Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work
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Edit: Nvm I got it:)
1 answer:
Answer:
Step-by-step explanation:
Question
Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work.
perimeter of a triangle = AB+AC+BC
Using the distance formula
AB = sqrt(-2-5)²+(2-2)²
AB = sqrt(-7)²
AB =sqrt(49)
AB =7
BC = sqrt(-2+2)²+(2-5)²
BC = sqrt(0+3²)
BC =sqrt(9)
BC =3
AC= sqrt(-2-5)²+(2-5)²
AC= sqrt(-7)²+3²
AC =sqrt(49+9)
AC =sqrt58
Perimeter = 10+sqrt58
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