In square \(ABCD\), points \(E\) and \(F\) lie on \(AB\) such that \(AE=EF=FB=10\). Similarly, points \(G\) and \(H\) lie on \(BC\) such that \(BG=GH=HC=10\). Let \(J\) be the intersection of line segments \(DH\) and \(CF\). The areas of \(\triangle DFJ\) and \(\triangle CJH\) are then shaded.
Determine the fraction of the area of square \(ABCD\) that is shaded.
Solution 1
In this solution, we set square \(ABCD\) on the Cartesian grid with vertex \(D\) at the origin. Then, since \(AE=EF=FB=10\), it follows that \(AB=30\), so the side length of \(ABCD\) is \(30\). Thus, the coordinates of the vertices are \(A(0,30)\), \(B(30,30)\), \(C(30,0)\), and \(D(0,0)\). Since \(AF=20\) and \(AF\) is parallel to the \(x\)-axis, it follows that the coordinates of \(F\) are \((20,30)\). Since \(HC=10\) and \(HC\) is parallel to the \(y\)-axis, it follows that the coordinates of \(H\) are \((30,10)\).
Now we will find the coordinates of \(J\) by finding the point of intersection of the lines through \(DH\) and \(CF\).
The line through \(DH\) has slope \(\frac{10}{30}=\frac{1}{3}\) and \(y\)-intercept \(0\). Thus, its equation is \(y=\frac{1}{3}x\).
The line through \(CF\) has slope \(\frac{0-30}{30-20}=\frac{-30}{10}=-3\). To determine its \(y\)-intercept we substitute \((30,0)\) into \(y=-3x+b\). This gives \(0=-3(30)+b\), so \(b=90\). Thus, its equation is \(y=-3x+90\).
We set these two equations equal to each other to solve for the point of intersection. \[\begin{aligned} \frac{1}{3}x &= -3x+90\\ \frac{10}{3}x &= 90\\ x &= 90 \times \frac{3}{10}=27 \end{aligned}\] Thus, \(y=\frac{1}{3}(27)=9\), so the coordinates of \(J\) are \((27,9)\).
Since the slope of the line through \(DH\) is \(\frac{1}{3}\) and the slope of the line through \(CF\) is \(-3\), and these are negative reciprocals, it follows that these lines are perpendicular. Thus, both \(\triangle DFJ\) and \(\triangle CJH\) are right-angled triangles. So to determine the area of \(\triangle DFJ\) we will consider the base to be \(DJ\) and the height to be \(JF\). Using the Pythagorean theorem,
\[\begin{aligned} DJ &= \sqrt{27^2 + 9^2}\\ &= \sqrt{810}\\ &= 9 \sqrt{10} \end{aligned}\]
\[\begin{aligned} JF &= \sqrt{(27-20)^2 + (9-30)^2}\\ &= \sqrt{7^2+(-21)^2}\\ &=\sqrt{490}=7 \sqrt{10} \end{aligned}\]
Thus, \[\text{Area }\triangle{DFJ} = \frac{1}{2} \times DJ \times JF =\frac{1}{2} \times 9 \sqrt{10} \times 7 \sqrt{10} = \frac{1}{2} \times 63 \times 10 = 315\] To determine the area of \(\triangle CJH\) we will consider the base to be \(CH=10\). To find the height, we drop a perpendicular from \(J\) to \(CH\) and call this point \(K\). Then the height is \(JK=30-27=3\). Thus, \[\text{Area }\triangle CJH = \frac{1}{2} \times CH \times JK= \frac{1}{2} \times 10 \times 3 = 15\] Then the total area shaded is \(315+15=330\).
The area of square \(ABCD\) is \(30 \times 30 = 900\). Therefore, the fraction of this area that is shaded is \(\frac{330}{900} = \frac{11}{30}\).
Solution 2
In this solution we use similar triangles to calculate the area indirectly.
Since \(AE=EF=FB=10\), and \(ABCD\) is a square, it follows that \(AB=BC=CD=AD=30\). Since \(AE=EF=10\) it follows that \(AF=20\).
We can then determine the areas of right-angled triangles \(\triangle DAF\) and \(\triangle CBF\). The area of \(\triangle DAF\) is \(\frac{1}{2} \times 30 \times 20 = 300\) and the area of \(\triangle CBF\) is \(\frac{1}{2} \times 30 \times 10 = 150\).
Next we determine the area of \(\triangle CJH\). First we notice that \(\triangle CBF\) and \(\triangle DCH\) are congruent, and \(\triangle DCH\) has been rotated by \(90\degree\). Thus, \(DH \perp CF\). Next we notice that \(\angle HCJ = \angle BCF\) and \(\angle CJH = \angle CBF = 90\degree\). Thus, \(\triangle CJH \sim \triangle CBF\).
Using the Pythagorean theorem, \(CF = \sqrt{30^2+10^2}=\sqrt{1000}=10 \sqrt{10}\). Since the corresponding side in \(\triangle CJH\) is \(CH=10\), it follows that the side lengths in \(\triangle CBF\) are \(\sqrt{10}\) times the length of their corresponding sides in \(\triangle CJH\). Thus, the area of \(\triangle CBF\) is \(\sqrt{10} \times \sqrt{10} = 10\) times the area of \(\triangle CJH\). Since the area of \(\triangle CBF\) is \(150\), it follows that the area of \(\triangle CJH\) is \(15\).
Finally, we can determine the total area shaded. \[\begin{aligned} \text{Shaded area} &= \text{Area }ABCD - \text{Area }\triangle DAF - 2 \times \text{Area }\triangle CBF + 2 \times \text{Area }\triangle CJH\\ &= 30 \times 30 - 300 - 2\times 150 + 2 \times 15\\ &= 330 \end{aligned}\] Therefore, the fraction of the area of square \(ABCD\) that is shaded is \(\frac{330}{900} = \frac{11}{30}\).