Clickteam Fusion 2.5 Flash Application. Click on the Application to set the focus.
Deriving Quadratic Formulations From
Areas of The Square
Let the quote symbol (") indicate the squaring operation.
The first display in the grid to the right includes the red
perimeter of a square partitioned into four sub-areas by intersecting
blue lines. The intersection separates the blue lines into two segments.
The length of the longer segment is designated (a) and the length of the
shorter segment is designated (b). Therefore the four sub-areas will be
designated (a"), (b") and (ab) and the total area of the square bounded
by red lines (a + b)" equals (a" + b" + 2ab). This analysis of the
area of a square gives us the quadratic formulation (a + b)" = (a" + 2ab
+ b"). Note areas designated 2ab comprise four yellow right triangles.
Click the NEXT AREA button. Now the four yellow
right triangles are each nested in the four corners of the red lined
square. Note that the areas of the yellow triangles do not overlap, and
the longest sides of the triangles form the perimeter of an inner
square. Since the areas of the four triangles equal 2ab, the area of the
inner square must equal the total area of the red lined square minus 2ab
or (a + b)" - 2ab. In expanded terms (a" + b" + 2ab -2ab) = (a" + b").
If we designate the lengths of the long sides of the right triangle (c)
then we have (a" + b") = (c"). This formulation is the pythagorean
theorem for right triangles.
Click the NEXT AREA button. Now the four yellow
right triangles have been rotated into the (c") area in such a manner
that there areas do not overlap. Note that the lengths of the sides of
the inner most square equal the length (a) of the longest leg of the
right triangles minus the length (b) of the shortest leg of the right
triangles, (a - b). Since the areas of the four triangles equal 2ab, the
area of the inner most square must equal the area of the middle inner
square (a" + b") - 2ab. So we have the quadratic formulation (a - b)" =
a" - 2ab + b".
The first frame displays blue intersecting lines that bound B".
Press NEXT
AREA.
The second frame displays blue and red intersecting lines that
bound the
areas associated with (b"), 2C1 and (a"), C1=b*a.
Press NEXT
AREA.
The third frame displays red intersecting lines, yellow
intersecting lines that bisect B" and blue and green intersecting
lines that bound the
areas associated with (f"), 2C2 and (e"), C2=f*e.
Press NEXT
AREA.
The fourth frame displays all intersecting lines without labels.
STEP1: Left Click Application. Press Enter/Return.
STEP2:Counting From the Rightmost Gold Bead, Left Click a Gold Bead
at a position equal any number. Press Right Arrow (PRA). This step generates
(B), which equals the displaced gold beads. The blue and lower right
corner lines bound the area B" STEP3: Left click another
gold beads in or out of the displaced group (B). PRA. This step
generates (a), the bead clicked in step 3 plus the total number of gold
beads to its right, while (b) equals (B) - (a). If (B)>(a), the
quadrilateral of brown beads above and to the left of of the a-group
equals the product of (a) times (b), equals (C1) in the equation X" - BX
+ C1 = Y and C1 is positive. If (B)<(a) the quadrilateral of brown beads
above and to the right of the displaced group plus the square of the
number of gold beads in the displaced group equals (C1) and (C1) is
negative. The red, blue and lower right corner lines bound the
areas associated with (b"), 2C1 and (a"). STEP4: Left click
another gold bead in any group. PRA. This step generates (e), the
bead clicked in step 4 plus the total number of gold beads to its right,
while (f) equals (B)-(e). If (B)>(e), the quadrilateral of brown beads
above and to the left of of the a-group equals the product of (e) times
(f), equals (C2) in the equation X" - BX + C2 = 0 and C2 is positive. If
(B)<(e) the quadrilateral of brown beads above and to the right of the
displaced group plus the square of the number of gold beads in the
displaced group equals (C2) and (C2) is negative. The green, blue and
lower right corner lines bound the areas associated with (f"), 2C2 and
(e"). Press down arrow key to see step by step solution to the
equation. For more details about generating various configurations of parabolic equations on the
pythabacus
click here.
By the Application of Areas Gathering the
Information Necessary to Graph a Parabola in the Cartesian Plane
The roots of the parabola are given by the
lengths or sum of the segments of the intersecting green lines (f) and (e) and the y-intercept equals the area f*e which equals C2.
The
x-coordinate (h) of the vertex equals the bisection
of the length B or B/2, and each quadrant of B" defined by the
intersecting yellow lines includes the area (h") .
