1. Points D(-4, 6), E(5, 3), and F(3, -2) are the vertices of triangle DEF. Find the perimeter of the triangle.
2007-11-09 13:14:16 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
√(5+4)^2+(3-6)^2=√90
√(3+4)^2+64=√113
√(5-3)^2+25=√29
√90+√113+√29= PERIMETER
2007-11-09 13:21:21 · answer #1 · answered by UNIQUE 3 · 0⤊ 0⤋
Points D(-4, 6), E(5, 3), and F(3, -2) are the vertices of triangle DEF. Find the perimeter of the triangle.
EF^2 = (4 + 25) = 29
FD^2 = (7^2 +8^2) = 113
DE^2 = ( 81 + 9) =90
Perimeter = sqrt(29) +sqrt(113) +sqrt(90) = 25.50
2007-11-09 13:21:35 · answer #2 · answered by Anonymous · 0⤊ 1⤋
this one is easy first you plot the coordinates , then after that you find the distance the three vertices using the distance formula. Do it in three steps first distance between D and E
second E and F then lastly F and D. Just add all the distance to get the perimeter.
2007-11-09 13:23:36 · answer #3 · answered by Rakiztah 2 · 0⤊ 0⤋
DE = √(9² + 3²) = √90 = 9.49
EF = √(2² + 5²) = √29 = 5.39
FD = √(7² + 8²) = √113 = 10.63
perimeter = 25.51
2007-11-09 13:21:35 · answer #4 · answered by Philo 7 · 0⤊ 1⤋
find the distance between the points using the distance formula:
d=rad( (x2-x1)^2+(y2+y1)^2)
D-->E
rad (-4-5)^2+(6-3)^2
rad (-9)^2 + (3)^2
rad 81+9
rad 90
E-->F
rad (5-3)^2+(3+2)^2
rad (2)^2 + (5)^2
rad 4+25
rad 29
F-->D
rad (3+4)^2 + (-2-6)^2
rad (7)^2 + (-8)^2
rad (49) + (64)
rad 113
Perimeter= side + side + side
P= rad 90 + rad 29 + rad 113
2007-11-09 13:35:16 · answer #5 · answered by Anonymous · 0⤊ 0⤋
For each pair of points, find the vertical and horizontal distances, and use the Pythagorean theorem.
2007-11-09 13:22:32 · answer #6 · answered by chesler.geo 2 · 0⤊ 0⤋
(4x+8)+(27) = 6x 4x+35=6x -4x -4x 35=2x /2 /2 17.5=x If D replaced into the MP, then the equation could exchange from 4x+8+27=6x to (4x+8=27) = 6x i'm unsure if it relatively is what your asking...
2016-10-15 23:18:37 · answer #7 · answered by ? 4 · 0⤊ 0⤋