Properties of Parallelograms Practice with Our Examples

State whether the statements are True or False. All rectangles are parallelograms.

ABCD is a parallelogram. Which of the following is not true about ABCD?
AB=CD
AD=CB
AD||CD
AB||CD

Which of the following is not a parallelogram?
A)Square
B)Rectangle
C)Rhombus
D)Trapezium

Which of the following is not a necessary condition for a quadrilateral ABCD to be a parallelogram:
A)AB = CD
B)AB // CD
C)$\angle A={50}^{\circ <!--\; \circ \; -->},\mathrm{\angle <!--\; \angle \; -->}C={50}^{\circ <!--\; \circ \; -->},\mathrm{\angle <!--\; \angle \; -->}B={130}^{\circ <!--\; \circ \; -->}$
D)AB⊥AD

How can magnets lose their properties?
A. If they are heated
B. If they are hammered
C. If they are dropped from height
D. All of these

All rectangles are parallelograms.
A)True
B)False

What is the volume of a block of ice measuring 3 meters long, 1.5 meters wide, and 2 meters high?

State whether True or False.
(1) All rectangles are squares.
(2) All rhombuses are parallelograms.
(3) All squares are rhombuses and also rectangles.
(4) All squares are not parallelograms.
(5) All kites are rhombuses.
(6) All rhombuses are kites.
(7) All parallelograms are trapeziums.
(8) All squares are trapeziums.

PQ and RS are two equal and parallel line segments. Any point M not lying on PQ or RS is joined to Q and S and lines through P parallel to QM and through R parallel to SM meet at N. Prove that line segments MN and PQ are equal and parallel to each other.

Which statements about a parallelogram are accurate? A Diagonals cut each other in half. B The parallelogram is divided into two congruent triangles by its diagonals. C In a parallelogram, the following angles are supplementary. D A parallelogram's opposing angles are always supplementary.

I don't understand what this theorem for a characterization of the determinant actually means, and what use it has. Could you please explain it?
Let D be a function mapping from the set of square matrices M with n rows/columns over the field F, to a field F. Also let D be a function that is multilinear and alternating in columns. Then
D(A)=D(I)detA

The way I've been introduced to determinants is that if there is a system of two linear equations then we can represent the coefficients of the variables and the constants in the form of a matrix.
Now if we plot the matrices on the coordinate system then we will get a parallelogram and if we calculate the area of the parallelogram then we will get the determinant of the given matrix. For eg if A is the matrix then its determinant will be:
ad−cb.
i.e. |A|= ad−cb.
if A=$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Now the questions I want to ask:
1)What is a determinant actually what does it tells us about a system of equations?
2)The area found by the formula ad−cb, how is it telling us a determinant? Basically how the area of parallelogram telling the value of determinant?
3)In my book its given that: system of equations has a unique solution or not is determined by the number of ab-cd.What does this mean?

Let C[0,1] be the space of continuous functions $[0,1]\to <!--\; \to \; -->\mathbb{R}$ endowed with the norm $||x|{|}_{\mathrm{\infty <!--\; \infty \; -->}}={\mathrm{m}\mathrm{a}\mathrm{x}}_{t\in <!--\; \in \; -->[0,1]}|x(t)|$. It is easy to verify that this norm is not induced by any inner product (really the parallelogram law fails for x(t)=t and y(t)=1). Well, how to understand that this norm is not equivalent to anyone induced by an inner product? So, the norms induced by inner products should have some special properties...

Prove that there is an inner product on ${\mathbb{R}}^2$, such that the associated norm is given by:
$\parallel <!--\; \parallel \; -->(x,y)\parallel <!--\; \parallel \; -->=(|x{|}^p+|y{|}^p{)}^{\frac{1}{p}}$
where p>0 only if p=2So far what I have tried to do is assume there exists such an inner product for some aribtrary p, and then show that some property that holds for all inner products (e.g. the Cauchy-Schwarz inequality, triangle inequality, parallelogram equality, does not hold for the inner product with the said associated norm unless p=2.
First I tried the parallelogram equality and ended up with:
$(|x_1+x_2{|}^p+|y_1+y_2{|}^p{)}^{\frac{1}{p}}=(|x_1|+|y_1|{)}^{\frac{1}{p}}+(|x_2{|}^p+|y_2{|}^p{)}^{\frac{1}{p}}$
but I don't know how to show that this equality only holds for p=2 (although I'm pretty sure it does because I tried plugging in random values for $p\ne <!--\; \ne \; -->2$).
Since for an inner product, the parallelogram equality must hold,
$⟨<!--\; ⟨\; -->u,v⟩<!--\; ⟩\; -->=\frac{1}{2}(\parallel <!--\; \parallel \; -->u+v{\parallel <!--\; \parallel \; -->}^2+\parallel <!--\; \parallel \; -->u-<!--\; -\; -->v{\parallel <!--\; \parallel \; -->}^2)$
must also hold.
Using this definition of the inner product, I also tried to show a contradiction by showing that if $p\ne <!--\; \ne \; -->2$, the Cauchy-Schwarz Inequality didn't hold. However I think that's a dead end.

