# What is the perimeter of a triangle with corners at (2 ,6 ), (9 ,2 ), and (3 ,1 )?

The perimeter of the triangle is $19.24 \left(2 \mathrm{dp}\right)$ unit
We know distance formula; $d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} \therefore$
SideA=sqrt((2-9)^2+(6-2)^2))=sqrt(49+16)=sqrt65=8.06
SideB=sqrt((9-3)^2+(2-1)^2))=sqrt(36+1)=sqrt37=6.08
SideC=sqrt((3-2)^2+(1-6)^2))=sqrt(1+25)=sqrt26=5.1 :.
Perimeter $P = \left(8.06 + 6.08 + 5.1\right) = 19.24 \left(2 \mathrm{dp}\right)$unit[Ans]