If MQ is 24 and PR is 10, what length of PM would make parallelogram MPQR a rhombus?

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,

And thus, PM should be 13 in length to make parallelogram MPQR a rhombus.

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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circle area = pi * radius^2

if diameter = 10 then radius = 5

circle area = pi * 5^2

circle area = pi * 25

circle area = 78.5398163397 square units

step-by-step explanation: