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mathmath333
 one year ago
Question maths/reasoning
mathmath333
 one year ago
Question maths/reasoning

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439124280221:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{In the figure: }\hspace{.33em}\\~\\ & \normalsize \text{K represents all Kites}\hspace{.33em}\\~\\ & \normalsize \text{Q represents all Quadrilaterals}\hspace{.33em}\\~\\ & \normalsize \text{R represents all Rhombus}\hspace{.33em}\\~\\ & \normalsize \text{P represents all Parallelogram}\hspace{.33em}\\~\\ & \normalsize \text{The statement "Rhombus is also a Kite"}\hspace{.33em}\\~\\ & \normalsize \text{can be described as}\hspace{.33em}\\~\\ & 1.) \normalsize \text{P and K is nothing but R}\hspace{.33em}\\~\\ & 2.) \normalsize \text{P or K is nothing but R}\hspace{.33em}\\~\\ & 3.) \normalsize \text{P and R is nothing but K}\hspace{.33em}\\~\\ & 4.) \normalsize \text{P or R is nothing but K}\hspace{.33em}\\~\\ \end{align}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439124864364:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we represent common region using "\(\cap\)" and read it out as "\(\text{and}\)"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2In above venn diagram, we have \[P \text{ and } K = R\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2therefore a \(R\)hombus is both a \(P\)arallelogram and a \(K\)ite

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Rhombus belongs to both the families of Parallelogram and Kite

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.11st option is correct ?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Given the Venn diagram in the question, I agree with @ganeshie8, but are we allowed to suspend the correct definitions of the terms used, so the problem works? A kite can never be a rhombus or a parallelogram, and a rhombus can never be a kite using the normal definitions of those quadrilaterals.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2A square is also a rhombus/rectangle/parallelogram/trapezoid/kite A rhombus is also a parallelogram/trapezoid/kite so, some kites are also squares/rhombii/parallelograms I don't see any conflict here, @mathstudent55

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1this is purely reasoning type question

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439128107280:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1didnt memtion kites

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2does a square satisfy the properties of a kite ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so, some kites are squares. does a rhombus satisfy the properties of a kite ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2since a square is also a rhombus, parallelogram and trapezoid, it follows that some kites are rhombii/parallelograms/trapezoids

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Then perhaps I never learned the definition of kite correctly. I thought a kite is a quadrilateral with a pair of two pairs of congruent adjacent sides, both pairs not being congruent to each other. If the definition of kite allows for the two pairs of adjacent sides to be congruent, making all sides congruent, then a rhombus is indeed a special case of a kite, and certainly a kite can be a square.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Now since you mention a trapezoid, there is another problem. My understanding of a trapezoid is that it's a quadrilateral with exactly one pair of parallel opposite sides. This means the second pair of opposite sides cannot be parallel, and therefore a trapezoid and a parallelogram are mutually exclusive quadrilaterals.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2i remember them as : A kite is just a quadrilateral with two pairs of congruent adjacent sides. A trapezoid is a quadrilateral with at least one pair of parallel sides I am also googling for correct definitions as we speak...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439129024111:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2some online materials do say that a trapezoid must have "exactly" one pair of parallel sides dw:1439129281970:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Allowing a kite to have all 4 sides congruent makes the original Venn diagram of the problem completely acceptable. I just have to get used to the correct definition of a kite.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Yes, and some websites state that a trapezoid has at least one pair of sides parallel. In addition, there seems to be different usage for trapezoid and trapezium in the U.S. and the UK.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I remember facing issues with trapezoid and trapezium before haha

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1U.S. trapezoid = UK trapezium (quadrilateral with either exactly or at least 1 pair of sides parallel) U.S. trapezium = UK trapezoid (quadrilateral with no sides parallel)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 As always, thanks for your insight.
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