SOLUTION: Find the area of the polygon whose vertices are at (1,-4),(4,-1),(4,5),(-1,4)&(-2,-1).

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Question 166152: Find the area of the polygon whose vertices are at (1,-4),(4,-1),(4,5),(-1,4)&(-2,-1).
Answer by Fombitz(13823) About Me  (Show Source):
You can put this solution on YOUR website!
First plot the points and see what you've got.
drawing%28+300%2C+300%2C+-6%2C+6%2C+-6%2C+6%2Cgrid%28+1+%29%2C%0D%0Acircle%28+1%2C+-4%2C+.2+%29%2C%0D%0Acircle%28+4%2C+-1%2C+.2+%29%2C%0D%0Acircle%28+4%2C+5%2C+.2+%29%2C%0D%0Acircle%28+-1%2C+4%2C+.2+%29%2C%0D%0Acircle%28+-2%2C+-1%2C+.2+%29%2C%0D%0Aline%281%2C-4%2C4%2C-1%29%2C%0D%0Aline%284%2C-1%2C4%2C5%29%2C%0D%0Aline%284%2C5%2C-1%2C4%29%2C%0D%0Aline%28-1%2C4%2C-2%2C-1%29%2C%0D%0Aline%28-2%2C-1%2C1%2C-4%29%2C%0D%0Alocate%281.1%2C-4.2%2CA%29%2C%0D%0Alocate%284.1%2C-1.2%2CB%29%2C%0D%0Alocate%284.2%2C5.2%2CC%29%2C%0D%0Alocate%28-1.5%2C4.2%2CD%29%2C%0D%0Alocate%28-2.5%2C-1.2%2CE%29%0D%0A%29
Define a couple of extra points that will help simplify the solution.
X at (4,4)
Y at (-1,-1)
drawing%28+300%2C+300%2C+-6%2C+6%2C+-6%2C+6%2Cgrid%28+1+%29%2C%0D%0Acircle%28+1%2C+-4%2C+.2+%29%2C%0D%0Acircle%28+4%2C+-1%2C+.2+%29%2C%0D%0Acircle%28+4%2C+5%2C+.2+%29%2C%0D%0Acircle%28+-1%2C+4%2C+.2+%29%2C%0D%0Acircle%28+-2%2C+-1%2C+.2+%29%2C%0D%0Acircle%284%2C4%2C.2%29%2C%0D%0Acircle%28-1%2C-1%2C.2%29%2C%0D%0Agreen%28line%284%2C4%2C-1%2C4%29%29%2C%0D%0Agreen%28line%28-2%2C-1%2C4%2C-1%29%29%2C%0D%0Agreen%28line%28-1%2C-1%2C-1%2C4%29%29%2C%0D%0Agreen%28line%284%2C-1%2C4%2C4%29%29%2C%0D%0Alocate%28-1.2%2C-1.2%2CY%29%2C%0D%0Alocate%284.2%2C4%2CX%29%2C%0D%0Aline%281%2C-4%2C4%2C-1%29%2C%0D%0Aline%284%2C4%2C4%2C5%29%2C%0D%0Aline%284%2C5%2C-1%2C4%29%2C%0D%0Aline%28-1%2C4%2C-2%2C-1%29%2C%0D%0Aline%28-2%2C-1%2C1%2C-4%29%2C%0D%0Alocate%281.1%2C-4.2%2CA%29%2C%0D%0Alocate%284.1%2C-1.2%2CB%29%2C%0D%0Alocate%284.2%2C5.2%2CC%29%2C%0D%0Alocate%28-1.5%2C4.2%2CD%29%2C%0D%0Alocate%28-2.5%2C-1.2%2CE%29%0D%0A%29
Now we can find the areas of the individual triangles and square and add them together.
Area of a square is s*s where s is the side length.
Area of a triagle is half the product of base and height.
.
.
.
Square BXDY : A=s%5E2=5%5E2=25
Triangle XCD :A=%281%2F2%29bh=%281%2F2%29%285%29%281%29=5%2F2
Triangle DEY :A=%281%2F2%29bh=%281%2F2%29%281%29%285%29=5%2F2
Triangle EAB :A=%281%2F2%29bh=%281%2F2%29%286%29%283%29=9
Now add all of the areas,
A%5Bp%5D=25%2B5%2F2%2B5%2F2%2B9
A%5Bp%5D=39