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# Find the area of the quadrilateral

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Find the area of the quadrilateral formed with vertices at (1,2) (5,8) (8,6), (4,1)

Dec 3, 2020

#1
+114592
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The easiest  way to solve  this is  with something known as "Pick's Theorem"

Definition :   Lattice point   =   point with  integer  coordinates

Area =

Number of  interior  lattice points in the  figure +  ( number of lattice points on the  border of the  figure ) / 2    -  1

So we  have

21   +  5/2  -  1   =

21  +  2.5  - 1  =

22.5    = Area

See the following  graph :

Dec 3, 2020
#2
+10843
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Find the area of the quadrilateral formed with vertices at (1,2) (5,8) (8,6), (4,1)

Hello Guest!

$$m_1=\frac{8-2}{5-1}=\frac{3}{2}\\ m_2=\frac{6-8}{8-5}=-\frac{2}{3}\\ m_3=\frac{1-6}{4-8}=\frac{5}{4}\\ m_4=\frac{2-1}{1-4}=-\frac{1}{3}$$

$$f(x)=m(x-x_P)+y_P$$

$$f_1(x)=\frac{3}{2}(x-1)+2=\frac{2}{3}x-\frac{3}{2}+2\\ f_1(x)=\frac{2}{3}x-\frac{1}{2}\\ f_2(x)=-\frac{2}{3}(x-5)+8=-\frac{2}{3}x+\frac{10}{3}+8\\ f_2(x)= -\frac{2}{3}x+\frac{34}{3}$$

$$f_3(x)=\frac{5}{4}(x-8)+6=\frac{5}{4}x-10+6\\ f_3(x)=\frac{5}{4}x-4\\ f_4(x)=-\frac{1}{3}(x-4)+1=-\frac{1}{3}x+\frac{4}{3}+1\\ f_4(x)=-\frac{1}{3}x+\frac{7}{3}$$

$$A=\int_{a}^{b} \! f_1(x) \, dx+\int_{a}^{b} \! f_2(x) \, dx -\int_{a}^{b} \! f_3(x) \, dx -\int_{a}^{b} \! f_4(x) \, dx$$

$$\int_{1}^{5} \! f_1(x) \, dx=\int_{1}^{5} \! (\frac{2}{3}x-\frac{1}{2}) \, dx=\left[\frac{2}{3}\cdot \frac{x^2}{2}-\frac{1}{2}x\right]_{5}^{1} \\ =\frac{1}{3}\cdot 5^2-\frac{5}{2}$$

$$\int_{5}^{8} \! f_2(x) \, dx=\int_{5}^{8} \! ( -\frac{2}{3}x+\frac{34}{3} ) \, dx=\left[-\frac{2}{3}\cdot \frac{x^2}{2} +\frac{34}{3}x\right]_{5}^{8}$$

$$\int_{4}^{8} \! f_3(x) \, dx=\int_{1}^{5} \! (\frac{5}{4}x-4) \, dx=\left[ \frac{5}{4}\cdot \frac{x^2}{2}-4x\right]_{4}^{8}$$

$$\int_{1}^{5} \! f_1(x) \, dx=\int_{1}^{5} \! (\frac{2}{3}x-\frac{1}{2}) \, dx=\left[\frac{2}{3}\cdot \frac{x^2}{2}-\frac{1}{2}x\right]_{5}^{1}$$

and so on

to be continued

!

Dec 3, 2020
edited by asinus  Dec 3, 2020
edited by asinus  Dec 3, 2020
edited by asinus  Dec 3, 2020