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.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116