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 -
[tex]MS = 1 / 2( 24 ) = 12 = QS,\\RS = 1 / 2( 10 ) = 5 = PS[/tex]
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,
[tex]( PM )^2 = ( MS )^2 + ( PS )^2,\\PM^2 = ( 12 )^2 + ( 5 )^2,\\PM^2 = 144 + 25 = 169\\-----\\PM = 13[/tex]
And thus, PM should be 13 in length to make parallelogram MPQR a rhombus.
