With two congruent right triangles and one more right triangle, form a right trapezoid and find its area, just like former United States President James A. Garfield did.
See solution.
Practice makes perfect
Let's begin by considering a right triangle with sides a, b, and c.
Next, we will consider a second right triangle with the same dimensions as the above triangle and we will place it as shown below.
Let's draw a segment connecting two vertices in such a way that we get a trapezoid. Also, let's label some angles.
By the Triangle Sum Theorem, we can write the equation below.
m ∠ 1 + m ∠ 2 + 90^(∘) = 180^(∘)
⇓
m ∠ 1 + m ∠ 2 = 90^(∘)
Also, by the Angle Addition Postulate we get the following equation for the straight angle.
m ∠ 1 + m ∠ 2^(90^(∘)) + m∠ 3 = 180^(∘)
⇓
m∠ 3 = 90^(∘)
In consequence, the right trapezoid we've constructed is formed by three right triangles.
Now, let's find the area of the trapezoid above by summing the areas of the triangles.
A_(trap) = 1/2 a b + 1/2 a b + 1/2 c^2
⇓
A_(trap) = 2 a b + c^2/2
Also, by using the formula to find the area of a trapezoid, we can write a second expression for the same area.
A_(trap) = (b_1+b_2)h/2
⇓
A_(trap) = ( a+ b)( a+ b)/2
Next, let's equate these two areas and simplify the resulting equation.
As we can see, we have obtained the relation established in the Pythagorean Theorem.
Extra
Note
The construction and proof we just did was made by James A. Garfield, who was the twentieth president of the United States. He made this proof in 1876, five years before he became president.
