Problem E and Solution

Stained Glass

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 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 \(\triangle FGJ\).