Question

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

Answer 1

Answer:

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: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

To obtain the midpoint of MJ:

$A=(\frac{-4+-2}{2},\frac{0+4}{2})=(-3,2)$

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;

$m=\frac{y_2-y_1}{x_2-x_1}$

To find the slope of AB = $\frac{4-2}{0--3}=\frac{2}{3}$

The slope of BC =$\frac{2-4}{3-0} =-\frac{2}{3}$

The slope of CD= $\frac{0-2}{0-3}=\frac{2}{3}$

The slope of AD=$\frac{0-2}{0--3}=-\frac{2}{3}$

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.