📐 Level 2 - Topic 6: Area of Triangles - Trigonometric Formulas & Heron's Formula 📏△

1) Expanding Area Calculation - Beyond Base and Height 📐

We all know the classic formula for the area of a triangle: $ \frac{1}{2} \times \text{base} \times \text{height} $. This formula works perfectly well when we know the base and the height of the triangle. However, in many situations, especially when working with oblique triangles, we might not readily know the height. Trigonometry offers us powerful alternative formulas to calculate the area of a triangle using different sets of information.

Why Trigonometric Area Formulas?

Versatility: Trigonometric formulas allow us to calculate the area of a triangle when we are given sides and angles, without needing to find the height directly.
Oblique Triangles: Especially useful for oblique triangles where finding the height can be less straightforward than in right triangles.
Applications: Essential in various fields like surveying, engineering, and geometry, where area calculations are needed based on angles and side lengths.

In this topic, we will explore two main trigonometric approaches to find the area of a triangle:

Formula 1: Using two sides and the included angle.
Formula 2: Heron's Formula - Using all three sides.

2) Area Formula 1 - Using Two Sides and the Included Angle 📏

Let's start with a formula that allows us to calculate the area of a triangle if we know the lengths of two sides and the measure of the angle included between them. Consider triangle $ ABC $ with sides $ a, b, c $ and angles $ A, B, C $.

Definition: Area Formula using Two Sides and Included Angle

The area of a triangle $ ABC $ can be calculated using any pair of sides and their included angle. The formulas are:

$ \text{Area} = \frac{1}{2}ab\sin(C) $

$ \text{Area} = \frac{1}{2}bc\sin(A) $

$ \text{Area} = \frac{1}{2}ac\sin(B) $

Understanding and Deriving the Formula:

Consider the standard area formula: $ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $. Let's take side $ b $ as the base of triangle $ ABC $. We need to express the height in terms of side $ a $ and angle $ C $ (or $ \sin(C) $).
Draw an altitude (height) from vertex $ A $ to side $ BC $ (or its extension), and let the foot of the altitude be $ H $. Then $ AH $ is the height, $ h $. In right triangle $ AHC $, $ \sin(C) = \frac{h}{b} $, so $ h = b \sin(C) $.
Substitute $ h = b \sin(C) $ and base $ = a $ into the area formula: $ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times a \times (b \sin(C)) = \frac{1}{2}ab\sin(C) $. (Note: We should have taken base as 'b', and then height would be related to side 'c' and angle 'B' - typo in initial thought process during writing - corrected to use side 'b' as base and height relative to side 'b'). Let's correct the derivation.
Correct Derivation: Let's take side $ c $ as the base. Draw altitude from vertex $ B $ to side $ c $ (or extension), let foot be $ D $. Height $ BD = h $. In right triangle $ BDC $, $ \sin(C) = \frac{h}{a} $, so $ h = a \sin(C) $. Area $ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times c \times (a \sin(C)) = \frac{1}{2}ac\sin(C) $. (Another typo in assumed base during derivation thought - corrected again). Let's use side 'b' as base finally for derivation to match formula order in definition.
Final Correct Derivation: Let's use side $ b $ as base. Draw altitude from vertex $ A $ to side $ b $ (or extension), let foot be $ H $. Height $ AH = h $. In right triangle $ ABH $, $ \sin(B) = \frac{h}{c} $, so $ h = c \sin(B) $. Area $ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times (c \sin(A)) = \frac{1}{2}bc\sin(A) $. Still not matching formula order. Let's restart derivation with angle C as target.
Restart Derivation (Targeting $ \frac{1}{2}ab\sin(C) $ ): Consider sides $ a $ and $ b $ and included angle $ C $. Let's take side $ a $ as base. Draw altitude from vertex $ B $ to side $ a $ (or extension), foot at $ D $. Height $ BD = h $. In right triangle $ BDC $, $ \sin(C) = \frac{h}{b} $, so $ h = b \sin(C) $. Area $ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times a \times (b \sin(C)) = \frac{1}{2}ab\sin(C) $. Finally correct derivation! Order in definition was based on angle given.
Thus, given two sides and the included angle, we can easily find the area using the sine of the angle and the product of the two sides.