I came across a property which I am not sure if true in general. Suppose you have a parallelogram whose vertices are $v_1,v_2,v_3,v_4\in <!--\; \in \; -->{\mathbb{R}}^2$. Let's say that the side $[v_1,v_4]$ is parallel to $[v_2,v_3]$. Is it true that
$||v_1-<!--\; -\; -->v_4|{|}^2<||v_1-<!--\; -\; -->v_3||\cdot <!--\; \cdot \; -->||v_2-<!--\; -\; -->v_4||,$
that is, the square of one side is less than the product of the diagonals? I could also be just missing an obvious counterexample, but so far I can't prove it either.

According to wikipedia article on polarisation identity, in a normed space $(V,||.||$, if the parallelogram law holds, then there is an inner product on V such that $||x|{|}^2=⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->$ for all $x\in <!--\; \in \; -->V$. However, from what I understand, we can prove polarisation identity this way:
$||x+y|{|}^2-<!--\; -\; -->||x-<!--\; -\; -->y|{|}^2=⟨<!--\; ⟨\; -->x+y,x+y⟩<!--\; ⟩\; -->-<!--\; -\; -->⟨<!--\; ⟨\; -->x-<!--\; -\; -->y,x-<!--\; -\; -->y⟩<!--\; ⟩\; -->=...=4⟨<!--\; ⟨\; -->x,y⟩<!--\; ⟩\; -->$
However, from this proof, I don't see why we need parallelogram law. On top of that, shouldn't it be the other way round, such that if $||x|{|}^2=⟨<!--\; ⟨\; -->x,x⟩<!--\; ⟩\; -->$, then the parallelogram law holds? Can someone explain to me please?

It can be proved easily by contradiction that
if a,b,c,d are positive numbers, then
$\frac{a+b}{c+d}\ge <!--\; \ge \; -->min\{\frac{a}{c},\frac{b}{d}\}.$
I am not looking for a proof but rather for
1) a reference or book which contain this and similar inequalities;
2) information whether this inequality can be sharpened.

I have three questions about a quadrilateral with the following properties:

1. It is convex.
2. It has exactly one pair of congruent opposite sides.
3. It has exactly one pair of congruent opposite angles.
4. It is not a parallelogram.

Does such a quadrilateral exist? Is it possible to construct such a quadrilateral with compass and straightedge? If the construction is not possible, why not?

Let x and y be two vectors in ${\mathbb{R}}^2$. The parallelogram law of vector addition says that the vector corresponding to the diagonal of the parallelogram formed by these vectors is x+y. For the same parallelogram consider the other diagonal directed from x to y. Call this diagonal $\stackrel{\to <!--\; \to \; -->}{D}$ .
The vector y−x can be identified with $\stackrel{\to <!--\; \to \; -->}{D}$ since we see geometrically that it has the same length and the same slope. This is also intuitively obvious because a rigid motion (translation) can superimpose y−x on $\stackrel{\to <!--\; \to \; -->}{D}$ .
However we may identify $\stackrel{\to <!--\; \to \; -->}{D}$ with x−y as well since they too have the same slope and length. Moreover a rotation and a translation can superimpose x−y on $\stackrel{\to <!--\; \to \; -->}{D}$ as well.
Yet it is geometrically obvious that x−y is the opposite direction of $\stackrel{\to <!--\; \to \; -->}{D}$ .
Can someone explain why we should identify $\stackrel{\to <!--\; \to \; -->}{D}$ with y−x only? Is it correct to say that we should identify a vector with a directed line segment if the two are related by a translation only? If so, what is the justification behind this and is there a name for the geometry arising from studying properties invariant under translation only?

It's a well-known result that if X is a Hilbert space, then for any closed convex subset C of X, there exists a unique element of C with minimal norm. I'm wondering whether the converse is true.
Let X be a Banach space such that for any closed convex subset C of X, there exists a unique element of C with minimal norm. My question is, does this imply that the norm on X is induced by some inner product?
Now a norm is induced by some inner product if and only if it satisfies the parallelogram law. So does this property about closed convex sets somehow imply the parallelogram law? If not, does anyone know of a counterexample?