Calculating the area of a triangle

 

The concept of calculating the area of a triangle has evolved over centuries, shaped by mathematical advancements and cultural influences. Here's a historical overview of how the calculation of the triangle's area has been approached across different eras:

1. Basic Formula (Using Multiplication):

Ø        Method: Egyptians, as early as 2000 BCE, used basic geometry to calculate areas for agricultural and construction purposes. They likely used a simplified version of the formula:

$$Area=\frac{1}{2} ×base×height$$

Ø        Significance: This approach would measure land for dividing fields along the Nile.

Example:

If the base b=5 and height h=15,

\(Area=\frac{1}{2} ×5×15=37.5\)

 

2. Using Heron’s Formula (Square Root and Multiplication):

Ø           Heron of Alexandria (10 CE) developed the formula to calculate the area of a triangle using its three sides:

For sides a, b, and c:

$$Area=\sqrt{s(s−a)(s−b)(s−c)}$$

where \(s = \frac{a+b+c}{2}\) is the semi-perimeter.

Ø         Significance: This method allowed for calculations without requiring the height of the triangle.

 

Example:

Sides are a=7, b=8, c=9:

\(s=\frac{7+8+9}{2}=12\)

$$Area=\sqrt{12(12−7)(12−8)(12−9)}$$

 \( Area=\sqrt{12(12−7)(12−8)(12−9)} =\sqrt{12×5×4×3} = \sqrt{720} ≈26.83\)

Fig2

3. Using trigonometry:

Advancements: Islamic mathematicians extended Greek and Indian methods.

Ø           Al-Biruni (973–1048 CE) used trigonometry to calculate areas for geographical measurements and surveying.

Ø            Development of spherical trigonometry enabled area calculations for triangles on curved surfaces (e.g., Earth or celestial spheres).

Ø            Method: Indian mathematicians developed sophisticated geometric and trigonometric methods. For example:

ü  The concept of right-angled triangles and the Pythagorean theorem is evident in the Sulbasutras (around 800 BCE).

ü  Early forms of trigonometric relationships were used to find areas, especially in astronomical applications.

 

To calculate the area of a triangle using trigonometry, we can use the formula:

$$Area=\frac{1}{2}absin(C)$$

Explanation:

1. a and b are the lengths of two sides of the triangle.
2. C is the angle between these two sides.
3. sin(C) is the trigonometric sine of the angle C.

Example:

Consider a triangle with:

ü  a=8, b=6, and angle \(C= 60^\circ\).

Using the formula:

\(Area=\frac{1}{2}⋅8⋅6⋅sin(60^\circ)=\frac{1}{2}⋅8⋅6⋅\frac{\sqrt{3}}{2}=\frac{48.\sqrt{3}}{2}=12\sqrt{3}≈20.78\)

  Applications:

ü  This formula is particularly useful when two sides and the included angle are known (SAS condition).

ü  It’s a direct application of trigonometric principles in geometry.

 

Fig3

 

4. Using Determinants (Coordinate Geometry):

Ø           Method: The development of coordinate geometry by René Descartes (1596–1650) introduced formulas using the vertices of a triangle:

 

For vertices \( (x_{1},y_{1}), (x_{2},y_{2}), and (x_{3},y_{3})\):

$$Area=\frac{1}{2}∣x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})∣$$

Ø         Significance: Made area calculations more precise for irregularly shaped land and in engineering.

 

Example:

Vertices are (0,0), (4,0), and (0,3):

\(Area=\frac{1}{2}∣x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})∣ = \frac{1}{2}∣0(0-3)+4(3-0)+0(0-0)∣ = \frac{1}{2}×12=6\)

 

Fig4

5. Using Integration (Calculus):

Ø          Method: The introduction of vectors and calculus enabled new ways to calculate the area:

When the triangle is under a curve y=f(x), the area can be found by:

$$Area=\int_{x_{1}}^{x_{2}}∣f(x)∣dx$$

Ø          Applications: Widely used in physics, computer graphics, and geospatial analysis.

Example:

Triangle under y=2x from x=0 to x=3:

\(Area=\int_{x_{1}}^{x_{2}}∣f(x)∣dx = \int_{0}^{3}∣2x∣dx = \int_{0}^{3}2xdx=\left[x^{2}\right]_{0}^{3} = 3^{2}-0^{2}=9\)

Divide by 2 (since it’s a triangle):

\(Area= \frac{9}{2} =4.5\)

 

Fig5

6. Using the Cross Product (Vectors):

Ø            Method: The introduction of vectors and calculus enabled new ways to calculate the area:

For vectors \(\overrightarrow{u}=⟨u_{x},u_{y}⟩ , \overrightarrow{v}=⟨v_{x},v_{y}⟩\)

$$Area=\frac{1}{2}∣u_{x}v_{y} − u_{y}v_{x}∣$$

Ø  Applications: Widely used in physics, computer graphics, and geospatial analysis.

Example:

If \(\overrightarrow{u}=⟨4,0⟩ , \overrightarrow{v}=⟨0,3⟩\)

\(Area=\frac{1}{2}∣u_{x}v_{y} − u_{y}v_{x}∣ = \frac{1}{2}∣4⋅3−0⋅0∣=\frac{1}{2}×12=6\)

 

Fig6

7. Using the Limit Definition (Derivative Concept):

Approximate the area of a triangle by slicing it into infinitely small rectangles (limit of summation):

$$Area=\lim_{Δx\to 0} \sum_{}^{}\frac{1}{2}f(x)Δx$$

Ø  Applications: Widely used in physics, computer graphics, and geospatial analysis.

Example:

For a triangle with base along x-axis, height given by y=3x, and limits x=0 to x=4:

\(Area=\int_{0}^{4}3xdx=\left[\frac{3x^{2}}{2}\right]_{0}^{4}=\frac{3.16}{2}=24\)

Divide by 2:

Area=12

Fig7

These methods demonstrate the flexibility of mathematical principles (multiplication, geometry, integration, and limits) in solving a single problem. Let me know if you need more examples or detailed derivations!