Question

Find the area of a polygon with the given vertices? (2,5) (7,1) (3,-4) (-2,3) Please show workprocess

Answer 1

`color(brown)(Area " " ABCD = 20.48 + 19 = 39.48 " sq. units"`

Explanation:

`A (2,5), B (7,1), C (3, -4), D(-2,3)`

Using distance formula,

`bar (AB) = sqrt((7-2)^2 + (1-5)^2) ~~ 6.4`

`bar (BC) = sqrt((3-7)^2 + (-4-1)^2) ~~ 6.4`

`bar (CD) = sqrt((-2-3)^2 + (3+4)^2) ~~ 8.6`

`bar (AD) = sqrt((-2-2)^2 + (3-5)^2) ~~ 4.47`

`bar (AC) = sqrt((3-2)^2 + (-4-5)^2) ~~ 9.06`

`"Area of " Delta " ABC " = sqrt(s * (s - a) * (s - b) * s - c)), " where " s = (a + b + c) / 2`

`s = (6.4 + 6.4 + 9.06) / 2 = 10.93`

`Delta_(ABC) = sqrt(10.93 * (10.93 - 6.4) * (10.93 - 6.4) * (10.93 - 9.06)) ~~ 20.48`

`"Area of " Delta " ADC " = sqrt(s * (s - a) * (s - d) * s - c)), " where " s = (a + d + c) / 2`

`s = (4.47 + 8.6 + 9.06) / 2 = 11.065`

`Delta_(ADC) = sqrt(11.065 * (11.065 - 4.47) * (11.065 - 8.6) * (11.065 - 9.06)) ~~ 19`

`"Area of quadrilateral " ABCD = " Area of " Delta_(ABC) + " Area of " Delta_(ADC)`

`color(brown)(Area " " ABCD = 20.48 + 19 = 39.48 " sq. units"`

Answer 2

`1/2 | 2(1)-5(7) quad + 7(-4)-1(3) quad + 3(3)-(-4)(-2) quad + -2(5) - 3(2) |`

`= 79/2`

Explanation:

I like looking through the featured answers that are more complicated than they need to be and redoing them.

The area of any polygon whose vertices are given by a list of 2D coordinates is given by the Shoelace Theorem. It says the area is half the absolute value of the sum of cross products for each side, order preserved.

The example illustrates it well. To keep track we list the vertices on top of a shifted copy:

(2,5) (7,1) (3,-4) (-2,3)
(7,1) (3,-4) (-2,3) (2,5)

Then the area is

`A = 1/2 | 2(1)-5(7) quad + 7(-4)-1(3) quad + 3(3)-(-4)(-2) quad + -2(5) - 3(2) | = 79/2`

Done exactly, with very little effort. Feature that.