Equivalence Theorem Between a Triangle and a Trapezoid

A trapezoid is equivalent to a triangle that has the same height (h) and a base equal to the sum of the two bases (B+b) of the trapezoid.
the theorem of equivalence between a trapezoid and a triangle

In other words, a trapezoid and a triangle have the same area if:

  • their heights are equal
  • the base of the triangle is equal to the sum of the bases of the trapezoid

In this case, they are equivalent polygons.

Two polygons are called equivalent when they do not overlap and cover the same area.

A Practical Example

In this example, we have a triangle and a trapezoid with the same height, h=4.

an example

The triangle has a base of 9, which is equal to the sum of the bases (6+3) of the trapezoid, so we can apply the equivalence criterion.

The area of the triangle is half the product of its base and height.

$$ A_t = \frac{b \cdot h}{2} = \frac{9 \cdot 4}{2} = \frac{36}{2} = 18 $$

Since they are equivalent polygons, the trapezoid also has the same area.

$$ A_z = 18 $$

Therefore, we can skip the calculation.

To verify this result, we can also calculate the area of the trapezoid using the geometric formula. The area of the trapezoid is half the product of the sum of its bases and the height. $$ A_z = \frac{(B_z + b_z) \cdot h}{2} = \frac{(6 + 3) \cdot 4}{2} = \frac{9 \cdot 4}{2} = \frac{36}{2} = 18 $$ The area is the same.

The Proof

Let's consider a triangle and a trapezoid.

example of a triangle and a trapezoid

Initially, the triangle and the trapezoid have the same height.

$$ \overline{CH} \cong \overline{GK} $$

Also, the sum of the two bases AB and CD of the trapezoid is equal to the base EF of the triangle.

$$ \overline{AB} + \overline{CD} \cong \overline{EF} $$

We need to prove that the polygons are equivalent, meaning they have the same area.

First, extend the base AB of the trapezoid with a segment BL equal in length to the shorter base CD.

extend the longer base AB of the trapezoid with the shorter base CD

Then, draw the segment DL between points D and L.

the segment DL

The segment AL is equal to the base EF of the triangle because

  • BL≅CD by construction
  • AB+CD≅EF by the initial hypothesis, therefore AB+BL≅EF

Therefore, the segment AL≅EF is equal to the base of the triangle

$$ \overline{AL} \cong \overline{EF} $$

According to the triangle congruence criterion, the triangle ADL is equivalent to the triangle EFG because they have the same height CH≅GK and the same base AL≅EF.

the triangle ADL is congruent with EFG

The triangles MCD and MBL are also congruent because

  • A trapezoid always has parallel bases AB||CD, and the segment AL is simply the extension of AB.
  • The segments BL≅CD are equal by construction
  • The angles adjacent to the bases BL and CD are equal by the parallel line theorem, because they are alternate interior angles of a transversal intersecting two parallel lines.

Therefore, the triangles MCD and MBL are congruent based on the second triangle congruence criterion.

$$ MCD \cong MBL $$

Since the triangles MCD and MBL are congruent, they are also equivalent, meaning they have the same area.

$$ MCD \doteq MBL $$ 

As a result, we can view the trapezoid ABCD as the sum of the quadrilateral ABMD and the triangle CDM.

$$ ABCD \doteq ABMD + MCD $$

the trapezoid

Similarly, we can view the triangle ADL as the sum of the quadrilateral ABMD and the triangle MBL.

$$ ADL \doteq ABMD + MBL $$

the triangle EFG

Since the triangles MCD and MBL are equivalent, we conclude that both the trapezoid and the triangle are composed of the same equivalent areas.

$$ ABCD \doteq ABMD + MCD \doteq ABMD + MBL \doteq ADL $$

Therefore, being equidecomposable, the trapezoid ABCD and the triangle ADL are equivalent polygons.

$$ ABCD \doteq ADL $$

Knowing that the triangle ADL is equivalent to the triangle EFG, we deduce that the trapezoid ABCD and the triangle EFG are equivalent.

$$ ABCD \doteq ADL \doteq EFG $$

$$ ABCD \doteq EFG $$

In this way, we have proven the initial theorem.

Alternative Proof

Let's try to prove the theorem using logical-mathematical reasoning.

The area of a triangle is half the product of its base and height.

$$ A_t = \frac{b_t \cdot h_t}{2} $$

The area of a trapezoid is half the product of the sum of its bases and the height.

$$ A_z = \frac{(B_z+b_z) \cdot h_z}{2} $$

We need to determine the conditions under which the triangle and the trapezoid have the same area.

$$ A_t = A_z $$

$$ \frac{b_t \cdot h_t}{2} = \frac{(B_z+b_z) \cdot h_z}{2} $$

Assuming the two polygons have the same height ht=hz

$$ \frac{b_t \cdot h_t}{2} = \frac{(B_z+b_z) \cdot h_t}{2} $$

Additionally, assuming the sum of the bases of the trapezoid Bz+bz is equal to the base of the triangle bt

$$ \frac{b_t \cdot h_t}{2} = \frac{b_t \cdot h_t}{2} $$

The two formulas are the same.

This proves that the triangle and the trapezoid have the same area, meaning they are equivalent, when they have the same height and the sum of the bases of the trapezoid is equal to the base of the triangle.

And so on.

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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Surfaces (Geometry)