# SOLUTION: I don't know how to do this. Directions: The vertices of an irregular figure are given. Find the area of the figure. T(-4,-2), U(-2,2), V(3,4), W(3,-2)

Algebra ->  Algebra  -> Surface-area -> SOLUTION: I don't know how to do this. Directions: The vertices of an irregular figure are given. Find the area of the figure. T(-4,-2), U(-2,2), V(3,4), W(3,-2)       Log On

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 Geometry: Area and Surface Area Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Surface-area Question 274213: I don't know how to do this. Directions: The vertices of an irregular figure are given. Find the area of the figure. T(-4,-2), U(-2,2), V(3,4), W(3,-2) Answer by Alan3354(30993)   (Show Source): You can put this solution on YOUR website!The vertices of an irregular figure are given. Find the area of the figure. T(-4,-2), U(-2,2), V(3,4), W(3,-2) -------------------------------------- There's more than one way. The most obvious is to make it 2 triangles, solve for the lengths and find the areas. A better way is: ....T. .U. .V. .W ..T x -4 -2 +3 +3 -4 y -2 +2 +4 -2 -2 ------------------- Multiply the elements with a slope of -1, ie, Tx*Uy, Ux*Vy, etc and add them Multiply the elements with a slope of +1, ie, Ty*Ux, Uy*Vx, etc and add them The absolute value of the difference is 2 x the area, divide by 2. --------------------- (-4*2) + (-2*4) + (3*-2) + (3*-2) = -8 - 8 - 6 - 6 = -28 --------- ((-2*-2) + (2*3) + (4*3) + (-2*-4) = 4 + 6 + 12 + 8 = 30 Difference = 58 Area = 29 sq units ------------ This method is good for any number of points. Note that the 1st and last points are the same.