What is the most precise classification of the quadrilateral formed by connecting in order the midpoints of the figure below? Show your work.

[tex]What is the most precise classification of the quadrilateral formed by connecting in order the midpo[/tex]

The midpoints of the sides of the trapezoid are:

M ( - 1 , 1 ), N ( 1, - 1 ), P ( 3, 1 ), Q ( 1, 3 ).

MN = NP = PQ = QM = √ ( 2² + 2² ) = √ 8 = 2√2

∠MNP = ∠NPQ = ∠PQM = ∠QMN = 90°

The quadrilateral formed by joining the midpoints of the sides of the trapezoid is a square.

Rhombus

Step-by-step explanation:

The given trapezoid has vertices at M(-4,0), J(-2,4), K(2,4), and L(4,0).

Use the midpoint formula: [tex](\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})[/tex]

To obtain the midpoint of MJ:

[tex]A=(\frac{-4+-2}{2},\frac{0+4}{2})=(-3,2)[/tex]

Similarly, the midpoint of of JK is B=(0,4), the midpoint of KL is C=(3,2), and the midpoint of LM is D=(0,0).

Now use the slope formula;

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

To find the slope of AB = [tex]\frac{4-2}{0--3}=\frac{2}{3}[/tex]

The slope of BC =[tex]\frac{2-4}{3-0} =-\frac{2}{3}[/tex]

The slope of CD= [tex]\frac{0-2}{0-3}=\frac{2}{3}[/tex]

The slope of AD=[tex]\frac{0-2}{0--3}=-\frac{2}{3}[/tex]

We can see that the slope of the opposites sides are parallel, CD is parallel to AB and AD is parallel to BC.

The product of the slopes of the adjacent sides is not -1.

Hence the shape formed by connecting the midpoint of the isosceles trapezoid is a parallelogram.

Obviously the side lengths are equal since we connected all midpoints.

The parallelogram is a rhombus.

[tex]What is the most precise classification of the quadrilateral formed by connecting in order the midpo[/tex]

We have to form a quadrilateral by connecting the midpoints of the square.

The side of the square: a.

The side of the new quadrilateral: x² = (a/2)² + (a/2)²

x² = a²/4 + a²/4 = a²/2

x = a√2 / 2

All sides of the quadrilateral are equal: a√2 / 2.

So it could be another square or a rhombus. But all interior angles of the new quadrilateral are the same: 90°.

This is the square.

A square again

Lets take a is side of a square.

if u take parts of square from mid point and measure distance or length of any sides that would be a/2

then use Pythagorean theorem to solve hypotenuse u will get all length of small quadrilateral a/√2.same for all sides.

I see that will be square again