Find the perimeter & area of triangle ABC, whose vertices are A(-4,-2), B(8,-2) & C(2,6). Find the length of the perpendicular from A to BC.

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A(-4,-2)  B(8,-2)  C(2,6)

let us calculate the length of the sides:

AB = sqrt[(8--4)^2 + (-2--2)^2]= sqrt(12^2)= 12

AC= sqrt[(2--4)^2 + (6--2)^2]= sqrt(100) = 10

BC= sqrt[(2-8)^2 + (6--2)^2]= sqrt(100)= 10

We notice that ABS is an isosceles triangle.

The perimeter (P) = AB + AC + BC= 12+ 10...

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A(-4,-2)  B(8,-2)  C(2,6)

let us calculate the length of the sides:

AB = sqrt[(8--4)^2 + (-2--2)^2]= sqrt(12^2)= 12

AC= sqrt[(2--4)^2 + (6--2)^2]= sqrt(100) = 10

BC= sqrt[(2-8)^2 + (6--2)^2]= sqrt(100)= 10

We notice that ABS is an isosceles triangle.

The perimeter (P) = AB + AC + BC= 12+ 10 + 10 = 32

Since it is an isosceles, where AC = BC

==> let the perpendicular line from C on AB divides AB in midpoint

Let D be the midpoint,

==> AD = BD = 12/2= 6

But :

AD^2 = AC^2 - CD^2

         = 100- 36 = 64

==> The perpendicular line from C to AB (AD) = 8

The area of the triangle = (1/2)*AD* AB

                                       = (1/2)*8*12= 48

The area (a) = 48

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