Lauren Mofield
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InterMath
Title
Triangle Inside a Rectangle
Problem Statement
A triangle has two shared vertices and one shared side with a rectangle. The third vertex is
anywhere on the side opposite of the shared side (see figures above).
Problem setup
In this problem, I want to find the find the relationship that exists between the area of a rectangle
and the area of a triangle that is inside the rectangle. The triangle and rectangle share a side, and
the triangle’s third vertex is positioned on the opposite side of the rectangle.
When I student taught at Oak Hill Middle school, my seventh grade class used geo-boards to
discover the area of a rectangle and triangle. The students had to divide a triangle into a right
triangle and form a rectangle around it. The students then found the area of both figures to
determine the existing relationship.
Here are three examples of how to surround a right triangle with a rectangle:
You can also divide a triangle into right triangles, form rectangles around each triangle, and then calculate
the areas of the rectangles:
In each case, the area of the triangle is half the area of the rectangle that surrounds it. (Annenberg Media)
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Plans to Solve/Investigate the Problem
Prediction: The triangle is exactly ½ of the rectangle because, visually, it looks like ½.
To begin the problem, I will construct, first a square, then the triangle inside the square. Once
the figures are constructed, I will determine of each area separately. Constructing these figures
in GSP allows me to use area formulas to determine each area. The two formulas I will use are
(for the triangle) A = ½ x base x height and (for the rectangle) A= length x width. Knowing the
two areas will allow me to determine whether or not the ratio between the area of the triangle and
rectangle is constant. Lastly, I will prove my point.
Investigation/Exploration of the Problem
I. Constructing a rectangle
1. Use the segment, ray, or line tool to construct a line segment.
2. Use the select tool to select one vertex and the line segment; construct a perpendicular
line; unselect everything and re-select the line segment and the other vertex—construct
another perpendicular line
3. make a point on one of the perpendicular lines
4. Select the line segment and the point made in #3—construct a parallel line and then add
the last point to complete the rectangle
D
C
A
l ine segmen t AB
B
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5. The rectangle is made up of three lines and one line segment. Individually select two
points that make up one of the three lines to form segments. Using the tool bar, construct
a segment (three different times).
6. Hide the parallel line and the perpendicular lines to make the picture look like a rectangle
(Display, hide parallel line, display, hide perpendicular lines)
D
C
l ine segme nt AB
A
B
II. Constructing a triangle (within the rectangle)
1. make point E on line segment CD
2. highlight point A, E, and B and Construct, segments
C
E
D
A
l ine segme nt AB
B
Moving point E around helps create a visual explanation to reveal that the triangle is in fact,
½ the area of the rectangle.
C
E
D
C
A
l ine segme nt AB
B
A
l ine segme nt AB
DE
E
C
B
A
D
l ine segme nt AB
B
III. Finding Area
1. To find the area of the rectangle: highlight points C and A or points D and B (for the
width) and Measure  distance. Do the same for one of the line segments.
CA = 2.06 cm
DB = 2.06 cm
AB = 4.71 cm
CD = 4.71 cm
CAAB = 9.72 cm 2
Line segments CA and DB are parallel, as are line segments AB and CD. Using one of each
of the elements for the length and the width (because they are parallel) will give the area of
the rectangle.
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2. Measure  calculate…click on the first distance found in #1 (for ex., CA=2.06cm) and
press the multiplication button (*) and insert the second distance found (AB= 4.71cm).
Press equals and the area for the rectangle is complete
CAAB = 9.72 cm 2
3. To find the area of the triangle, make a perpendicular line through the triangle.
C
E
D
A
l ine segme nt AB
B
The perpendicular line is parallel to line segment CA and line segment DB, meaning
that the width of the rectangle is equal to the height of the triangle. The length of the
rectangle is the same as the base of the triangle.
A= ½ (b x h)
CAAB 
2
= 4.86 cm 2
Using the area formula for a triangle reveals an answer of 4.86 square cm, which is half the
area of the rectangle, 9.72 square cm.
III. Ratio Explored
1. The ratio between the area of the rectangle and the area of the triangle proves the
prediction: the triangle is exactly ½ of the rectangle because, visually, it looks like ½.
However, using formulas to calculate the area has mathematically proven the
prediction true.
2. Selecting a vertex to make the rectangle smaller or larger illustrates how the area of
the rectangle and area of the triangle continue to prove the prediction true.
Lauren Mofield
Write Up 1
CA = 4.58 cm
DB = 4.58 cm
AB = 4.71 cm
CD = 4.71 cm
CAAB = 21.56 cm 2
C
CAAB 
2
E
5
D
= 10.78 cm 2
A
CA = 0.98 cm
DB = 0.98 cm
AB = 1.01 cm
CD = 1.01 cm
CAAB = 0.98 cm 2
l ine segme nt AB
B
C E D
A l ineBsegme nt AB
CAAB 
2
= 0.49 cm 2
Looking at both examples above, the area of the triangle remains half the area of the rectangle,
no matter what the change in length and width of the figures.
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Extensions of the Problem
What if we extend the sides of the rectangles to lines HI and DF rather than line segments HI and
DF.
H
E
D
I
G
j
F
Q: If the vertex of the triangle can be anywhere on line HI, will the same relationship hold?
A: Yes. The area inside the rectangle remains the same throughout the transformations. The
lines are still parallel, meaning the new vertex is on the same exact line, meaning the area
formulas still hold true. The width of the rectangle is equal to the height of the triangle. The
length of the rectangle is the same as the base of the triangle.
Author & Contact:
Lauren Mofield
Junior student at Georgia College and State University
Middle Grades Education Major
Darlnlulu4@yahoo.com