The y-coordinate (K)
of the vertex equals the sum of the areas with shared corners at the
intersection of the green lines, if one of the areas equals h" a quadrant of B"
with a corner at the intersection of
the yellow lines. Or the y-coordinate (K) of the vertex equals the
area of the square with a shared corner at the intersection of green
lines and the its opposite corner at the intersection of the yellow
lines. For example select the YAxis, go to the third
frame and see a large square (K=-201/4)
squares equal the sum of areas with a
shared corner at the intersection of green lines, including h" a
quadrant of B". Then click NEXT AREA and see a square (K =-1/4) squares with a shared corner at the
intersection of green lines and an opposite corner at the intersection
of yellow lines, included in h" a quadrant of B". The y-coordinate (K) of the vertex will always be
negative in these examples associated with the pythabacus.
The y-intercept
(C2=K+h") for K=-201/4 (third frame of YAxis) will equal the area of (K) that elbows
around an area equal a quadrant of B". In this case the elbow C2 equals
(-201/4 + 61/4) = -14
squares. Or (C2=K+h") for K =1/4 (fourth frame
of YAxis) will equal an area of a
quadrant of B" that elbows around the area equal (K). In this case the
elbow C2 equals (-1/4 + 61/4)
= 6 squares.
Therefore the y-coordinate (K) will also equal
the area (C2-h"). Now by the application of areas, all the
information necessary to graph a parabola in the cartesian plane, the
roots, y-intercept and coordinates of the vertex can be visually derived. Use the
Pythabacus above to explore more equations and areas.
Select an equation configuration and go to
the fourth frame.
If the sum of the segments or parts of the
green line is equal the length of the blue line, then the roots will
equal the lengths of the segments of the green line and the values
of both roots are positive.
If the sum of the segments or parts of the
green line is greater than the length of the blue line, then the
first root will equal the length of the segment by which the sum of
the green line is greater then the blue and its value is negative,
and the second root will equal the sum of the lengths of the
segments of the green line and its value is positive
Enter the value of the smallest root into
the top input box of the cartesian plane application (cpa) to the
right, and
enter the value of the largest root into the middle input box of the
cpa to the right.
As noted above, the y-coordinate (K)
of the vertex equals the sum of the areas with shared corners at the
intersection of the green lines if one area equals h" a quadrant of
B" with a corner at the intersection of
yellow lines.
Or the
y-coordinate (K) of the vertex equals the area of the square with a
shared corner at the intersection of green lines and its opposite
corner at the intersection of yellow lines.
The K value always has a
negative value for these examples.
Enter the value of the y-coordinate of the
vertex into the bottom input box of the cpa.
The value of the y-intercept equals root1
times root2 and is calculated by the application.
Press the GRAPH button to display graph of
parabola. (A length of each square in the plane equals (5) units)
The display will have a tolerance of +/- 1
unit length.
Explore other examples with the pythabacus
and application of areas above, then graph in the cartesian plane.
Identifying the Algebraic Elements of
Completing the Square by the Application of Areas
Click application and press enter.
Counting From the Rightmost Gold Bead, Left
Click a Gold Bead at a position equal any Number. Press Right Arrow
(PRA). This step generates (B), which equals the displaced gold
beads. Let B" designate B-squared.
Left click another gold beads in or out of
the displaced group (B). PRA. This step generates (a), the bead
clicked plus the total number of gold beads to its right.
Left click another gold bead in any group.
PRA. This step generates (e), the bead clicked in step 4 plus the
total number of gold beads to its right, while (f) equals (B)-(e)
C1 the third term of the equation will
equal the areas that elbow around an area equal a quadrant of B".
These areas will share corners at the intersection of the red lines
Or C1 will equal areas of a quadrant of
B"that elbows around the area of the square with a shared corner at
the intersection of red lines and its the opposite corner at the
intersection of yellow lines.
Below you will Press Down Arrow (PDA) to enter numbers in
equations.
And you will Press Up Arrow (PUA) to enter variables.
PDA twice.
C2 the third term of the equation equal (0)
will equal the areas that elbow around an area equal a quadrant of
B". These areas will share corners at the intersection of the green
lines
Or C2 will equal areas of a quadrant of
B"that elbows around the area of the square with a shared corner at
the intersection of the green lines and its opposite corner at the
intersection of the yellow lines.
In the next equation, after X"-BX+/-C2=0, h" equals a quadrant
of B" and h"-C2 equals (-K).
If C2 is negative, the y-coordinate (K)
of the vertex equals the sum of the areas with shared corners at the
intersection of the green lines if one area equals h" a quadrant of
B" with a corner at the intersection of
yellow lines.
If C2 is positive the y-coordinate (K) of
the vertex equals the area of the square with a shared corner at the
intersection of green lines and its opposite
corner at the intersection of yellow lines.
PDA twice.
(X-h) equals the square root of (-K) where X
is the length of a segment of the intersecting green lines and h is
a length of a side of a quadrant of B"
Since K always has a negative value in these
examples, (-K) kas a positive value.
Remember the X that equals the length of the
segment by which a green line exceeds the length of the blue
line has a negative value, otherwise the value of X is positive,
while the value of h is always positive.