Find the area of a triangle whose vertices are A(-3,-1), B(5,3), and C(2,-8).?

Question

thanks

Answer 1

Well, there is a special formula called the shoelace formula for solving area problems on the coordinate plane, and it's very slick: It is very hard to explain, so try using this site:

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf

Well, if I set it up like this

(-3,-1)
(5,3)
(2,-8)
(-3,-1)

my products on the rightwill be -9-40-2=-51
and on the left -5+6+24=25.

The positive difference is 25-(-51)=76.

76/2=38 square units.

this method/algorithim works for any polygon on the coordinate plane.

Alternative methods for this problem could have used hero(n)'s formula, and lots of algebra to use bh/2.

Answer 2

The area can be found of any polygon, given the coordinates of the vertices. In your case you have a triangle.
First graph the three vertices, points A, B and C. and the x and y axes. DO NOT GO TO THE TROUBLE OF FINDING THE LENGTH OF THE SIDES OF THE TRIANGLE. Instead draw the rectangle that contains triangle ABC, with horizontal lines through B and C and a vertical line through A and B. The base of that rectangle is 9 and the height is 11. You can find these numbers by simply counting squares. To find the area of triangle ABC, find the area of that rectangle, 99. and SUBTRACT the right triangles not needed, the base and height of each can be found by counting squares. The area of these right triangles is one half the product of the legs, 18, 11 and 24.5. The area of triangle ABC is 99-(18+11+24.5).
With a little ingenuity the solution of this interesting problem can be extended to any polygon given the coordinates of the vertices.
Peter m

Answer 3

Here's a trick for finding the area of ANY polygon.

1. Write the points in a column.
2. Copy the first point to the bottom of the column.
3. Multiply diagonally in both directions. In other words...
- multiply Ay * Bx and write the result to the left of Cx
- multiply By * Cx and write that result below the first
- multiply Cy * Ax and write that result below the second
- multiply Ax * By and write the result to the right of Cy
- multiply Bx * Cy and write that result below the previous one.
- multiply Cx * Ay and write that result below the previous one.
4. Add up each of the two columns of products.
5. Subtract the the sums and divide by 2, and the absolute value of that is the area of the polygon.

(The following digram uses periods to make the spacing come out almost right. Copy and paste it into something like Notepad for a better looking version.)

... -3, -1
... +5. +3
.-5 +2, -8 -9
.+6 -3, -1 -40
+24 ...... -2
--- ...... ---
+25 ...... -51
.......... -25
.......... ---
.......... -76 / 2 = -38
Area of the triangle is 38.

Exercise for the interested reader:
Find the area of the front of my house. It looks like a polygon from the front: (0,0), (10, 0), (10, 10), (9, 11), (9, 16), (7, 16), (7, 13), (5, 15), (0, 10).

Answer 4

2A=det
-3 -1 1
5 3 1
2 -8 1

=(-40-6)-(24+2)+(-9+5)
=(-46)-26-4
=-72-4=-76

Area can not be negative

Area=76/2=38

Answer 5

web site
http://www.mste.uiuc.edu/dildine/heron/triarea.html

use the formula
Area=SQRT(s(s-a)(s-b)(s-c)),
where s=(a+b+c)/2 or perimeter/2.

Given A(-3,-1), B(5,3), C(2,-8)
a = BC = sqrt( (3+8)^2 + (5 - 2)^2)
= sqrt( 121 + 9) = sqrt(130) = 11.4
b = AC = sqrt( (-1+8)^2 + (-3-2)^2)
= sqrt( 49 + 25) = sqrt (74) = 8.6
c = AB = sqrt ( (3+1)^2 +(5+3)^2)
= sqrt( 16 + 64) = sqrt (80) = 8.9

s = (a+b+c)/2 = (11.4+8.6+8.9)/2 = 28.9/2 = 14.45
Area=SQRT(s(s-a)(s-b)(s-c))
= sqrt( (14.15)(3.05)(5.85)(5.55) )
= sqrt(1430.9) = 37.83 - - - - Ans

Answer 6

AB = SQRT[ (-3-5)^2 + (1-3)^2 ] = 8.246211251
BC = SQRT[ (5-2)^2 + (3--8)^2 ] = 11.40175425
AC = SQRT[ (-3-2)^2 + (1--8)^2 ] = 10.29563014

S = Perimeter/2 = (8.246211+11.40175+10.29563)/2 = 14.97179782

Area = SQRT[ S*(S-A)*(S-B)*(S-C) ] = 41

Answer 7

You might want to checkout my area calculater tool on math tool section of my website.

Just use lenght AB,AC,BC as input where
AB=sqrt( (yb-ya)^2+(xb-xa)^2) , ...........

http://www.mycoolaids.com/MathTools.aspx