ABCD-quadrilateral A (-3; -1), B (1; 2), C (5; -1), D (1; -4). Prove that ABCD is a rhombus?
| July 8, 2021 | education
Given the coordinates of the vertices of the quadrangle, we determine the lengths of their sides.
AB ^ 2 = (X2 – X1) ^ 2 + (Y2 – Y1) ^ 2 = (1 – (-3)) ^ 2 + (2 – (-1)) ^ 2 = 16 + 9 = 25.
AB = 5 cm.
ВС ^ 2 = (5 – 1) ^ 2 + (-1 – 2) ^ 2 = 16 + 9 = 25.
BC = 5 cm.
СD ^ 2 = (1 – 5) ^ 2 + (-4 – (-1)) ^ 2 = 16 + 9 = 25.
CD = 5 cm.
AD ^ 2 = (1 – (-3)) ^ 2 + (-4 – (-1)) ^ 2 = 16 + 9 = 25.
AD = 5 cm.
All four sides are equal to 5 cm, quadrilateral square or rhombus. Let us determine the lengths of the diagonals.
AC ^ 2 = (5 – (-3)) ^ 2 + (-1 – (-1)) ^ 2 = 64 + 0 = 64.
AC = 8 cm.
BD ^ 2 = (1 – 1) ^ 2 + (-4 – 2) ^ 2 = 0 + 36 = 36.
ВD = 6 cm.
The diagonals are of different lengths, therefore, the quadrilateral is a rhombus, as required.
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