If two vertices of a parallelogram are (3, 2), (-1, 0) and the diagonals cut at (2, -5), find the other vertices of the parallelogram.
Answers
Answer:
Draw (with a ruler) a long skinny rectangle, say base 6, height 1. Now draw the diagonals. You will see that the two diagonals definitely do not meet at right angles, not even close!
It turns out that a parallelogram has its diagonals meeting at right angles if and only if the parallelogram is a rhombus (all sides equal). Note that a square is a special case of a rhombus.
Proof: It is fairly easy to prove that the diagonals of a parallelogram (and therefore of the special parallelogram called a rectangle) bisect each other.
Let the two diagonals of a parallelogram have length 2p and 2q respectively. If the angle at which they meet is 90∘, then by the Pythagorean Theorem each side of the rectangle has length p2+q2−−−−−−√. So in particular all the sides of the parallelogram are equal, that is, we have a rhombus.
Conversely, by a congruent triangles argument, if we have a rhombus, its diagonals meet at right angles.
Draw (with a ruler) a long skinny rectangle, say base 6, height 1. Now draw the diagonals. You will see that the two diagonals definitely do not meet at right angles, not even close!
It turns out that a parallelogram has its diagonals meeting at right angles if and only if the parallelogram is a rhombus (all sides equal). Note that a square is a special case of a rhombus.
Proof: It is fairly easy to prove that the diagonals of a parallelogram (and therefore of the special parallelogram called a rectangle) bisect each other.
Let the two diagonals of a parallelogram have length 2p and 2q respectively. If the angle at which they meet is 90∘, then by the Pythagorean Theorem each side of the rectangle has length p2+q2−−−−−−√. So in particular all the sides of the parallelogram are equal, that is, we have a rhombus.
Conversely, by a congruent triangles argument, if we have a rhombus, its diagonals meet at right angles...
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