Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (gokuporter):

http://prntscr.com/3i95cc

OpenStudy (ksaimouli):

I go with C but no sure if it is a //gram

OpenStudy (mathmale):

Would you please look up the vocabulary words used here and then choose the likeliest answer.

OpenStudy (calculusxy):

If all sides are congruent, then it is a square. All sides are parallel, so parallelogram. A square can be a rectangle, but a rectangle cannot be a square.

OpenStudy (mathmale):

@ksaimouli: A reminder: on OpenStudy we give suggestions and guidance aimed at helping the questioner to find his/her own answers. We do not say what we "think" is the answer. If you are unsure, it'd make more sense for you to stay in the background of this discussion.

OpenStudy (ksaimouli):

" A square can be a rectangle, but a rectangle cannot be a squar" true plz disregard my answr

OpenStudy (mathstudent55):

@calculusxy In your last statement you meant to say "A square is always a rectangle, but a rectangle is not always a square."

OpenStudy (mathmale):

Note how calculusxy has provided helpful (and correct) information, but has not given away the answer to this question.

OpenStudy (mathstudent55):

Also, @calculusxy , in your second statement, you meant "A quadrilateral with opposite sides parallel is a parallelogram."

OpenStudy (mathstudent55):

@calculusxy Sorry to do this, but in your first statement, you meant "If all sides are congruent, then it is a rhombus."

OpenStudy (ksaimouli):

square can be a rhombus, so if u think it is a square it is a rhombus for sure (90 angle)

OpenStudy (mathstudent55):

How about some definitions. Congruent sides: sides of equal length. Quadrilateral: a 4-sided polygon. Rhombus: a quadrilateral with 4 congruent sides. Parallelogram: a quadrilateral with exactly 2 pairs of opposite sides parallel. Rectangle: a parallelogram with 4 right angles. Square: a rectangular rhombus.

OpenStudy (mathstudent55):

@ksaimouli A square is always a rhombus.

OpenStudy (ksaimouli):

yup

OpenStudy (mathmale):

@Gokuporter: would you please involve yourself in this conversation. After all, it's still YOUR question. In the end YOU need to choose the correct answer.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!