(1/30/11)
(Page 42 #4)
Is it possible for the diagonals of a parallelogram to have the same length? Of a trapezoid? of a non-isosceles trapezoid?

It is possible for the diagonals of a parallelogram to have the same length; for example, a rectangle or square.
It is possible for the diagonals to be the same for an isosceles triangle, but not for a non-isosceles triangle.

Parallelogram Proof Screencast

1/21/11

Complex Numbers GeoGebra/Screencast Winter Break

(1/2/11)
Geogebra Applet:

Screencast:

Box Screencast GeoGebra3D WP5

(12/13/10)

Point Reflection WP4

(12/2/10)

Four Isometries WP3

(11/22/10)

Sequence Construction WP2

Triangle WP1

Recent Homework

P.46 #2

P.46 #12

Old Classwork

Intersection with two parametric points:

Problem 5 Page 3 (Phillips Exeter Academy)

Let A = (1, 5) and B = (3,-1). Verify that P = (8,4) is equidistant from A and B. Find at least two more points that are equidistant from A and B. Describe all such points.

Formula for Hypotenuse

Square Root of 50

Twice

Square Root of 50

Find at least two more points equidistant. Describe them.

The point C was found using the midpoint command through Geogebra. Combined with the other points, this forms a kite. You can manipulate the kite at will, but it will keep the hypotenuse of the two triangles the same length (assuming you don't move P,B, or A), thus verifying that the points are equidistant.

## Table of Contents

Guten tag! Ave! Welcome to my wiki page## Portfolio Problems

Problem 1:Link

It has been 1600 years since the first sacking of Rome (during the Roman Empire) by Alaric the Visigoth. 8/24/10Fun fact## Entry

Exterior Angle TheorumWeekly Problems## Parallelogram Problem Review

(1/30/11)(Page 42 #4)

Is it possible for the diagonals of a parallelogram to have the same length? Of a trapezoid? of a non-isosceles trapezoid?

It is possible for the diagonals of a parallelogram to have the same length; for example, a rectangle or square.

It is possible for the diagonals to be the same for an isosceles triangle, but not for a non-isosceles triangle.

## Parallelogram Proof Screencast

1/21/11## Complex Numbers GeoGebra/Screencast Winter Break

(1/2/11)Geogebra Applet:

Screencast:

## Box Screencast GeoGebra3D WP5

(12/13/10)## Point Reflection WP4

(12/2/10)## Four Isometries WP3

(11/22/10)## Sequence Construction WP2

## Triangle WP1

Recent Homework## P.46 #2

## P.46 #12

## Old Classwork

Intersection with two parametric points:

Problem 5 Page 3 (Phillips Exeter Academy)

Let A = (1, 5) and B = (3,-1). Verify that P = (8,4) is equidistant from A and B. Find at least two more points that are equidistant from A and B. Describe all such points.

Find at least two more points equidistant. Describe them.

The point C was found using the midpoint command through Geogebra. Combined with the other points, this forms a kite. You can manipulate the kite at will, but it will keep the hypotenuse of the two triangles the same length (assuming you don't move P,B, or A), thus verifying that the points are equidistant.

Helpful comment box (credit goes to Burke):