# The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

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## Question:

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

Best is to draw a sketch of the three points.
Next step is to find the distances BC, CD, DB.
The perimeter is the sum of the three distances.

The distances are found using the distance formula:
D=sqrt((y2-y1)^2+(x2-x1)^2)
order of (x1,y1), (x2,y2) is not important.

Given A(2,8),B(16,2),C(6,2)
we calculate
AB=sqrt((16-2)^2+(2-8)^2)=sqrt(14^2+6^2)=sqrt(232);
BC=sqrt((6-16)^2+(2-2)^2)=sqrt(10^2+0)=10
CA=sqrt((2-6)^2+(8-2)^2)=sqrt(4^2+6^2)=sqrt(16+36)=sqrt(52)

Perimeter=AB+BC+CA=32.443 units

For the area, we note that BC is horizontal (parallel to the x-axis), so
area = (1/2)bast * height
=(1/2)10*(ya-yb)
=(1/2)10*(8-2)
=(1/2)10*6
=30 unit^2

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