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Question

Find the perimeter of a triangle whose vertices have the coordinates (3,10),(5,2) and (14,12).

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Solution

Solution:
Let A(3,10),B(5,2) and C(14,12) be the coordinates of the vertices of a triangle.
AB=(35)2+(102)2
=4+64
=68
=217 units =8.25 units
BC=(145)2+(122)2
=81+100
=181
=13.45 units
AC=(314)2+(1012)2
=121+4
=125
=55 units
=11.180 units
So, perimeter of the required triangle =AB+BC+AC
=8.25+13.45+11.180 units
=32.880 units

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