Area of a Triangle

Introduction

The area of a triangle, denoted by $\Delta$ or sometimes S, is a fundamental property. Trigonometry provides several powerful methods to calculate this area, connecting the sides, angles, and special radii of the triangle.

1. Area using Two Sides and the Included Angle

This is the most direct trigonometric formula for the area. The area is half the product of two sides and the sine of the angle between them. $$\Delta = \frac{1}{2}ab\sin C$$

$$\Delta = \frac{1}{2}bc\sin A$$

$$\Delta = \frac{1}{2}ac\sin B$$

2. Area using One Side and All Angles

By using the Sine Rule (e.g., $b = a \frac{\sin B}{\sin A}$) to substitute into the formula above, we can express the area in terms of just one side and all three angles.

$$\Delta = \frac{a^2 \sin B \sin C}{2\sin A}$$

Similar formulas can be written using sides b and c.

3. Heron's Formula (using all three sides)

When the lengths of all three sides are known, the area can be found using Heron's formula. It relies on the semi-perimeter, $s = \frac{a+b+c}{2}$.

$$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$

4. Area in Terms of Circumradius (R)

The circumradius R is the radius of the circle passing through all three vertices of the triangle. The area is related to it by:

$$\Delta = \frac{abc}{4R} \quad \implies \quad R = \frac{abc}{4\Delta}$$

This formula is crucial for problems involving the circumcircle.

5. Area in Terms of Inradius (r)

The inradius r is the radius of the circle inscribed within the triangle, tangent to all three sides. The area is given by the simple formula:

$$\Delta = rs$$

where s is the semi-perimeter.

Example 1: Area with Two Sides and Included Angle

Question: In a $\triangle ABC$, if $a=4, b=5$ and angle $C=30^\circ$, find the area of the triangle.

Solution:

Using the formula $\Delta = \frac{1}{2}ab\sin C$:

$$\Delta = \frac{1}{2}(4)(5)\sin 30^\circ = \frac{1}{2}(20)\left(\frac{1}{2}\right) = 5 \text{ sq. units}$$

Example 2: Area with Heron's Formula

Question: The sides of a triangle are 13, 14, and 15. Find its area.

Solution:

First, find the semi-perimeter: $s = \frac{13+14+15}{2} = \frac{42}{2} = 21$.

Now use Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.

$$\Delta = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6}$$

$$= \sqrt{(3 \cdot 7) \cdot (2^3) \cdot 7 \cdot (2 \cdot 3)} = \sqrt{2^4 \cdot 3^2 \cdot 7^2} = 2^2 \cdot 3 \cdot 7 = 84 \text{ sq. units}$$

Example 3: Finding Circumradius (R)

Question: For a triangle with sides 13, 14, 15, find the circumradius R.

Solution:

From the previous example, we know the area $\Delta = 84$ and the product of sides $abc = 13 \times 14 \times 15 = 2730$. We use the formula $R = \frac{abc}{4\Delta}$.

$$R = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{2730}{336} = \frac{65}{8} = 8.125$$

Example 4: Finding Inradius (r)

Question: For a triangle with sides 13, 14, 15, find the inradius r.

Solution:

We know $s=21$ and $\Delta=84$. Using the formula $\Delta = rs$:

$$84 = r(21) \implies r = \frac{84}{21} = 4$$

Example 5: Area from One Side and Angles

Question: In a triangle ABC, $A=60^\circ, B=45^\circ$ and side $b=2\sqrt{3}$. Find the area.

Solution:

First, find angle $C = 180^\circ - (60^\circ+45^\circ) = 75^\circ$. We use the formula $\Delta = \frac{b^2 \sin A \sin C}{2\sin B}$.

We need $\sin 75^\circ = \sin(45^\circ+30^\circ) = \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$. Note that $\frac{\sqrt{3}+1}{2\sqrt{2}}$ is an equivalent form.

$$\Delta = \frac{(2\sqrt{3})^2 \sin 60^\circ \sin 75^\circ}{2\sin 45^\circ} = \frac{12 (\sqrt{3}/2) ((\sqrt{3}+1)/2\sqrt{2})}{2(1/\sqrt{2})} = \frac{3\sqrt{3}(\sqrt{3}+1)}{2} = \frac{9+3\sqrt{3}}{2}$$

Example 6: Proving an Area-related Identity

Question: In any triangle ABC, prove that $\Delta = 2R^2 \sin A \sin B \sin C$.

Solution:

Start with the formula $\Delta = \frac{abc}{4R}$. From the Sine Rule, we have $a=2R\sin A, b=2R\sin B, c=2R\sin C$. Substitute these into the formula:

$$\Delta = \frac{(2R\sin A)(2R\sin B)(2R\sin C)}{4R} = \frac{8R^3\sin A\sin B\sin C}{4R} = 2R^2 \sin A \sin B \sin C$$

This gives a useful relation between area, circumradius, and angles.

Example 7: Area using Coordinates

Question: Find the area of the triangle with vertices A(0,0), B(5,0), and C(3,4).

Solution:

We can find side lengths: $c=AB=5$, $b=AC=\sqrt{3^2+4^2}=5$, $a=BC=\sqrt{(5-3)^2+(0-4)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$.

It's an isosceles triangle. We find angle A using the Cosine Rule:

$$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{5^2+5^2-(2\sqrt{5})^2}{2(5)(5)} = \frac{25+25-20}{50} = \frac{30}{50} = \frac{3}{5}$$

If $\cos A = 3/5$, then $\sin A = \sqrt{1-(3/5)^2} = 4/5$. (Since A is an angle in a triangle, sin A > 0).

Area $\Delta = \frac{1}{2}bc\sin A = \frac{1}{2}(5)(5)(4/5) = 10$ sq. units. (Also note: the base is 5 and height is 4, so area is $1/2*5*4=10$)

Example 8: Area from Half-Angle Formulas

Question: Show that $\Delta = s(s-a)\tan(A/2)$.

Solution:

We know the half-angle formula: $\tan(A/2) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.

RHS = $s(s-a) \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \sqrt{s^2(s-a)^2 \frac{(s-b)(s-c)}{s(s-a)}} = \sqrt{s(s-a)(s-b)(s-c)}$.

By definition, this is Heron's formula for the area, so RHS = $\Delta$ = LHS.

Example 9: Ratio of Areas

Question: A median divides a triangle into two smaller triangles. What is the ratio of their areas?

Solution:

Let AD be the median from vertex A to side BC. This means D is the midpoint of BC, so BD=DC. The two triangles $\triangle ABD$ and $\triangle ADC$ share the same vertex A and their bases BD and DC lie on the same straight line. Therefore, they share the same altitude (height) from A to BC.

Area($\triangle ABD$) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(BD)(h)$.

Area($\triangle ADC$) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(DC)(h)$.

Since BD=DC, the areas are equal. The ratio is 1:1. A median bisects the area of a triangle.

Example 10: Finding a Side from Area

Question: In a triangle, the area is 6, side a=3, side b=5. Find the value of $\sin C$.

Solution:

Using the formula $\Delta = \frac{1}{2}ab\sin C$.

$$6 = \frac{1}{2}(3)(5)\sin C = \frac{15}{2}\sin C$$

$$\sin C = \frac{6 \times 2}{15} = \frac{12}{15} = \frac{4}{5}$$