InterMath
Title
Triangle Inside a Rectangle
Problem Statement
A
C
E
B
D
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 figure above).
Problem setup
I am given a rectangle with a triangle in the middle. One side of the triangle shares the entire side
of the rectangle. The other vertex of the triangle is located anywhere on the parallel line segment,
which is on the opposite side of the rectangle from the shared side. I am trying to figure out how
the area of the rectangle and triangle compare.
Plans to Solve/Investigate the Problem
Prediction: The area of the triangle seems to be 1/2 that of the rectangle because by looking at
the rectangle 1/2 seems to be correct.
In order to determine whether or not the prediction is correct, first I plan to construct a replica of
the figure in GSP. After I have made the rectangle and triangle, I will make sure all the lengths
of the opposite sides of the rectangle are congruent. Then I measure the areas by using the
calculate feature in GSP. In order to find the area, I will use the measurements of the needed
sides for the formula. To find the area of a rectangle the formula is A= length * width, where
“A” equals the area in units squared, or in this case we are using centimeters squared. The area of
a triangle is A= ½ * base * height. The base will be the same as the length of the rectangle due to
the fact the triangle shares the side with the rectangle. The height is found by constructing a right
angle at a vertex and measuring the new constructed length from the vertex to where it crosses a
perpendicular line that is run from a different vertex. Measuring proves how the areas compare. I
will also show how different cases like moving a vertex of the triangle or making the rectangle
larger will also be true.
Investigation/Exploration of the Problem
To form a rectangle, I first place two points on a blank sketchpad. Then I connect the
points with a segment and named the points A and B. By selecting the segment, I then construct
two perpendicular lines from the line segment AB. I place a point on the perpendicular line going
through point A. I name the point C. Then I make another perpendicular line through point C.
When the new perpendicular line is formed, the intersection where it crosses through the line that
runs through B is where point D is located.
A
B
A
B
C
D
A
C
E
A
B
C
D
After all the perpendicular lines are formed, I make line segments
between AC, CD, and BD and hide my perpendicular lines until
later. After doing this, I have rectangle ABCD.
B
D
To construct a triangle inside the rectangle, I will need to decide
which side the triangle can share with the rectangle. I choose to
use line segment CD. To make a triangle, I put a point on line
segment AB and name it E. Then I construct line segments from
C to E and D to E.
A
E
B
Vertex E can be moved anywhere along line segment AB.
C
m BA = 4.23 cm
m CD = 4.23 cm
m AC = 1.24 cm
m BD = 1.24 cm
m CD = 4.23 cm
m EC = 3.70 cm
m ED = 1.44 cm
D
After the figure is first created, with E in its place that I originally constructed,
the lengths must be measured. Congruent lengths on opposite sides of the
rectangle prove the rectangle is indeed a rectangle. A rectangle has two pairs of
parallel sides, and the parallel sides are congruent. Line segments AB and CD
are congruent as well as parallel. Line segments AC and BD are also congruent
and parallel, but they are also perpendicular to the above line segments, which
makes the rectangular shape.
The triangle has different lengths of the sides depending on where point E is
located. It is important to note that the measurement of line segment CD is equal
in measure to the line segment CD in the rectangle, since they share the segment.
To prove the area of the triangle is ½ of the rectangle, the areas of both shapes must be
calculated next. The area of a rectangle is length times the width. Using segments AB and CD as
the length and segments AC and BD for the width are vital. The area of the given rectangle is
4.23 centimeters times 1.24 centimeters, which is 5.24 centimeters squared. The calculate
function of GSP aides in finding the area. Area=
2
m BAm AC = 5.24 cm
The area of a triangle equals .5 times the base times the height. The base of the triangle in the
given case is segment CD, so it is 4.23 centimeters. From point E, the height is found by
constructing a 90 degree angle perpendicular to the base. The height is the same as line segments
AC and BC since the point lies on the same line segment as A and B and extends to the line
segment C and D lie on. The line segment EF measures to be 1.24 centimeters. The area of the
triangle is found by multiplying .5, 4.23 centimeters, and 1.24 centimeters. Using GSPs calculate
0.5m CDEF = 2.62 cm 2
function I was able to find the area of the triangle.
Area=
A
A
E
C
F
B
E
B
F
D
F
C
D
EF = 1.24 cm
The area of the rectangle, 5.24 centimeters squared is two times the area of the triangle, which is
2.62 centimeters squared. The areas can be set up as a proportion or divided to determine the
relationship between the shapes. Dividing proves the rectangle is two times as large as the
triangle and that the triangle is half the size of the rectangle.
5.23
2.62
A
C
B
E
F
D
2.62
= 2.00
5.23
= 0.50
My prediction that the triangles area is half the size of the
rectangle is correct. Even if point E moves along line segment
AB, the area will not change because the point remains on the line
segment that is parallel to the base. Since the point remains on the
same line segment, the height does not change, and so therefore
the area stays the same. If the length or height of the rectangle
changes, the areas change proportionally, but the triangle still
remains to be half the size of the rectangle.
EF = 1.24 cm
0.5m CDEF = 2.62 cm 2
m CDm AC = 5.23 cm 2
A
E
B
EF = 1.47 cm
0.5m CDEF = 3.69 cm 2
m CDm AC = 7.39 cm 2
2.62
5.23
2.62
= 2.00
5.23
= 0.50
F
C
D
Extensions of the Problem
What if we extend the sides of the rectangles to lines AB and CD rather than line segments AB
and CD? If the vertex of the triangle can be anywhere on line AB, will the same relationship
hold?
E
A
C
B
F
D
If the line segments are extended, the triangle will still be half the area of the rectangle. The
extended lines make the triangle proportional to the rectangle. As long as the rectangle extends
or reduces in length or width, so does the base and height of the triangle. Even if the triangle is
extended past the original rectangle ABCD, the base, or line segment CD, will be the same. The
triangle will still lie on the perpendicular line AB, so the height of the triangle will remain the
same no mater how far it is stretched. Place the vertex outside the rectangle and name it point F.
To get the height of the triangle beyond the length of the rectangle, draw a perpendicular line to
line CD from point F. The height is the length of the line segment from point F to the extension
of the line segment into a line, where the base lies. Measuring in GSP displays how the length,
height, and area remain the same even after the extension. Any extension will bring about the
same result. Any triangle that shares the base of one side of a rectangle and has a vertex on the
parallel line opposite will result in an area half that of the rectangle no matter the translation,
enlargement, or reduction.
A
B
F
m CD = 4.23 cm
m FG = 1.24 cm
C
Author and Contact
Laura Sims
Email
D
G
0.5m CDm FG = 2.62 cm 2