Question 824253: Prove that a quadrilateral ABCD with vertices A(-1,4), B(-3,3), C(1,2) and D(3,3) is a parallelogram
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! quadrilateral
edge lengths
| (sqrt(5) | sqrt(17) | sqrt(5) | sqrt(17))
=(2.23607 | 4.12311 | 2.23607 | 4.12311)
diagonal lengths
| (2 sqrt(2) | 6)
=(2.82843 | 6)
area | 6
perimeter | 2 (sqrt(5)+sqrt(17))=12.7183
interior angles
| ((180 (pi-tan^(-1)(1/4)-tan^(-1)(1/2)))/pi° | (180 (tan^(-1)(1/4)+tan^(-1)(1/2)))/pi° | (180 (pi-tan^(-1)(1/4)-tan^(-1)(1/2)))/pi° | (180 (tan^(-1)(1/4)+tan^(-1)(1/2)))/pi°)=
(2.43297 radians | 0.708626 radians | 2.43297 radians | 0.708626 radians)
interior angle sum | 360° = 2 pi rad
exterior angle sum | 1080° = 6 pi rad
| (-1, 4) | (-3, 3) | (1, 2) | (3, 3)
(-1, 4) | 0 | sqrt(5)=2.23607 | 2 sqrt(2)~~2.82843 | sqrt(17)~~4.12311
(-3, 3) | sqrt(5)~~2.23607 | 0 | sqrt(17)~~4.12311 | 6
(1, 2) | 2 sqrt(2)~~2.82843 | sqrt(17)~~4.12311 | 0 | sqrt(5)~~2.23607
(3, 3) | sqrt(17)~~4.12311 | 6 | sqrt(5)~~2.23607 | 0
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