The same plot can appear as two triangles or as a rectangle between two wedges. With the same measurements and no overlapping pieces, both calculations must agree. That is more useful than guessing a formula from a silhouette.
Area adds when we partition a figure without overlap; moving or rotating a piece preserves its area. Many area formulas therefore follow from rectangles and triangles. We use quadrilaterals, trigonometry and the sine and cosine laws.
The unit is a square, not a segment
One square metre is the area of a 1×1 m square. A rectangle with sides a,b has S=ab. For whole-number lengths we count unit squares; for rational lengths we subdivide the grid. The formula also holds for irrational lengths, although a finite grid alone does not prove that extension.
Since 1 m=100 cm, 1 m²=10,000 cm²: we apply the factor 100 twice. Perimeter instead has a length unit. Rectangles 4×3 and 6×2 have equal area 12, but perimeters 14 and 16.
Equal-area figures need not be congruent as whole figures. Equidecomposable figures can be cut into finitely many pairwise congruent pieces and rearranged into one another. Such a rearrangement explains equal areas. AMSI, dissection.
From rectangles to parallelograms and triangles
Cut the corner triangle off the illustrated parallelogram and move it to the other end. The resulting rectangle has the same base a and height h, so S=ah. The sloping side does not replace h: we need the perpendicular distance between the base lines.
Two congruent copies of any triangle make a parallelogram with the same base and height. Thus the triangle occupies half:
S_(△) = (ah)/(2).
An altitude may fall outside the triangle. The formula still works: the distance from the vertex to the line containing the base remains the height.
Cut and translate
Trapezoid: 12 + 6 = 18
Same trapezoid: 3 + 12 + 3
Heron and height: 7–8–9
A trapezoid, two partitions
A trapezoid’s diagonal divides it into two triangles. Their bases are the parallel sides a,b; their height is the same h. Therefore:
S = (ah)/(2) + (bh)/(2) = (a + b)/(2)h.
Our own example has bases 8 m and 4 m, height 3 m. The diagonal partition gives 12+6=18 m². The vertical partition gives a 4×3 rectangle and two right triangles with legs 2 and 3: 12+3+3=18 m². These describe one area, rather than a coincidence between formulas.
For a general simple polygon, seek a partition into triangles with disjoint interiors. Subtract any hole once. A hidden trap is choosing different base lines for two parts while assigning them the same numerical height without justification.
A rhombus: height or diagonals
A rhombus is a parallelogram, so S=ah. Its perpendicular diagonals d₁,d₂ bisect each other and form four right triangles. Adding their areas yields:
S_(rhombus) = (d_(1)d_(2))/(2).
Omitting the half doubles the answer. Perpendicularity matters: do not transfer this formula to another quadrilateral’s diagonals without checking it.
Two sides and the included angle
In a triangle, sides a,b enclose angle C. The height to a is b sinC, hence:
S = (1)/(2)absinC.
This works for 0°<C<180°, including obtuse angles: the sine is positive. Scaling all lengths by k scales area by k² and perimeter by k.
Likewise, four triangles around the diagonal intersection of a convex quadrilateral give S=d₁d₂sinφ/2, where φ is the angle between the diagonals. A rhombus has φ=90°, recovering half their product.
For a regular n-gon, add n congruent central triangles. With perimeter P and apothem r, S=Pr/2. It is the same principle, rather than an unrelated formula.
Three sides: Heron’s formula
First check the triangle inequalities. Positive a,b,c must each be less than the sum of the other two. Let p=(a+b+c)/2 be the semiperimeter. Then:
S = √(p(p − a)(p − b)(p − c)).
For sides 7,8,9 m, p=12 m and S=√(12·5·4·3)=12√5≈26.83 m². An independent check uses base 9 m and height 8√5/3 m: 9·(8√5/3)/2=12√5 m². Four lengths multiplied under the root have units m⁴; taking the square root gives m². OpenStax, Heron’s formula.
Other known dataOptional · safe to skip
If a triangle’s circle radius is known
Joining the incenter to the vertices partitions the triangle into three triangles of height r, the inradius. Thus S=(a+b+c)r/2=pr. For circumradius R, the sine law gives sinC=c/(2R), hence S=abc/(4R). The two circle centers generally differ, as explained in G009.
Our 7–8–9 triangle has area 12√5, so r=S/p=√5 and R=abc/(4S)=10.5/√5. Each formula describes the same area using different known data.
Derivation detailOptional · safe to skip
Heron from the cosine law
From S=ab sinC/2 and cosC=(a²+b²−c²)/(2ab), we obtain:
16(S)^(2) = 4(a)^(2)(b)^(2) − ((a)^(2) + (b)^(2) − (c)^(2))^(2).
The difference of squares factors into (a+b+c)(a+b−c)(a−b+c)(−a+b+c), or 16p(p−a)(p−b)(p−c). Divide by 16 and take the positive root. Trigonometric derivation, University of Georgia.