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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