Find the perimeter of a triangle whose vertices have the coordinates
(3,10) (5,2) (4,12)
Answers
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Answer:
32.88 units
Step-by-step explanation:
Answered by
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A = ( 3 ,10)
B = ( 5,2)
C = ( 4,12)
AB = √[( 3-5)² + (10 -2)²]
= √ [ -2² + 8 ²]
= √(4 +64)
= √68
AC = √[(3-4)² + (10 - 12)²]
= √ [ -1² + -2²]
= √ 1 + 4
= √ 5
BC = √[(5-4)²+(2-12)²]
= √[(-1)² + ( -10)²]
= √[1 + 100]
= √ 101
perimeter
= √68 + √5 + √101
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