Let's think about this. MQ is given to be a length of 24 units, PR a length of 10 whilst we must determine what length PM must be in order to satisfy the criteria of parallelogram MPQR to be a rhombus.
Assume this figure is a rhombus, rhombus MPQR. If that is so, all sides must be congruent, and the diagonals must be perpendicular ( ⊥ ) by " Properties of a Rhombus. " That would make triangle( s ) MRQ and say RMP isosceles, and by the Coincidence Theorem, MS ≅ QS, and RS ≅ PS. Therefore -
PS and MS are legs of a right triangle, so by Pythagorean Theorem we can determine the hypotenuse, or in other words the length of PM. This length would make parallelogram MPQR a rhombus,
<u><em>And thus, PM should be 13 in length to make parallelogram MPQR a rhombus.</em></u>
True, because perpendicular means crossing at a 90° angle
Let H (6, 7) be (x₁, y₁) and I (-7, -6) be (x₂, y₂)
Mid point =
Mid point = [
Mid point =
Santiago is not correct, the mid point is (-0.5, 0.5)
Answer:
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Step-by-step explanation:
Answer:
Step-by-step explanation:
You think about it or go search it up my guy.