SOLUTION: prove the following parallelograms is not rectangle? C(-9,4) D(-4,8) E(2,6)F (-3,2) ? Prove that A PARALLELOGRAM WITH VERTICES T(-2,3) E(-5,-4) A(2,-1) M (5,6) is a rhombus but

Question 1010962: prove the following parallelograms is not rectangle? C(-9,4) D(-4,8) E(2,6)F (-3,2) ?
Prove that A PARALLELOGRAM WITH VERTICES T(-2,3) E(-5,-4) A(2,-1) M (5,6) is a rhombus but not a square
Prove that R(2,1) E (10,7) C(7,11) T (-1,5) are the vertices of a rectangle but not a square

Prove that M(-2,1) A(1,6) T(8,3) H (5,-4) are the vertices of a square I really need the help please!
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
I assume that you have followed the labeling convention of starting with the lower left point and then proceed clockwise.
********************************************************************************
1) a rectangle has two pairs of equal sides and four 90 degree interior angles
the sides are CD, DE, EF, FC
note that the diagonals bisect the interior angles
CD = sqrt((-4-(-9))^2 + (8-4)^2) = 6.403124237
DE = sqrt((2-(-4))^2 + (6-8)^2) = 6.32455532
EF = sqrt((-3-2)^2 + (2-6)^2) = 6.403124237
FC = sqrt((-9-(-3))^2 + (4-2)^2) = 6.32455532
side CD = EF and side DE = FC
the adjacent sides of a rectangle are perpendicular to each other which means that the slopes of adjacent sides must be the negative reciprocal of each other
slope of FC = (2-4) / (-3-(-9) = -2 / 6 = -1/3
slope of CD = (8-4) / (-9-(-4)) = 4 / -5 = -4/5
therefore CDEF is not a rectangle
*******************************************************************************
2) note that all squares are rhombus but not all rhombus are squares
a square has all four sides equal and four 90 degree interior angles
TE = sqrt((-5-(-2))^2 + (-4-3)^2) = 7.615773106
EA = sqrt((2-(-5))^2 + (-1-(-4)^2) = 7.615773106
AM = sqrt((5-2)^2 + (6-(-1))^2) = 7.615773106
MT = sqrt((5-(-2))^2 + (6-3)^2) = 7.615773106
side TE=EA=AM=MT
the diagonals of a rhombus are perpendicular to each other and bisect each other
let X be the point of intersection of the diagonals
TA = sqrt((2-(-2))^2 + (-1-3)^2) = 5.656854249
EM = sqrt((5-(-5))^2 + (6-(-4))^2) = 11.180339887
sin of angle XEA = (TA/2) / EA = 0.371390676
inverse sin of 0.371390676 = 21.801409465 degrees
angle XEA = 21.801409465
now angle TEA = 2 * 21.801409465 = 43.602818929 degrees
TEAM is a rhombus but not a square
******************************************************************************
3) RE = sqrt((10-2)^2 + (7-1)^2) = 10
EC = sqrt((7-10)^2 + (11-7)^2) = 5
CT = sqrt((-1-7)^2 + (5-11)^2) = 10
TR = sqrt((-1-2)^2 + (5-1)^2 = 5
side RE=CT and EC=TR
RECT is not a square
******************************************************************************
4) MA = sqrt((1-(-2))^2 + (6-1)^2) = sqrt(34)
AT = sqrt((8-1)^2 + (3-6)^2) = sqrt(58)
TH = sqrt((5-8)^2 + (-4-3)^2) = sqrt(58)
HM = sqrt((5-(-2))^2 + (-4-1)^2) = sqrt(74)
the coordinates are not correct for a square
*****************************************************************************

RELATED QUESTIONS
Given: triangle ABC with vertices A(-6,-2), B(2,8), and C(6,-2). Line segment AB has... (answered by solver91311)

I've been trying to use the same formula that I use to find the area of a triangle but it (answered by Alan3354,ikleyn)

This is from the geometry regents August 2011, number 38. Given: Triangle ABC with... (answered by KMST)

Prove that the points A(-2,-3), B(6,2), C(8,7) and D(0,2) are the vertices of a... (answered by math_helper)

parallelogram abcd is not a rectangle the points of a b c and d are (-2,2) (6,5) (4,0)... (answered by vleith)

Prove that the points A(-4,-1), B(-2,-3), C(4,3), and D(2,5) are the vertices of a... (answered by Alan3354)

Prove that the points A(-4,-1),B(-2,-3),C(4,3), and D(2,5) are the vertices of a... (answered by Alan3354)

Find the area of the parallelogram with vertices A(2, -3), B(7, -3), C(9, 2), D(4,... (answered by ikleyn,MathTherapy)

Suppose that Parallelogram ABCD has vertices A(2, 4), B(8, 4), C(5, -3) and D(-1, -3).... (answered by KMST)