Area of a Quadrilateral by Triangulation

A quadrilateral can have any shape , even highly irregular ones. There is no single area formula that works for all quadrilaterals. The universal strategy is triangulation: draw one diagonal, split the quadrilateral into two triangles, find each area separately (often using Heron's formula or $\tfrac{1}{2} b h$), and add. This lesson teaches the technique.

The procedure

Given a quadrilateral $ABCD$ with side lengths and at least one diagonal length known:

1. Draw a diagonal , say $AC$ , that divides $ABCD$ into $\triangle ABC$ and $\triangle ACD$.
2. Apply Heron's formula (or any suitable formula) to find Area$(\triangle ABC)$ using sides $AB, BC, AC$.
3. Apply Heron's formula to find Area$(\triangle ACD)$ using sides $AC, CD, DA$.
4. Add the two areas: Area$(ABCD)$ $=$ Area$(\triangle ABC)$ $+$ Area$(\triangle ACD)$.

What if no diagonal length is given? Then you cannot apply Heron's formula directly. You will need to find the diagonal length using other geometric facts (e.g. Pythagoras, parallelogram properties) before applying Heron's formula.

A worked walk-through

A quadrilateral $ABCD$ has $AB = 9, BC = 40, CD = 28, DA = 15$, and diagonal $AC = 41$.

Step 1. Diagonal $AC = 41$ divides the quadrilateral into $\triangle ABC$ (sides $9, 40, 41$) and $\triangle ACD$ (sides $41, 28, 15$).

Step 2. For $\triangle ABC$ with sides $9, 40, 41$: Note $9^2 + 40^2 = 81 + 1600 = 1681 = 41^2$, so this is a right triangle. Area $= \tfrac{1}{2} \cdot 9 \cdot 40 = 180$ sq units.

Step 3. For $\triangle ACD$ with sides $41, 28, 15$: semi-perimeter $s = 42$. $s - 41 = 1, s - 28 = 14, s - 15 = 27$. Product $= 42 \cdot 1 \cdot 14 \cdot 27 = 15876$. Area $= \sqrt{15876} = 126$ sq units.

Step 4. Total area $= 180 + 126 = 306$ sq units.

Special quadrilaterals

For some shapes there are dedicated formulas , no triangulation needed.

• Square of side $a$: Area $= a^2$.
• Rectangle of sides $a$ and $b$: Area $= a \times b$.
• Parallelogram of base $b$ and height $h$: Area $= b \times h$.
• Rhombus with diagonals $d_1$ and $d_2$: Area $= \tfrac{1}{2} d_1 d_2$.
• Trapezium with parallel sides $a$ and $b$ and distance $h$ between them: Area $= \tfrac{1}{2} (a + b) h$.
• Kite with diagonals $d_1$ and $d_2$: Area $= \tfrac{1}{2} d_1 d_2$.

For a general (non-special) quadrilateral, triangulation is the method.

Worked examples

Example 1. Find the area of a rhombus with diagonals $24$ and $10$.

Area $= \tfrac{1}{2} \cdot 24 \cdot 10 = 120$ sq units.

Example 2. A trapezium has parallel sides $8$ and $12$ cm with distance $5$ cm between them. Find the area.

Area $= \tfrac{1}{2}(8 + 12) \cdot 5 = 50$ sq cm.

Example 3. A quadrilateral $ABCD$ has $AB = 3, BC = 4, CD = 4, DA = 5$, and $\angle ABC = 90^\circ$. Find the area.

$\triangle ABC$ has legs $3$ and $4$ and hypotenuse $AC$. By Pythagoras, $AC = 5$. Area of $\triangle ABC = \tfrac{1}{2} \cdot 3 \cdot 4 = 6$ sq units. $\triangle ACD$ has sides $5, 4, 5$. Heron's: $s = 7$, $s - 5 = 2, s - 4 = 3, s - 5 = 2$. Product: $7 \cdot 2 \cdot 3 \cdot 2 = 84$. Area: $\sqrt{84} = 2\sqrt{21}$ sq units. Total: $6 + 2\sqrt{21}$ sq units.

Example 4. A field is in the shape of a rectangle with semicircles attached on two opposite sides. The rectangle has dimensions $20 \times 14$. Find the total area. (Use $\pi \approx \tfrac{22}{7}$.)

Two semicircles together make one full circle of diameter $14$ (= width of the rectangle), so radius $7$. Circle area $= \pi r^2 = \tfrac{22}{7} \cdot 49 = 154$. Rectangle area $= 20 \cdot 14 = 280$. Total: $434$ sq units.

Example 5. Find the area of a quadrilateral $ABCD$ where $AB = 7, BC = 6, CD = 6, DA = 4$, and diagonal $AC = 8$.

$\triangle ABC$ sides $7, 6, 8$: $s = 10.5$. $s - 7 = 3.5, s - 6 = 4.5, s - 8 = 2.5$. Product: $10.5 \cdot 3.5 \cdot 4.5 \cdot 2.5 = 413.4375$. Area: $\sqrt{413.4375} \approx 20.33$.

$\triangle ACD$ sides $8, 6, 4$: $s = 9$. $s - 8 = 1, s - 6 = 3, s - 4 = 5$. Product: $9 \cdot 1 \cdot 3 \cdot 5 = 135$. Area: $\sqrt{135} \approx 11.62$.

Total: $\approx 31.95$ sq units. (Exact: $\sqrt{413.4375} + \sqrt{135}$.)

Try it yourself

1. Find the area of a rhombus with diagonals $12$ and $16$.
2. Find the area of a trapezium with parallel sides $10, 14$ and height $7$.
3. A quadrilateral $ABCD$ has $AB = 5, BC = 12, CD = 14, DA = 15$, and diagonal $AC = 13$. Find the area.
4. A field is in the shape of a rectangle of $50 \times 30$ m, with a triangular plot of base $30$ m and height $20$ m attached. Find the total area.
5. Find the area of a square of side $7$.
6. A quadrilateral has all sides $5$ and one angle $90^\circ$. Find the area.
7. Find the area of a rectangle of length $25$ and breadth $14$.
8. A quadrilateral $ABCD$ has $AB = 9, BC = 12, CD = 5, DA = 8$, and diagonal $AC = 15$. Find the area.
9. Find the area of a parallelogram of base $14$ and height $8$.
10. The roof of a temple is shaped like a quadrilateral with sides $12, 35, 13, 36$ m and one diagonal $37$ m. Find the area (use Heron's formula on each half).

Pitfalls / Insight

• Always identify which diagonal to draw. Sometimes only one diagonal makes the triangle sides nice.
• Use the right formula for each half. Right triangles use direct $\tfrac{1}{2} ab$; general triangles use Heron.
• Check the data. Triangle inequality must hold for each half.

Insight. Any quadrilateral, however irregular, becomes manageable once you split it into two triangles. This is the universal area technique , and once it is in your toolkit, no exam question on quadrilateral areas can stump you.