Area of a Triangle: Half Base Times Height

A proof that the area of any triangle equals half the base times the height, using an enclosing rectangle argument.

Keywords: area, triangle, altitude, height, base, rectangle

Difficulty: beginner

The area of a triangle is given by the famous formula:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
    

Proof Strategy

Drop the altitude from vertex $A$ to base $BC$, meeting it at foot $F$. Construct a rectangle $BCDE$ with base $BC$ and height equal to $AF$.

The altitude splits the triangle into two right triangles. Each right triangle is exactly half of a sub-rectangle. So the total triangle area is half the rectangle area.

The Proof

Step 1: Drop the Altitude

Start with triangle $ABC$ and drop the altitude from $A$ perpendicular to base $BC$. The foot of the altitude is $F$.

Step 2: Half Base Times Height

Claim: The area of any triangle equals half the base times the height: $$\text{Area}(\triangle ABC) = \frac{1}{2} \cdot BC \cdot AF.$$ Here $BC$ is the base and $AF$ is the perpendicular height.

Proof sketch. Enclose the triangle in a rectangle of base $BC$ and height $AF$. The altitude $AF$ splits both the triangle and the rectangle into two halves; each sub-triangle is half its sub-rectangle (the diagonal cuts the rectangle in two). So the triangle is half the rectangle.

Step 3: Construct the Enclosing Rectangle

Construct rectangle $BDEC$ with base $BC$ and height equal to $AF$. The altitude $AF$ divides the rectangle into two smaller rectangles.

Step 4: Each Triangle is Half a Rectangle

Triangle $ABF$ is half of the left sub-rectangle $BDAF$: side $AB$ is the diagonal that splits the rectangle into two equal triangles.

Similarly, triangle $ACF$ is half of the right sub-rectangle $FAEC$: side $AC$ is the diagonal.

So the total triangle $ABC$ = half the left rectangle + half the right rectangle = half the full rectangle.

Step 5: Area $= ½ \times BC \times AF$

The full rectangle has area $BC \times AF$. The triangle is half of this:

\text{Area}(\triangle ABC) = \frac{1}{2} \times BC \times AF
    

This formula holds for any triangle and any choice of base.

Step 6: Apex on a Parallel Line

Corollary: If the apex $A$ slides along any line parallel to base $BC$, the triangle's area is unchanged.

The base $BC$ is fixed, and the perpendicular height (the distance between the two parallel lines) is constant, so $\text{Area} = \frac{1}{2} \cdot BC \cdot h$ stays the same.

Step 7: Same Apex, Collinear Bases

Corollary: Two triangles that share an apex and whose bases lie on the same line have area in proportion to their base lengths.

Both triangles have the same perpendicular height from the apex to the shared line, so $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot h$ scales linearly with the base.

Conclusion

The area of triangle $ABC$ with base $BC$ and altitude $AF$ is:

\text{Area}(\triangle ABC) = \frac{1}{2} \times BC \times AF
    

This formula works for any choice of base and corresponding height.