How do you find the area of a 4 sided polygon?

Question

All four sides are a different length and no two are parallel.

All I know are the lengths of the four sides. Its an odd shape with no parallel lines. It is a lot and I'm trying to figure out the acreage.

The front of the lot is 65' back is 166', left side is 231' and the right side is 122'.

Answer 1

divide your quadilateral into 2 triangles. Use the 1/2*b*h, to find the area of each triangle and add them together. This will work for convex quadilaterals also. For a convex quadilateral, choose the vertex of the "indent" and draw a line to the opposite vertex, this will separate the quadilateral.

Answer 2

i'm going to be incorrect, yet i don't think of that the circumstances of the question are sufficiently defined. If the parabola have been a short elongated one then might the polygon fill a similar proportion as a parabola it fairly is greater just about 1 / 4 circle? additionally, do you recommend any 7-sided polygon or a commonly happening one? If any 7-sided polygon is authorized, then 2 aspects must be the axes from the beginning to the element of intersection. the different 5 aspects might shop on with the parabola as heavily as conceivable. Why a 7-sided polygon and no different? EDIT. i admire the respond 2 decrease than in spite of the incontrovertible fact that it is composed of an blunders which does not ensue to electrify the effect. The factors on the parabola are actually not (a, a^2), (b, b^2), (c, c^2) yet (a, a million - a^2), (b, a million - b^2), (c, a million - c^2). this suggests that the area of the pentagon A = (a million/2)(b - a)(2 - a^2 - b^2) + (a million/2)(c - b)(2 - b^2 - c^2) in spite of the incontrovertible fact that, dA/db = 0 finally ends up in a similar b = (a + c)/2 so the in simple terms real answer is valid. P.S. To make working greater handy you need to invert the diagram and use the curve y = x^2. you will possibly then be attempting to minimise the area decrease than the polygon utilising 4 factors on the curve between (0, 0) and (a million, a million). With factors (0, 0), (a, a^2), (b, b^2), (c, c^2), (d, d^2) and (a million, a million) the area A decrease than the polygon is 2A = a^3 + (a^2 + b^2)(a - b) + (b^2 + c^2)(c - b) + (c^2 + d^2)(d - c) + (a million + d^2)(a million - d) 2dA/da = 2ab - b^2 ----> b = 2a 2dA/db = -2ab + a^2 + 2bc - c^2 Substituting b = 2a finally ends up in c = 3a. persevering with like this might bring about d = 4a and a million = 5a so this confirms answer 2 decrease than.

Answer 3

You can always split a quadrilateral into 4 triangles by connecting the two pairs of points that are opposite each other (I'm assuming that this is a convex quadrilateral). So, in general, split those and find the area of the four traingles. There are formulas for specific types of quadrilaterals, such as squares, rectangles, kites, rhombuses and the like.

Answer 4

Divide the polygon into triangles.
For square and rectangle: (2)(1/2)(b)h). There will always be two (2) large triangles.
For parallelogram: 1/2(b1)(h1) + 1/2(b2)(h2) + 1/2(b3)(h3) + 1/2(b4)(h4). There will always be four (4) triangles.
Please Note:
ADDITIONAL INFORMATION PER
YOUR SPECIFIC MEASUREMENTS:
Measurements: (Going around)
Front = 65'
Right = 122'
Back = 166'
Left = 231'
Draw the polygon using a scale of 1 mm:1 ft
Solution: Combination of Graphical and Pythagorean theorem
From point A,
Draw a horizontal line AB (Front) going to the right.
From point A,
Draw a vertical line AD (Left) going upward.
Front point D,
Draw a downward line DC going towards the right .
Front point C,
Draw a downward line CB going towards the left.
Your polygon is now ABCD.
Given: AB=65'
BC=122' (hypotenuse)
CD=166' (hypotenuse)
AD=231'
Then,
Draw a perpendicular line (towards the left) from point C and intersecting line AD and label this as point E.
Draw another perpendicular line (upward) from point B intersecting line CE and label this as F.
You now have a rectangle ABFE, triangle BCF (small), and triangle CDE (large).
By measurement (So many unknowns to use equations)
AE=110 cm or ft
Therefore,
DE=231-AE=121
BF=AE=110
EF=AB=65
CE=113.64 (Using triangle CDE) - (Base)
CF=CE-EF=48.64 (Base)
Area of Square=AExAB=7150
Area of small triangle=(1/2)(CF)(FB)=5351.5
Area of large triangle=(1/2)(CE)(DE)=6875.2
Total=19,376.7 sq. ft.
What did you come up?
If you use CE=114 by graphical method, you are only off by around 82 sq. ft. (very small error, indeed).

I hope this will help you.
I will fax you my diagram if you need it.

Answer 5

I'm working on it. This will take a minute.

It depends on the information you are given. There are eight pieces of information that you can have. If you only have 4 sides, then you can't solve it.

I'm going to make an equation using four sides and an angle.

Answer 6

Depends on the information you have... you need to know six pieces of information; for instance, the lengths of three of the sides and three of the angles.

I will give the answer in case you know the coordinates of all four points.

Given: four points (x1,y1), (x2,y2), (x3,y3), (x4,y4).

Calculate:
(x1.y2 - x2.y1) + (x2.y3 - x3.y2) +
(x3.y4 - x4.y3) + (x4.y1- x1.y4)

If the answer is negative, drop the minus sign. Divide by two. This is the area.

For instance, for the quadrilateral through points (2,1), (3,6), (8,5), (5,0) we find

... (2 x 6 - 3 x 1) + (3 x 5 - 8 x 6) +
... (8 x 0 - 5 x 5) + (5 x 1 - 2 x 0) =
= 9 + (-33) + (-25) + 5 = -44

so the area is 44/2 = 22.

Answer 7

Depends on the polygon...if it's a square it's the length of the side squared; if it's a rectange it's length x width; if it's a parallelogram it's base x height, etc.

Answer 8

Plain and simple it's base (b) multiplied by height (h). This works for all four sided polygons. Just make sure the height is pendicular to the base. I hope that makes sense.

Answer 9

For all figures with four sides ( quadrilaterals, they are generally known as), the area can be obtained by the following formula.

A = 1/2 d (h1 + h2) Sqare Measures, where

'A' is the Area, 'd' is the diameter, 'h1' & 'h2' are the lengths of the perpendicular distances from the opposite vertices on the diameter.

Suppose the quadrilateral is ABCD, AC is one of the diameters, h1 is the perpendicular distance of AC from 'B' & h2 .
perpendicular distance from D on AC.

With diagrams, it will be easy.

Hope, it's clear.

Answer 10

If it is a regular polygon (square or rectangle) just multiply the length times the height. If not, can you divide it into squares or rectangles, figure out their areas and add them up.