# I have a problem in the chapter, "Perimeter". Find the area of the triangle whose vertices are(-3, -5), (2, -5), and (-3, -1). What does this have to do with perimeter?

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### 2 Answers

Since the problem provides the coordinates of the vertices of triangle, you may also use, as alternative method, the following formula, such that:

`A = (1/2)|[(x_A,y_A,1),(x_B,y_B,1),(x_C,y_C,1)]|`

Substituting the coordinates of the vertices A,B,C in the formula of area, yields:

`A = (1/2)|[(-3,-5,1),(2,-5,1),(-3,-1,1)]|`

`A = (1/2)((-3)*(-5)*1 + 2*1*(-1) + 1*(-3)*(-5) - (-3)*(-5)*1 - 1(-1)*(-3) - 2*(-5)*1)`

`A = (1/2)(-2 + 15 - 3 + 10) = 10`

**Hence, evaluating the area of the given triangle `Delta ABC,` using determinant formula, yields **`A = 10.`

The area could be computed using many formulas. One of them is using the half-perimeter of triangle. The formula is known as Heron's formula.

S = sqrt[p(p-a)(p-b)(p-c)]

p = (a+b+c)/2

a,b,c are the lengths of the sides of the triangle.

a = sqrt[(2+3)^2 + (-5+5)^2]

a = 5

b = sqrt [(-3+3)^2 + (-1+5)^2]

b = 4

c = sqrt[(-3-2)^2 + (-1+5)^2]

c = sqrt41

**p = (9+sqrt41)/2**

p-a = (9+sqrt41)/2 - 5

p-a = (sqrt41 - 1)/2

p-b = (9+sqrt41)/2 - 4

p-b = (sqrt41 + 1)/2

p-c = (9+sqrt41)/2 - sqrt41

p-c = (9-sqrt41)/2

S = sqrt (9+sqrt41)/2*(9-sqrt41)/2* (sqrt41 - 1)/2* (sqrt41 + 1)/2

S = sqrt (81-41)/4* (41-1)/4

S = 40/4

**S= 10 square units**