Fully Worked Out Questions
- Take \(p_1=(1,2),p_2=(3,4),p_3=(2,5)\). Use the shoelace formula to determine the area of the triangle \(\Delta p_1p_2p_3\) and to determine whether or not the triangle is positively or negatively oriented.
Answer
Use the shoelace formula: \[\begin{align*}
\alpha(\triangle p_1p_2p_3)&=\frac{1}{2}\left(1\cdot 4+3\cdot 5+2\cdot 2-2\cdot 3-4\cdot 2-5\cdot 1\right)\\
&=\frac{1}{2}\cdot(4)\\
&=2.
\end{align*}\] So the area is \[\mathcal{A}(\Delta p_1p_2p_3)=|\alpha(\triangle p_1,p_2,p_3)|=|2|=2.\] Because \(\alpha(\triangle p_1p_2p_3)\) is positive, this means that the triangle is positively oriented.
- Take \(p_1=(1,2),p_2=(3,4),p_3=(2,5)\) Use Heron’s formula to determine the area of the triangle \(\Delta p_1p_2p_3\) by first finding the length of each of its sides.
Answer
First find the lengths of the triangle: \[a=\|p_1-p_2\|=\sqrt{8}=2\sqrt{2}\] \[b=\|p_2-p_3\|=\sqrt{2}\] \[c=\|p_3-p_1\|=\sqrt{10}.\] The quantity \(S\) is \[S=\frac{a+b+c}{2}=\frac{3\sqrt{2}+\sqrt{10}}{2}.\] The quantities \(S-a\), \(S-b\), and \(S-c\) are \[S-a=\frac{-\sqrt{2}+\sqrt{10}}{2},\] \[S-b=\frac{\sqrt{2}+\sqrt{10}}{2},\] and \[S-c=\frac{3\sqrt{2}-\sqrt{10}}{2}.\] Use Herron’s formula to get \[\begin{align*}\mathcal{A}(\triangle p_1p_2p_3)&=\sqrt{S(S-a)(S-b)(S-c)}\\
&=\sqrt{\frac{3\sqrt{2}+\sqrt{10}}{2}\cdot \frac{-\sqrt{2}+\sqrt{10}}{2}\cdot \frac{\sqrt{2}+\sqrt{10}}{2}\cdot \frac{3\sqrt{2}-\sqrt{10}}{2}}\\
&=\sqrt{ \left(\frac{-\sqrt{2}+\sqrt{10}}{2}\cdot \frac{\sqrt{2}+\sqrt{10}}{2}\right)\cdot\left( \frac{3\sqrt{2}+\sqrt{10}}{2}\cdot\frac{3\sqrt{2}-\sqrt{10}}{2}\right)}\\
&=\sqrt{\left(\frac{-2+10}{4}\right)\cdot\left(\frac{9\cdot 2-10}{4}\right)}\\
&=\sqrt{2\cdot 2}\\
&=2.
\end{align*}\]
- Take \(\square(1,2)(2,4)(4,3)(3,1)\) to be the parallelogram with the given ordered vertex set. Calculate \(\mathcal{A}(\square (1,2)(2,4)(4,3)(3,1)).\)
Answer
Here are two ways to do it.
Shoelace Formula
One method is to use the shoelace formula: \[\begin{align*}\alpha(\square (1,2)(2,4)(4,3)(3,1))&=\frac{1}{2}\cdot \left(1\cdot 4+2\cdot 3+4\cdot 1+3\cdot 2-2\cdot 2-4\cdot 4-3\cdot 3-1\cdot 1\right)\\
&=\frac{1}{2}(-10)\\
&=-5.
\end{align*}\] So the area is \[\mathcal{A}(\square (1,2)(2,4)(4,3)(3,1))=|\alpha(\square (1,2)(2,4)(4,3)(3,1))|=5.\]
Triangulation
One method is to create a triangulation and find the area of each triangle.
