Derivation of the Proof of the Area of a Rectangle

In getting the areas of figures, calculating the area of a square is probably the easiest. We can easily calculate the area of a square by squaring its side length . For example, a square with side length 5 units has area 25 square units. What about the area of a rectangle?

Based on the finding the area of a square, our intuition tells us that it is the product of the length and the width. However, how do we prove that it is indeed true? In this post, we are going to discuss the proof of the area of a rectangle. We show that it is the product of its length and its width.

Theorem: The area of a rectangle is the product of its length and width.

Consider the square below with side length $latex x + y$ units.  The square is divided into four parts: two squares and two rectangles. We already know that the area of the two squares are $latex x^2$ and $latex y^2$. We do not know the area of the rectangle yet because that is what we are trying to prove.

Derivation of the Proof of the Area of a Rectangle

Now, let $latex A$ be the area of each rectangle shown above. Clearly, the area of the largest square is the sum of the areas of the two smaller squares and the two rectangles. In equation form, we have

$latex (x + y)(x + y) = x^2 + y^2 + 2A$.

Expanding the left hand side, we have

$latex x^2 + 2xy + y^2 = x^2 + y^2 + 2A$.

Subtracting x^2 + y^2 from both sides results to

$latex 2xy = 2A$.

Solving for $latex A$ gives us

$latex A = xy$.

But $latex x$ and $latex y$ are the length and width of the rectangle, therefore, the area of any rectangle is the product of its length and its width.

Reference: Geometry by Edwin E Moise and Floyd L. Downs, Jr.

7 thoughts on “Derivation of the Proof of the Area of a Rectangle

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