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In order for the parallelogram tobe a square, x = [? ].4x+17 126-23

In order for the parallelogram tobe a square, x = [? ].4x+17 126-23-example-1
User Bsa
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1 Answer

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For the given parallelogram to be a square, its diagonals should be equal. Let's put more details in the given figure to better understand:

Naming the two diagonals to be AC and BD, for it to be a square,


\text{ AC = BD}

AC = 12x - 23

BD = 4x + 17

We get,


\text{ AC = BD}
\text{ 12x - 23 = 4x }+\text{ 17}
\text{ 12x - 4x = 17 + 23}
\text{ 8x = 40}
\text{ }\frac{\text{8x}}{8}\text{ = }\frac{\text{40}}{8}
\text{ x = 5}

For the diagonals to be equal, x = 5.

Therefore, for it (parallelogram) to be a square, x should be equals to 5.

Let's check,

AC = 12x - 23 = 12(5) - 23 = 60 - 23 = 37

BD = 4x + 17 = 4(5) + 17 = 20 + 17 = 37

In order for the parallelogram tobe a square, x = [? ].4x+17 126-23-example-1
User Matheus Jacob
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