The Investigation
Explain why rectangle DFGI has the same area as
triangle ABC.
First let’s measure the area of each figure.
Notice for this particular triangle the area of the
two figures we are investigating is 24 cm².
Next let’s construct an altitude of triangle ABC from
point B. This will be the height of
triangle ABC.
*Remember the formula for the area of a triangle is:
A = ½ base times height
Or
A = ½ (AC)(BJ)
And,
The formula for the area of a rectangle is:
A = length times width
Or
A = (FD)(DI)
You can clearly see from our construction that segment
BJ which we are calling the height of triangle ABC is equal in length to
segments FD and GI. Therefore we can
conclude that the height of triangle ABC is equal to the length of rectangle
DFGI. So,
BJ = FD
Let’s substitute this into the formula for the area of
rectangle DFGI:
A = (BJ)(DI).
Since we are trying to explain/show that the two areas
are equal we want to show that:
DI = ½ (AC),
if this is so we have shown that the two areas of each
figure are equal to each other.
If a mathematics teacher was doing this investigation
with their students they may want the students to create this sketch in GSP and
start to drag sides of triangle ABC, and then measure the sides to conclude
different postulates. Maybe even see if
the areas stay the same for different types of triangles.
Obtuse
What do you notice?
Acute
What do you notice?
Right
What do you notice?
Let’s create a segment from E to H.
In the figure above, we know from our construction
that EH is parallel to AC. Then the triangles ABC and EBH are similar by
Angle-Angle. This is true because angle BEH is congruent to angle BAC because
they are corresponding angles, and angle ABC and angle EBH are congruent since
they are identical.
Since point E and H are midpoints of segments AB and
BC then EB is ½ the length of segment AB and HB is ½ the length of segment
CB. Since triangles ABC and EBH are
similar (Click here
for a definition of similar figures) then EH has to be ½ the length of segment
AC, and we can clearly see that EH is equal in length to DI therefore,
DI = ½ (AC)
We have now shown that the areas are equal!
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