InterMath | Workshop Support | Write Up Template
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
I want to determine the relationship, if one exists, of the area of a triangle inscribed in a
rectangle to the area of that rectangle. It is meaningful to observe that the triangle shares
a side with the rectangle, and that the triangles third vertex lies somewhere on the
diametrically opposed side of the rectangle. This tells us that the triangle and rectangle
share a common base and height.
Plans to Solve/Investigate the Problem
For my investigation, I will use GSP to construct a rectangle with an embedded triangle.
I will then measure the areas of each and calculate the ratio of their areas so that I may
determine if a relationship does exist and if so what properties hold. I will do this by
using GSP’s measurement feature along with the area formulas for a triangle and a
rectangle, which are A=1/2*b*h and A=length*width, respectively. First I’ll move the
vertex of the triangle to see if that changes the area. After finding the areas of the
triangle and rectangle, I can find out if there is a constant ratio of their areas, which I
predict to be 2:1 (triangle:rectangle). I will then try to make a convincing argument as to
why this relationship will always be true in this case.
Investigation/Exploration of the Problem
I constructed a rectangle using parallel and perpendicular lines. In order to do this I first
had to create a line segment. I then constructed a perpendicular line to the segment (AB)
at point A. Then I constructed a perpendicular line to the segment (AB) at point B. I
then constructed a parallel line to the segment (AB). After that I created the triangle,
making sure that it fit the specifications of the problem.
AC = 5.03 cm
AB = 6.99 cm
Area of the
rectangle (l*w)
ABAC = 35.11 cm 2
Area of the
triangle (.5*b*h)
0.5ABAC = 17.56 cm 2
S
Ratio of the area of
the rectangle to the
area of the triangle.
D
C
ABAC
0.5ABAC
= 2.00
B
A
Area of shaded
in triangle.
If the shaded in triangle is
1/2 of the area of the
rectangle, then the two
triangles in the whitespace
of the rectangle should also
be 1/2 of the area of the
rectangle, which we can
see from the calculations.
Area of
triangle ACS
CS = 3.31 cm
0.5CSAC = 8.31 cm 2
SD = 3.68 cm
S
D
C
Area of
triangle BDS
0.5ABAC = 17.56 cm 2
0.5SDAC = 9.24 cm 2
Area of triangle
ACS + Area of
triangle BDS
0.5CSAC +0.5SDAC  = 17.56 cm 2
AC = 5.03 cm
B
A
AB = 6.99 cm
AC = 5.03 cm
AB = 6.99 cm
Area of the
rectangle (l*w)
Area of the
triangle (.5*b*h)
Ratio of the area of
the rectangle to the
area of the triangle.
ABAC = 35.11 cm 2
0.5ABAC = 17.56 cm 2
S
C
ABAC
0.5ABAC
D
= 2.00
A
B
As you can see, it did not matter where I moved my vertex, because the area of the
triangle was consistently half of the area of the rectangle. What makes this true is that the
height of the triangle never changed – the vertex of the triangle stayed on the line
segment CD. I only moved the vertex along the diametrically opposed side from the side
that shared the other two vertices. I can therefore conclude that the ratio will stay 2:1 no
matter how I manipulate the point (S) along the line segment. As we can see from the
second picture, the height of the two triangles in the white space of the rectangle will
always be the same as the height of the original triangle (in blue). Also, the base of the
original triangle will always be equal to the sum of the bases of the other two triangles no
matter where the vertex lies on the rectangle. Therefore we can conclude that a triangle
inscribed in a rectangle with two of its vertices shared and the third on the diametrically
opposed side of the rectangle will always have an area that is one-half that of the area of
the rectangle.
Extensions of the Problem
What if we extend the sides of the rectangles to lines CD and AB rather than line
segments CD and AB. If the vertex of the triangle can be anywhere on line CD, will the
same relationship hold?
Yes, the relationship will still hold as long as the third vertex stays somewhere on the line
CD. The third vertex does not have to stay in the rectangle; it just has to stay on the
extended line segment in order for the relationship to hold. The reason is because even if
the triangles third vertex is extended past the rectangle, the triangles area is still just half
of the rectangle because the triangle itself will be half of the rectangle.
C
E
A
D
G
B
Author & Contact
Tiffany Graham
tiffany_graham@ecats.gcsu.edu
Link(s) to resources, references, lesson plans, and/or other materials
Geometer Sketch Pad
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