One piece is \(\Delta (1,2)(2,4)(4,3)\) and the other is \(\Delta((1,2)(4,3)(3,1)).\) The area of each triangle is \[\mathcal{A}(\Delta(1,2)(2,4)(4,3))=\frac{5}{2}\quad\text{and}\quad\mathcal{A}(\Delta((1,2)(4,3)(3,1))=\frac{5}{2}.\] So the area of the parallelogram is \[\begin{align*}\mathcal{A}(\square (1,2)(2,4)(4,3)(3,1))&=\mathcal{A}(\Delta(1,2)(2,4)(4,3))+\mathcal{A}(\Delta((1,2)(4,3)(3,1))\\&=\frac{5}{2}+\frac{5}{2}\&=5.\end{align*}\]
Regardless of the method, the area is \(5.\)
- Take \(((-3,1),(-1,4),(1,3),(-4,8))\) to be the ordered vertex set of a polygonal path with epoch \((0,3,8,11,15)\). Identify the given path with a piecewise linear function, \(c\). Simulate a particle whose position at time \(t\) is \(c(t)\).
Answer
Based on the ordered vertex set and epoch, we need the particle to do the following:
- at \(t=0\), it needs to be at \((-3,1)\)
- at \(t=3\), it needs to be at \((-1,4)\)
- at \(t=8\), it needs to be at \((1,3)\)
- at \(t=11\), it needs to be at \((-4,8)\)
- at \(t=15\), it needs to be at \((-3,1).\) The piecewise linear function \(c\) is therefore \[c(t)=\begin{cases}\frac{t}{3}\langle 2,3\rangle
+(-3,1)&\text{if }0\leq t<3\\
\frac{(t-3)}{5}\langle 2,-1\rangle
+(-1,4)&\text{if }3\leq t<8\\
\frac{(t-8)}{3}\langle -5,5\rangle
+(1,3)&\text{if }8\leq t<11\\ \frac{(t-11)}{4}\langle 1,-7\rangle
+(-4,8)&\text{if }11\leq t\leq 15\\
\end{cases}.\]
- Take \(((-3,1),(-1,4),(1,3),(-4,8))\) to be the ordered vertex set of a polygon \(P\). Identify a positively oriented triangulation for \(P\) and identify all of the triangles in this triangulation. Calculate the area of your triangulation.
Answer
Here is one example of a positively oriented triangulation of \(P\). One triangle is \(\Delta (-3,1)(-1,4)(-4,8)\) and the other is \(\Delta (-1,4)(1,3)(-4,8).\) Calculate the area using the shoelace formula to get \[\mathcal{A}(\Delta (-3,1)(-1,4)(-4,8))=\frac{17}{2}\quad\text{and}\quad\mathcal{A}(\Delta (-1,4)(1,3)(-4,8))=\frac{5}{2}.\] The sum of these areas equals the area of \(P\); \[\mathcal{A}(P)=11.\]
- Take \(((-3,1),(-1,4),(1,3),(-4,8))\) to be the ordered vertex set of a polygon \(P\). Identify a negatively oriented triangulation for \(P\) and identify all of the triangles in this triangulation.
Answer
You can take the triangulation used in the previous question and change the orientation. One triangle is \(\Delta (-3,1)(-4,8)(-1,4)\) and the other is \(\Delta (-1,4)(-4,8)(1,3).\) Calculate the area using the shoelace formula to get \[\mathcal{A}(\Delta (-3,1)(-1,4)(-4,8))=\frac{17}{2}\quad\text{and}\quad\mathcal{A}(\Delta (-1,4)(1,3)(-4,8))=\frac{5}{2}.\] The sum of these areas equals the area of \(P\); \[\mathcal{A}(P)=11.\]
- Take \(((-3,1),(-1,4),(1,3),(-4,8))\) to be the ordered vertex set of a polygon \(P\). Use the shoelace formula to determine the area of \(P\) and the orientation of \(\partial_o P\).
