Area of triangles

A rectangle's area is straightforward , count rows of unit squares. A triangle, with a slanted side, is trickier. But here is the surprise: every triangle is exactly half of a rectangle that bounds it. So the area of a triangle is always

$\boxed{\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}.}$

This single formula handles all triangles , right, acute, even sprawling obtuse ones , and we will see why.

Concept

The half-rectangle picture. Draw a rectangle $ABCD$ and mark a point $X$ on side $AB$. Join $X$ to $D$ and to $C$. The triangle $\triangle XDC$ together with the two right triangles outside it ($\triangle AXD$ and $\triangle XBC$) fill the rectangle. By symmetry (cutting along the diagonals or by a careful argument), the area of $\triangle XDC$ is exactly half the rectangle's area:

$\text{Area}(\triangle XDC) = \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} \times \text{base} \times \text{height}.$

Base and height , meaning.

• The base is any side of the triangle (you may choose).
• The height (or altitude) is the perpendicular distance from the opposite vertex to the line through the base.

For a right triangle, the two legs serve as natural base and height. For a non-right triangle, you may need to drop a perpendicular from a vertex to the base line , and that perpendicular might fall outside the triangle (this happens for obtuse triangles). That is fine; the formula still works.

Why the formula works for all triangles.

• For an acute triangle, drop a perpendicular from a vertex; you cut the triangle into two right triangles, both inside a single rectangle. Their areas add up to half the rectangle.
• For a right triangle, two sides are the base and height; the rectangle is the obvious one with two legs.
• For an obtuse triangle (with a vertex sticking out past the base), the altitude falls outside the base. Using the difference of two right-triangle areas gives the same $\dfrac{1}{2} \times b \times h$ formula. So the formula is genuinely universal.

Two important corollaries.

• A triangle and the median to one of its sides divides the triangle into two triangles of equal area (same base half-length, same height).
• The four triangles formed by the two diagonals of a rectangle have equal areas in opposite pairs, and in fact all four are equal because the diagonals bisect each other and form base-pairs of equal length.

Worked examples

Example 1. A triangle has base $7$ cm and height $4$ cm. Find its area.

• Area $= \dfrac{1}{2} \times 7 \times 4 = 14 \mathrm{cm}^2$.

Example 2. A right triangle has legs $5$ cm and $12$ cm. Find its area.

• The legs serve as base and height. Area $= \dfrac{1}{2} \times 5 \times 12 = 30 \mathrm{cm}^2$.

Example 3. A diagonal of a rectangle splits it into two triangles. If the rectangle is $8 \times 5$ cm, find each triangle's area.

• Each triangle $= \dfrac{1}{2} \times 5 \times 8 = 20 \mathrm{cm}^2$.

Example 4. Triangle $ABC$ has $AB = 4$ cm, $BC = 5$ cm, and the altitude from $A$ onto $BC$ measures $3$ cm. Find the area. Then find the altitude from $B$ onto $AC$, given $AC = 4$ cm.

• Area $= \dfrac{1}{2} \times 5 \times 3 = 7.5 \mathrm{cm}^2$.
• Using $AC$ as the new base: $7.5 = \dfrac{1}{2} \times 4 \times h$ so $h = 3.75$ cm.

Try it yourself

1. Triangle with base $10$ cm, height $6$ cm. Area?
2. A right triangle has legs $9$ cm and $12$ cm. Area?
3. A triangle has area $24 \mathrm{cm}^2$ and base $8$ cm. Height?
4. A rectangle $6 \times 10$ has both diagonals drawn. Area of each of the four triangles?
5. A triangle inside a $5 \mathrm{cm} \times 5 \mathrm{cm}$ square has the same base as the square and its vertex on the opposite side. Area?
6. Triangle $\triangle XDC$ sits inside a rectangle $ABCD$ of $7 \times 4$. Area of $\triangle XDC$?
7. The median from $A$ in $\triangle ABC$ divides $BC$ into two halves. Show that the two smaller triangles have equal areas.
8. A triangular flag has area $48 \mathrm{cm}^2$ and base $16$ cm. Find the height.

Activity

Cut-and-paste proof. Cut a rectangle from paper. Draw a diagonal and cut along it. You now have two right triangles. Stack them on top of each other , they exactly overlap (they are congruent). Each has half the rectangle's area , that's the formula in action. Now try drawing any other triangle inside the rectangle with one side along the rectangle's base. Cut the triangle out. Can you rearrange the two leftover pieces to also form a triangle congruent to the one you cut? If yes, you've shown both halves of the rectangle's area belong to the triangle and its leftover , confirming the half formula.