Question

Triangle ABC has vertices at A(-2,-3), B(6,-3), C (-1,5). Answer the following and round your answers to

the nearest hundredth if necessary.
a. Find the perimeter of triangle ABC.
b. Find the area of triangle ABC.
a. Perimeter:
b. Area:

Answer 1

a) Perimeter
`= {\bf 8} + \sqrt {\bf 113} + \sqrt {\bf 65}`

b)Area
`={\bf 96}`

Explanation:

Given ABC is a triangle with vertices at A(-2,-3), B(6,-3) and C(-1,5)

The vertices A(-2,-3), B(6,-3) and C(-1,5) are represented by
`(x_A,y_A) ,(x_B,y_B), (x_C,y_C)` respectively

Now find the perimeter of the triangle ABC

The perimeter is found by first finding the three distances between the three vertices
`d_AB, d_BC\ and\ d_CA` given by

`d_(AB) = \sqrt {(x_A - x_B)^2 + (y_A - y_B)^2)}`

`d_(BC) = \sqrt {(x_B - x_C)^2 + (y_B - y_C)^2)}`

`d_(CA)= \sqrt {(x_C - x_A)^2 + (x_C - y_A)^2}`

The perimeter is given by

Perimeter
`=d_(AB) + d_(BC) + d_(CA)`

now find
`d_(AB) = \sqrt {(x_A - x_B)^2 + (y_A - y_B)^2)}`

`d_(AB)= \sqrt {(-2 - 6)^2 + (-3+3 )^2)}`

`d_(AB) = \sqrt {(-8)^2 + (0)^2)}`

`d_(AB) = \sqrt {8^2}`

`d_(AB)= \sqrt {64}`

`d_(AB) = 8`

Similarly we find
`d_(BC) = \sqrt {(x_B - x_C)^2 + (y_B - y_C)^2)}`

`d_(BC)= \sqrt {(6+1)^2 + (-3-5)^2)}`

`d_(BC) = \sqrt {(7)^2 + (-8)^2)}`

`d_(BC)= \sqrt {49 + 64}`

`d_(BC) = \sqrt {113}`

find
`d_(CA) = \sqrt {(x_C - x_A)^2 + (x_C - y_A)^2}`

`d_(CA) = \sqrt {(-1 +2)^2 + (5+3)^2}`

`d_(CA) = \sqrt {(1)^2 + (8)^2}`

`d_(CA) = \sqrt {1 + 64}`

`d_(CA) = \sqrt {65}`

Now adding the distances we get

Perimeter
`=d_(AB)+ d_(BC) + d_(CA)`

Perimeter
`= 8+ \sqrt {113} + \sqrt {65}`

b) Area of the given triangle ABC

The formula for the area of the triangle defined by the three vertices A, B and C is given by:

`Area= (1)/(2) {\det {\left[\begin{array}{ccc}x_A&x_B&x_C\\y_A&y_B&y_C\\1&1&1\end{array}\right]}}`

where det is the determinant of the three by three matrix.

`Area=(1)/(2){{\det \left[\begin{array}{ccc}-2&6&-1\\ -3& -3&5\\ 1 & 1 & 1\end{array}\right]}}`

`Area=(1)/(2)[-2(-3-5)-6(-3-5)-1(-3+3)+3(6+1)-3(-2+1)-5(-2-6)+1(30-3)-1(-10-3)+1(6+18)]`

`Area=(1)/(2)[-2(-8)-6(-8)-1(0)+3(7)-3(-1)-5(-8)+1(27)-1(-13)+1(24)]`

`Area=(1)/(2)[16+48+0+21+3+40+27+13+24]`

`Area=(1)/(2) (192)`

`Area=96`