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Answer to Question #12182 in Algebra for ann

Question #12182
Find the perimeter of a rectangle if three of its vertices are (5,-2), (-3, -2), and (-3, 3).
b) Find the area of a triangle whose vertices have coordinates (0, 9), (0, -4), and (5, -4).

Describe what you did to solve each problem. Looking at the ordered pairs, did you see a correlation of the y-coordinate? Could you have solved this without graphing? If so, how?

Expert's answer
a) Let's mark the given points by letters A, B and C so that their coordinates will be (5,-2), (-3, -2), and (-3, 3) respectfully. Let's find the coordinates of missing point D. It's coordinates are (5, 3). We can find them without graphing by looking through no-repeated coordinates of the given points. A and C are opposite vertices. So,

T = 2AB + 2AD = 2(5-(-3)) + 2(3-(-2)) = 26.

b) Let's apply the Heron's formula:

T = √[p(p-a)(p-b)(p-c)],

where

p = [a+b+c]/2

and a, b and c are the lenghts of the sides of the triangle.

a = AB = √[(5-(-3))²+(-2-(-2))²] = 8;
b = BC = √[(-3-(-3))²+(-2-3)²] = 5;
c = CA = √[(-3-5)²+(3-(-2))²] = 9;
p = [8+5+9]/2 = 11;

and so,

T = √[11(11-8)(11-5)(11-9)] = √396.

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