# Math problem

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• Oct 29th 2012, 10:01 AM
Petrus
Math problem
Consider the following amounts of plane figures: A = {All squares}, B = {all diamonds}, C = {all rectangles}, D = {all parallelograms}, E = {all trapezoidal with three equal sides}, F = {all trapezoidal with at least a straight corner angle}. Examine which of the following statements are true. Reply with the letter R (true) or F (false).
__________
__________
__________
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__________ (e) https://webwork.math.su.se/webwork2_...fbac1a7bf1.png

Tips how i shall think :)? I usually draw to solve this but dont know what to do because its not number anymore :/
• Oct 29th 2012, 07:56 PM
chiro
Re: Math problem
Hey Petrus.

One tip for you is to look at each combination at a time and think of whether anything is common to both: If not, then the intersection is the empty set but other-wise the intersection includes all the elements common to both.

Two sets are equal only if all the elements in each set are equal and you can not repeat elements in sets so they both need to contain exactly the same elements.
• Oct 30th 2012, 12:37 AM
Petrus
Re: Math problem
is it possible i can draw..? idk how to draw this so i can easy se... :/ i like to draw to solve this problem :)
• Oct 30th 2012, 03:43 AM
cac2008
Re: Math problem
Think about the definition of each:
Square: A plane figure with four equal straight sides and four right angles
Rectangle: A plane figure with four straight sides and four right angles
Diamond: ==Rhombus: A parallelogram with opposite equal acute and obtuse angles and four equal sides.
Parallelogram: A four-sided plane rectilinear figure with opposite sides parallel
Trapezoid: A quadrilateral with only one pair of parallel sides

Therefore a square is a member of the set of rectangles for example, as it has four straight sides and four right angles, but has the special condition that all sides are equal.

Just looking at the first one: B union C = D, as both diamonds and rectangles are four sides with opposite sides parallel, this statement is true. The rest are just as easy.

I think you might complicate matters by trying to draw this one out, as you are thinking about the properties of each set, not its appearance.
• Nov 2nd 2012, 07:36 AM
Petrus
Re: Math problem
ty ty :D i alredy solved this^^ ty anyway