What is the perimeter of a triangle with corners at (1 ,4 ), (6 ,7 ), and (4 ,2 )?

1 Answer

Perimeter =sqrt(34)+sqrt(29)+sqrt(13)=3.60555

Explanation:

A(1,4) and B(6,7) and C(4,2) are the vertices of the triangle.

Compute for the length of the sides first.

Distance AB

d_(AB)=sqrt((x_A-x_B)^2+(y_A-y_B)^2)

d_(AB)=sqrt((1-6)^2+(4-7)^2)

d_(AB)=sqrt((-5)^2+(-3)^2)

d_(AB)=sqrt(25+9)

d_(AB)=sqrt(34)

Distance BC

d_(BC)=sqrt((x_B-x_C)^2+(y_B-y_C)^2)

d_(BC)=sqrt((6-4)^2+(7-2)^2)

d_(BC)=sqrt((2)^2+(5)^2)

d_(BC)=sqrt(4+25)

d_(BC)=sqrt(29)

Distance BC

d_(AC)=sqrt((x_A-x_C)^2+(y_A-y_C)^2)

d_(AC)=sqrt((1-4)^2+(4-2)^2)

d_(AC)=sqrt((-3)^2+(2)^2)

d_(AC)=sqrt(9+4)

d_(AC)=sqrt(13)

Perimeter =sqrt(34)+sqrt(29)+sqrt(13)=3.60555

God bless....I hope the explanation is useful.