SOLUTION: find the are of a quadrilateral having points (5,2), (4,3), (2,4), (-8,-1) as consecutive vertices.

Question 1197252: find the are of a quadrilateral having points (5,2), (4,3), (2,4), (-8,-1) as consecutive vertices.
Found 3 solutions by math_tutor2020, ikleyn, Alan3354:
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Plot the given points and form a quadrilateral by joining up adjacent vertices
Eg: (5,2) connects to (4,3)

After doing that, mark lattice points on the boundary and lattice points inside the polygon.
A lattice point has both coordinates as integer values. It's where the dashed grid lines intersect.

Each point marked has integer coordinates x,y
Red points are on the boundaries
Blue points are interior

b = number of boundary lattice points = number of red points = 8
i = number of interior lattice points = number of blue points = 15

Use Pick's Theorem
https://mathworld.wolfram.com/PicksTheorem.html
https://en.wikipedia.org/wiki/Pick%27s_theorem
to get this
A = area of the polygon
A = i + 0.5*b - 1
A = 15 + 0.5*8 - 1
A = 15 + 4 - 1
A = 19 - 1
A = 18

Answer: The area of the quadrilateral is 18 square units

Keep in mind that Pick's Theorem only applies if all vertex or corner points are lattice points.

Another approach you could use is the shoelace formula (see this example problem )

Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.
find the area of a quadrilateral having points (5,2), (4,3), (2,4), (-8,-1) as consecutive vertices.
~~~~~~~~~~~~~~~~

Place your quadrilateral, whose vertices are the lattice points, into the minimal rectangle,
whose boundary lines are parallel to x- and y-axis.

Find the area of this rectangle as the product of its dimensions.

From this area, subtract the areas of all triangles and rectangles (squares) that are inside the rectangle
but outside the quadrilateral.

Calculating these areas is easy: it requires multiplication of integer numbers, only
(sometimes, dividing by 2 is needed).

Doing this way, every 4-th grade student can complete such job on his or her own.

Another way is to plot vertices to make sure that the quadrilateral is convex;
then to draw a diagonal and to calculate the areas of two triangles using the Heron's formula;
finally to add the areas of the triangles.

Third way is to use the "shoelace formula".
On the "shoelace formula" to calculate the area of a polygon see these sources
https://en.wikipedia.org/wiki/Shoelace_formula
https://www.101computing.net/the-shoelace-algorithm/
https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem
https://www.tuitionkenneth.com/a-maths-area-shoelace-method

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
find the are of a quadrilateral having points A(5,2), B(4,3), C(2,4), D(-8,-1) as consecutive vertices.
-------------------

A B C D A 5 4 2 -8 5 2 3 4 -1 2

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Add the diagonal products starting at the upper left:
5*3 + 4*4 + 2*-1 + -8*2 = 15 + 16 - 2 - 16 = 13
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Add the diagonal products starting at the lower left:
2*4 + 3*2 + 4*-8 + -1*5 = 8 + 6 -32 - 5 = -23
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The difference between the sums is 36.
1/2 of 36 is the area
Area = 18 sq units

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