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Find the perimeter of the triangle whose vertices are (0,1,2)(2,0,4) and (-4,-2,7). |
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Answer» Solution :Let A =(0,1,2)B =(2,0,4) and C =(-4,-2,7) ' "Then" AB = sqrt((2-0)^2+(0-1)^2+(4-2)^2)=sqrt(4+1+4) =3 ` `BC =sqrt((-4-2)^2+(-2-0)^2+(7-4)^2) =sqrt(36+4+9) =7` ` AC=sqrt((-4-0)^2 +(-2-1)^2+(7-2)^2) =sqrt(16+9+25)= sqrt(50) =5sqrt2` `:. "Perimetre of the TRIANGLE"=3 +7 +5 sqrt2 =10+5sqrt2` |
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