3) Area Formula 2 - Heron's Formula (Using Three Sides) 📏

What if we know all three sides of a triangle, but no angles? We can still find the area using Heron's Formula. This formula is particularly useful in SSS (Side-Side-Side) cases.

Definition: Heron's Formula

For a triangle with side lengths $ a, b, c $, the area can be calculated as follows:

First, calculate the semi-perimeter, $ s $, of the triangle:

$ s = \frac{a + b + c}{2} $
Then, the Area is given by:

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

Understanding and Applying Heron's Formula:

Heron's Formula only requires the lengths of the three sides of the triangle. It's incredibly useful when angles are not known or are hard to calculate first.
The semi-perimeter $ s $ is half the sum of the sides. It's a convenient intermediate value to simplify the formula.
To apply Heron's formula, you simply need to calculate $ s $ first, and then substitute $ s, a, b, c $ into the square root expression.
Note: Heron's Formula works for all types of triangles - acute, obtuse, and right-angled.

4) Calculating Triangle Area - Examples Using Trigonometric Formulas 🚀

Example 1: Area using two sides and included angle

In triangle $ ABC $, $ a = 10 $ cm, $ b = 15 $ cm, and $ \angle C = 30^\circ $. Find the area of triangle $ ABC $.

Use Area Formula 1: $ \text{Area} = \frac{1}{2}ab\sin(C) $.
Substitute the values: $ \text{Area} = \frac{1}{2} \times 10 \times 15 \times \sin(30^\circ) $.
Calculate: $ \text{Area} = \frac{1}{2} \times 10 \times 15 \times \frac{1}{2} = \frac{150}{4} = 37.5 $ square cm.

Solution: Area of triangle $ ABC $ is $ 37.5 $ square cm.

Example 2: Area using Heron's Formula (SSS case)

Find the area of a triangle with sides $ a = 13 $ units, $ b = 14 $ units, and $ c = 15 $ units.

Calculate the semi-perimeter $ s $:
$ s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 $.
Apply Heron's Formula:
\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \] \[ = \sqrt{21(21-13)(21-14)(21-15)} \]
Calculate:
\[ \text{Area} = \sqrt{21 \times 8 \times 7 \times 6} \] \[ = \sqrt{(3 \times 7) \times (2 \times 4) \times 7 \times (2 \times 3)} \] \[ = \sqrt{3^2 \times 7^2 \times 2^2 \times 4} \] \[ = 3 \times 7 \times 2 \times \sqrt{4} \] \[ = 3 \times 7 \times 2 \times 2 = 84 \]

Solution: Area of the triangle is $ 84 $ square units.

Example 3: Combining with Law of Sines (ASA case)

In triangle $ ABC $, $ \angle A = 40^\circ $, $ \angle B = 60^\circ $, and side $ c = 20 $ cm. Find the area of triangle $ ABC $.

Find angle $ C $: $ C = 180^\circ - A - B = 180^\circ - 40^\circ - 60^\circ = 80^\circ $.
Find side $ a $ using Law of Sines: $ \frac{a}{\sin(A)} = \frac{c}{\sin(C)} \Rightarrow a = \frac{c \sin(A)}{\sin(C)} = \frac{20 \times \sin(40^\circ)}{\sin(80^\circ)} $.
Find side $ b $ using Law of Sines: $ \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \Rightarrow b = \frac{c \sin(B)}{\sin(C)} = \frac{20 \times \sin(60^\circ)}{\sin(80^\circ)} $.
Use Area Formula 1 with sides $ a $ and $ b $ and angle $ C $: $ \text{Area} = \frac{1}{2}ab\sin(C) = \frac{1}{2} \times \left( \frac{20 \sin(40^\circ)}{\sin(80^\circ)} \right) \times \left( \frac{20 \sin(60^\circ)}{\sin(80^\circ)} \right) \times \sin(80^\circ) $.
Simplify and Calculate: $ \text{Area} = \frac{1}{2} \times \frac{20 \times 20 \times \sin(40^\circ) \sin(60^\circ)}{\sin(80^\circ)} = \frac{200 \times \sin(40^\circ) \sin(60^\circ)}{\sin(80^\circ)} \approx \frac{200 \times 0.6428 \times 0.8660}{0.9848} \approx 113.4 $ square cm.

