Find the area A of polygon CDEFGH with the given vertices. C(0,5), D(2,5), E(2,3), F(3,2), G(-1,2), H(0,3) 
                enot [183]             
         
        
Answer:
<em>The area of the polygon CDEFGH is 7</em>
Step-by-step explanation:
<u>Area of a Polygon</u>
The area of a polygon is generally calculated as the sum of the smaller areas that form its full shape, give each partial area has a known shape, like a square, rectangle, triangle, circle, etc.
The six points given in the question are plotted in the image below. They form a polygon whose area can be divided into two smaller shapes:
The area CDHE is a square of length side 2. Area of a square:

The area HEFG is a trapezoid with bases lengths 4 and 2,  and height 1. Area of a trapezoid:

Calculate both areas:


Total Area=4+3=7
The area of the polygon CDEFGH is 7
 
        
             
        
        
        
Greater than 5.4 i hate the 20 characters rule
        
                    
             
        
        
        
A quadtriatel is said to contain perpendicular diagnols if four go 90 degree angles are formed at the intersection of these lines
        
             
        
        
        
<span>So, what you want is the biggest square that can evenly tile a space 240*300. The biggest possible number that can divide 240 and 300. Does that ring any bells? Maybe you should divide 240 and 300</span>
        
             
        
        
        
The given matrix equation is,
![1.5\left[\begin{array}{cc}x&6\\8&4\end{array}\right] +y\left[\begin{array}{cc}1&4\\3&2\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right]](https://tex.z-dn.net/?f=%201.5%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%266%5C%5C8%264%5Cend%7Barray%7D%5Cright%5D%20%2By%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%264%5C%5C3%262%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20) .
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Multiplying the matrices with the scalars, the given equation becomes,
![\left[\begin{array}{cc}1.5x&9\\12&6\end{array}\right] +\left[\begin{array}{cc}y&4y\\3y&2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right]  \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%269%5C%5C12%266%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dy%264y%5C%5C3y%262y%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20%20)
Adding the matrices,
![\left[\begin{array}{cc}1.5x+y&9+4y\\12+3y&6+2y\end{array}\right]  =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right]  \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%2By%269%2B4y%5C%5C12%2B3y%266%2B2y%5Cend%7Barray%7D%5Cright%5D%20%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20)
Matrix equality gives,

Solving the equations together,

We can see that the equations are not consistent.
There is no solution.