Question

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.

Accepted Answer

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

Answer

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.

Math

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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