Points \(A(7,11)\), \(B(9,9)\), \(C(6,2)\), and \(D(2,6)\) form quadrilateral \(ABCD\). Determine the area of \(ABCD.\)
Solution 1
Consider points \(E(2,11)\), \(F(9,11)\), \(G(9,2)\), and \(H(2,2).\) Draw in \(EFGH.\)
Since \(E\) and \(F\) both have \(y\)-coordinate \(11\), \(EF\) is a horizontal line which passes through \(A.\)
Since \(F\) and \(G\) both have \(x\)-coordinate \(9\), \(FG\) is a vertical line which passes through \(B.\)
Since \(G\) and \(H\) both have \(y\)-coordinate \(2\), \(GH\) is a horizontal line which passes through \(C.\)
Since \(E\) and \(H\) both have \(x\)-coordinate \(2\), \(EH\) is a vertical line which passes through \(D.\)
Thus, \(EFGH\) is a rectangle that encloses \(ABCD.\) Also, \[\text{area }ABCD = \text{area }EFGH - \text{area }\triangle AED- \text{area }\triangle AFB - \text{area }\triangle BGC - \text{area }\triangle CHD\]
In rectangle \(EFGH\), \(EF=9-2=7\) and \(EH=11-2=9\). The area of rectangle \(EFGH=EF\times EH=7\times 9=63\text{ units}^2.\)
Since \(EFGH\) is a rectangle, \(\triangle AED\) is right-angled at \(E.\) Since \(AE=7-2=5\) and \(ED=11-6=5\), the area of \(\triangle AED =\frac{AE\times ED}{2}=\frac{5\times 5}{2}=\frac{25}{2} = 12.5\text{ units}^2.\)
Since \(EFGH\) is a rectangle, \(\triangle AFB\) is right-angled at \(F.\) Since \(FA=9-7=2\) and \(FB=11-9=2\), the area of \(\triangle AFB =\frac{FA\times FB}{2}=\frac{2\times 2}{2}=\frac{4}{2} = 2\text{ units}^2.\)
Since \(EFGH\) is a rectangle, \(\triangle BGC\) is right-angled at \(G.\) Since \(GC=9-6=3\) and \(BG=9-2=7\), the area of \(\triangle BGC =\frac{GC\times BG}{2}=\frac{3\times 7}{2}=\frac{21}{2} = 10.5\text{ units}^2.\)
Since \(EFGH\) is a rectangle, \(\triangle CHD\) is right-angled at \(H.\) Since \(CH=6-2=4\) and \(DH=6-2=4\), the area of \(\triangle CHD =\frac{CH\times DH}{2}=\frac{4\times 4}{2}=\frac{16}{2}=8\text{ units}^2.\)
Therefore, \[\begin{aligned} \text{area }ABCD &= \text{area }EFGH - \text{area }\triangle AED- \text{area }\triangle AFB - \text{area }\triangle BGC - \text{area }\triangle CHD \\ &=63 -12.5 - 2- 10.5 -8\\ &=30\text{ units}^2 \end{aligned}\]
Solution 2
Consider points \(E(6,10)\), \(F(7,9)\), \(G(6,6)\), and \(H(6,9)\). Draw in \(AF\), \(EC\), \(HB\), and \(DG\).
Since \(A\) and \(F\) both have \(x\)-coordinate \(7\), \(AF\) is a vertical line.
Since \(E\), \(H\), \(G\), and \(C\) have \(x\)-coordinate \(6\), \(EC\) is a vertical line which passes through \(H\) and \(G.\)
Since \(H\), \(F\), and \(B\) have \(y\)-coordinate \(9\), \(HB\) is a horizontal line which passes through \(F.\)
Since \(D\) and \(G\) both have \(y\)-coordinate \(6\), \(DG\) is a horizontal line.
To determine the area of \(ABCD\), we find the areas in the interior shapes, \(\triangle AFB\), \(\triangle BHC\), \(\triangle DGC\), \(\triangle DGE\), and trapezoid \(EAFH\), and calculate their sum.
Since \(AF\) is vertical and \(HB\) is horizontal and passes through \(F\), \(\triangle AFB\) is right-angled at \(F.\) Since \(AF = 11-9=2\) and \(FB = 9-7 = 2\), area \(\triangle AFB=\frac{AF\times FB}{2}= \frac{2\times 2}{2}=\frac{4}{2}=2\text{ units}^2.\)
Since \(HB\) is horizontal and \(EC\) is vertical and passes through \(H\), \(\triangle BHC\) is right-angled at \(H.\) Since \(HB = 9-6=3\) and \(HC = 9-2 = 7\), area \(\triangle BHC=\frac{HB\times HC}{2}= \frac{3\times 7}{2}=\frac{21}{2}=10.5\text{ units}^2.\)
Since \(DG\) is horizontal and \(EC\) is vertical and passes through \(G\), \(\triangle DGC\) is right-angled at \(G.\) Since \(DG = 6-2=4\) and \(GC = 6-2 = 4\), area \(\triangle DGC=\frac{DG\times GC}{2}= \frac{4\times 4}{2}=\frac{16}{2}=8\text{ units}^2.\)
Since \(DG\) is horizontal and \(EC\) is vertical and passes through \(G\), \(\triangle DGE\) is right-angled at \(G.\) Since \(DG = 6-2=4\) and \(EG = 10-6 = 4\), area \(\triangle DGE=\frac{DG\times EG}{2}= \frac{4\times 4}{2}=\frac{16}{2}=8\text{ units}^2.\)
Since trapezoid \(EAFH\) has side lengths \(EH = 10-9 = 1\) and \(AF = 11-9=2\), and height \(HF = 7-6=1\), area trapezoid \(EAFH=\frac{EH + AF}{2} \times HF= \frac{1 + 2}{2} \times 1= \frac{3}{2} \times 1=1.5\text{ units}^2.\)
Therefore, \[\begin{aligned} \text{area } ABCD &= \text{area } \triangle AFB+ \text{area } \triangle BHC + \text{area } \triangle DGC + \text{area } \triangle DGE + \text{area trapezoid } \triangle EAFH \\ &= 2+ 10.5+8+8+1.5\\ &=30\mbox{ units}^2 \end{aligned}\]