Is this TopScore math question a little off?

[FROGGBUSTER]

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Given a rectangle, square, and parallelogram, if only one of these quadrilaterals is selected at random, what is the probability that its diagonals bisect each other?

A. 1/3
B. 1/9
C. 1/27
D. 1/32
E. 1

Answer: "All parallelograms have diagonals that bisect each other."

The answer is E, but I chose A. I thought bisect was cut into 2 EQUAL parts, and Google seems to agree, lol. If it's cut into 2 equal parts, the quadrilateral has to be either a square or a rhombus (a parallelogram too).



A little weird or no?

 

[LetsGo2DSchool]

The answer seems right. Each diagonal will bisect the other diagonal into two equal parts.

 

[FROGGBUSTER]

The answer seems right. Each diagonal will bisect the other diagonal into two equal parts.

That's only for squares and rhombuses though I think. Like draw a really long rectangle and its diagonals, they don't bisect into equal segments.

 

[FROGGBUSTER]

Oops, yeah you're right. It's correct. 😳

 

[LetsGo2DSchool]

I don't think the question is saying that all four segments (two from each diagonal) after the bisection have to be equal in length, but instead, that the two segments of each diagonal have to be equal to each other.

 

[FROGGBUSTER]

Thanks.