Find the perimeter of a triangle with coordinate of the vertises (0, 4) (0, 0) (3, 0)
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Let the three vertices be :
A = (0,4) O = (0,0) C = (3,0)
side A = √ 0²-4²= √16 = 4
B = √0²-0² = 0
C = √3²- 0²=9 = 3
PERIMETER OF A TRIANGLE = 4+0+3 = 7
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A = (0,4) O = (0,0) C = (3,0)
side A = √ 0²-4²= √16 = 4
B = √0²-0² = 0
C = √3²- 0²=9 = 3
PERIMETER OF A TRIANGLE = 4+0+3 = 7
Hope it helps... :)
pls rate my ans as brainliest if u find it helpful...
thnks..:)
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