Answer
Use the shoelace formula to get \[\begin{align*}\alpha(P)&=\frac{1}{2}\cdot \left(-3\cdot 4+(-1)\cdot 3+1\cdot 8+(-4)\cdot 1-1\cdot (-1)-4\cdot (1)-3\cdot (-4)-8\cdot (-3)\right)\\
&=\frac{1}{2}\cdot(22)\\
&=11.
\end{align*}\] Because \(\alpha(P)\) is positive, \(\partial_oP\) is positively oriented. The area of \(P\) is \(11.\)
Questions
- For each triangle \(\Delta p_1p_2p_3\), use the shoelace formula to describe the area of the triangle and determine whether or not the triangle is positively or negatively oriented.
- \(p_1=(-1,3),p_2=(1,5),p_3=(0,7)\)
- \(p_1=(0,-3),p_2=(3,0),p_3=(-1,3)\)
- \(p_1=(3,5),p_2=(2,5),p_3=(2,9)\)
- For each parallelogram \(\square p_1p_2p_3p_4\), calculate its area.
- \(p_1=(2,4),p_2=(6,7),p_3=(8,12),p_4=(4,9)\)
- \(p_1=(1,-1),p_2=(3,2),p_3=(8,4),p_4=(6,1)\)
- \(p_1=(3,1),p_2=(4,6),p_3=(9,7),p_4=(8,2)\)
- For each ordered vertex set of a polygon path and epoch, identify the given path with a piecewise linear function, \(c\).
- ordered vertex set \(((1,4),(3,6),(2,7),(1,5),(-1,3),(-2,6))\), epoch \((0,2,4,5,7,8,10).\)
- \(((0,0),(2,3),(2,6),(0,6),(-3,6),(-3,0))\), epoch \((0,3,6,8,10,12,15).\)
Answers
- \(\mathcal{A}(\Delta p_1p_2p_3)=3\), positively oriented.
- \(\mathcal{A}(\Delta p_1p_2p_3)=\frac{21}{2}\), positively oriented.
- \(\mathcal{A}(\Delta p_1p_2p_3)=2\), negatively oriented.
- \(\mathcal{A}(\square p_1p_2p_3p_4)=14\)
- \(\mathcal{A}(\square p_1p_2p_3p_4)=11\)
- \(\mathcal{A}(\square p_1p_2p_3p_4)=24\)
- \(c(t)=\begin{cases}\langle 2,-2 \rangle\frac{t}{2}+(1,4) &\text{ if } 0\leq t<2\\ \langle -1,-1 \rangle\frac{t-2}{2}+(3,6) &\text{ if } 2\leq t<4\\ \langle -1,2 \rangle(t-4)+(2,7) &\text{ if } 4\leq t<5\\\langle -2,2 \rangle\frac{t-5}{2}+(1,5) &\text{ if } 5\leq t<7\\ \langle -1,-3 \rangle(t-7)+(-1,3) &\text{ if } 7\leq t<8\\ \langle 2,6 \rangle\frac{t-8}{2}+(-2,6) &\text{ if } 8\leq t\leq 10\\ \end{cases}\)
- \(c(t)=\begin{cases}\langle 2,-3 \rangle\frac{t}{3}+(0,0) &\text{ if } 0\leq t<3\\ \langle 0,-3 \rangle\frac{t-3}{3}+(2,3) &\text{ if } 3\leq t<6\\ \langle -2,0 \rangle\frac{t-6}{2}+(2,6) &\text{ if } 6\leq t<8\\\langle -3,0 \rangle\frac{t-8}{2}+(0,6) &\text{ if } 8\leq t<10\\ \langle 0,6 \rangle\frac{t-10}{2}+(-3,6) &\text{ if } 10\leq t<12\\ \langle 3,0 \rangle\frac{t-12}{3}+(-3,0) &\text{ if } 8\leq t\leq 10\\ \end{cases}\)