Solution: Area of triangle $ ABC $ is approximately $ 113.4 $ square cm.

5) Practice Questions 🎯

5.1 Fundamental – Area Calculations using Trigonometric Formulas

1. In triangle $ ABC $, $ a = 7 $ cm, $ b = 9 $ cm, and $ \angle C = 45^\circ $. Find the area.

2. In triangle $ PQR $, $ p = 12 $ m, $ r = 15 $ m, and $ \angle Q = 120^\circ $. Find the area.

3. Find the area of a triangle with sides $ a = 5 $, $ b = 12 $, and $ c = 13 $ using Heron's Formula. Verify if it's a right triangle and calculate area using $ \frac{1}{2} \times \text{base} \times \text{height} $.

4. For a triangle with sides $ 6, 6, 6 $, calculate the area using Heron's Formula.

5. Calculate the area of triangle $ XYZ $ where $ XY = 8 $, $ YZ = 5 $, and $ \angle Y = 60^\circ $.

6. Find the area of a triangle with sides $ 17, 25, 28 $ units using Heron's Formula.

7. Triangle $ ABC $ has $ a = 11 $, $ c = 14 $, and $ \angle B = 30^\circ $. Find its area.

8. Calculate the area of an equilateral triangle with side length 10 cm using $ \frac{1}{2}ab\sin(C) $. (Verify all angles are $ 60^\circ $).

9. For a triangle with sides $ 20, 21, 29 $, find the area using Heron's Formula. Verify if it's a right triangle.

10. Find the area of triangle $ RST $ given $ RS = 9 $, $ ST = 10 $, and $ TR = 17 $. (Use Heron's Formula).

5.2 Challenging – Problem Solving and Combined Concepts 💪🚀

1. A triangular garden plot has sides of length 40 feet and 60 feet, and the angle between these sides is $ 55^\circ $. Calculate the area of the garden plot. If fertilizer costs $2 per square foot, what will be the cost to fertilize the garden?

2. A surveyor is measuring a triangular field. They find that two sides are 250 meters and 320 meters, and the angle opposite the 320-meter side is $ 70^\circ $. Use the Law of Sines to find the angle opposite the 250-meter side, and then use the area formula $ \frac{1}{2}ab\sin(C) $ to find the area of the field.

3. A rhombus has diagonals of length 12 cm and 16 cm. Find the area of the rhombus. (Hint: Diagonals of a rhombus are perpendicular bisectors of each other. Consider the triangles formed by the diagonals).

4. The area of triangle $ ABC $ is 50 square cm. If side $ a = 10 $ cm and side $ b = 20 $ cm, find the possible values for angle $ C $.

5. (Conceptual) Can Heron's formula be derived from the formula $ \text{Area} = \frac{1}{2}ab\sin(C) $ using the Law of Cosines? (This is a more advanced challenge, think about expressing $ \sin(C) $ in terms of sides using Law of Cosines and trigonometric identities).

6) Summary - Area of Triangles - Multiple Approaches 🎉

Trigonometric Area Formula 1: $ \text{Area} = \frac{1}{2}ab\sin(C) $ (and variations) - Useful when two sides and included angle are known (SAS case, also works for ASA/AAS after finding sides).
Heron's Formula: $ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} $ where $ s = \frac{a+b+c}{2} $ - Useful when all three sides (SSS case) are known.
Versatile Area Calculation: These formulas allow area calculation without directly finding the height, using sides and angles.
Applications: Widely used in geometry, surveying, and engineering for area determination in various scenarios.
Choice of Formula: Choose based on given information. Formula 1 for SAS or if angle is easily available, Heron's for SSS.

Fantastic! You've now expanded your toolkit for calculating the area of triangles using trigonometry. You can now find the area of triangles in various scenarios, given sides and angles, using either the $ \frac{1}{2}ab\sin(C) $ formula or Heron's Formula. Understanding these methods is crucial for a wide range of applications in mathematics and practical fields. In the upcoming levels, we will continue to build upon these trigonometric principles and explore even more exciting mathematical concepts! Keep up the great work! 📐📏△🌟