# 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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### 2 Answers

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**

A(-4,-2), B(8,-2) and C(2,6)

To find area of ABC and Perimeter.

Solution:

We know that the distance d between (x1,y1) and (x2,y2) is given by:

d^2 = (x2-x1)^2+(y2-y1)^2.

AB^2 = (8--4)^2+(-2- -2)^2 = 12^2+0^2. So AB = c = 12

AC^2 = (2--4)^2+((6- -2)^2 = 36+64 =100. So AC=b =10.

BC^2 = (2-8)^2+(6- -2)^2 = 36+64 =100, BC = 10 = a = 10

Therefore perimeter p = a+b+c = 12+10+10 = 32.

Area of ABC by Heron's formula = sqrt{s(s-a)(s-b)(s-c)}, wgere s = (a+b+c)/2 = 32/2 =16 units.

Area of triangle ABC = sqrt{16(16-12)(16-10)^2} = sqrt{16*4*10^2

=4*2*10 = 80 sq units.