{25}-0\right)^2 + \left(\frac{28}{25}-4\right)^2} =\sqrt{\frac{9216}{625} + \frac{5184}{625}} = \sqrt{\frac{14\,400}{625}} = \sqrt{\frac{576}{25}} =\frac{24}{5}$ Therefore, the area of $$\triangle FGJ$$ is equal to $$\frac{1}{2} \times 5 \times \frac{24}{5} = 12 \text{ cm}^2$$.

Extension: Suppose $$AB=p$$ and $$BC=q$$, for some real numbers $$p$$ and $$q$$. Determine the area of $$\triangle FGJ$$.