G23: Area of Triangle Formula
Know and use the formula: Area = ½ab sin C
📋 Key Concepts
The formula Area = ½ab sin C is used to find the area of any triangle when you know two sides and the included angle.
Advantage: Works for all triangles, not just right-angled ones. No need to find the height.
📝 The Formula
Area = ½ab sin C
Where:
- a and b are two sides of the triangle
- C is the angle between sides a and b
Important: The angle MUST be between the two sides (included angle).
📝 Finding the Area
Example 1
Find the area of a triangle with sides 8 cm and 10 cm, and included angle 45°.
Solution:
Area = ½ab sin C
= ½ × 8 × 10 × sin 45°
= 40 × 0.707
= 28.3 cm² (to 1 d.p.)
Example 2
Find the area of triangle ABC where a = 12 cm, b = 15 cm, and angle C = 60°.
Solution:
Area = ½ab sin C
= ½ × 12 × 15 × sin 60°
= 90 × 0.866
= 77.9 cm² (to 1 d.p.)
Example 3
Find the area of triangle PQR with sides PQ = 7 cm, PR = 9 cm, and angle P = 30°.
Solution:
Area = ½ × PQ × PR × sin P
= ½ × 7 × 9 × sin 30°
= 31.5 × 0.5
= 15.75 cm²
📝 Using Exact Values
Example 4
Find the exact area of a triangle with sides 6 cm and 8 cm, and included angle 60°.
Solution:
Area = ½ × 6 × 8 × sin 60°
= 24 × √3/2
= 12√3 cm²
Example 5
Find the exact area of a triangle with sides 10 cm and 10 cm, and included angle 45°.
Solution:
Area = ½ × 10 × 10 × sin 45°
= 50 × √2/2
= 25√2 cm²
📝 Finding a Side
Example 6
A triangle has area 30 cm², two sides 10 cm and x cm, with included angle 30°. Find x.
Solution:
Area = ½ab sin C
30 = ½ × 10 × x × sin 30°
30 = 5x × 0.5
30 = 2.5x
x = 12 cm
📝 Finding an Angle
Example 7
A triangle has area 24 cm², two sides 8 cm and 10 cm. Find the included angle.
Solution:
Area = ½ab sin C
24 = ½ × 8 × 10 × sin C
24 = 40 × sin C
sin C = 24/40 = 0.6
C = sin⁻¹(0.6) = 36.9°
📝 Comparing with Right-Angled Triangle Formula
For right-angled triangles: Area = ½ × base × height still works and gives the same result.
Example 8
Verify: a right-angled triangle with legs 6 cm and 8 cm has area 24 cm².
Using base × height:
Area = ½ × 6 × 8 = 24 cm² ✓
Using ½ab sin C:
Angle between 6 and 8 is 90°
Area = ½ × 6 × 8 × sin 90° = 24 × 1 = 24 cm² ✓
❓ Practice Questions
Q1: Find the area of a triangle with sides 5 cm and 9 cm, included angle 40°.
Q2: Find the exact area when sides are 12 cm and 12 cm, angle is 30°.
Q3: A triangle has area 40 cm², sides 10 cm and 12 cm. Find the included angle.
Q4: Find the area of a triangle with sides 7 cm and 11 cm, included angle 120°.
Q5: A triangle has area 50 cm², one side 15 cm, included angle 45°. Find the other side.
✅ Answers
- ½ × 5 × 9 × sin 40° = 14.4 cm²
- ½ × 12 × 12 × sin 30° = 36 cm²
- sin C = 40/60 = 2/3, C = 41.8°
- ½ × 7 × 11 × sin 120° = 33.3 cm²
- 50 = ½ × 15 × b × sin 45°, b = 50/(7.5 × 0.707) = 9.43 cm
🎯 Exam Tips
- Make sure the angle is between the two sides
- Check if the answer should be exact or decimal
- Rearrange the formula to find sides or angles
- For exact values, use surds instead of decimals
- Always check units are consistent
- sin 90° = 1, so for right angles this formula reduces to ½ab
🧠 Problem-Solving Strategies
Problem-Solving
Area of triangle = 1/2 x a x b x sinC (where C is the angle between sides a and b). This works for ANY triangle, not just right-angled ones. Combine with the sine rule to find missing sides or angles. The formula can also be rearranged to find an angle when the area and two sides are known.
Multi-Step Problem
Triangle PQR has PQ = 9 cm, QR = 7 cm and area = 21 cm squared. Find the two possible values of angle PQR.
Solution: Area = 1/2 x 9 x 7 x sin(angle) = 21. 31.5 x sin(angle) = 21. sin(angle) = 21/31.5 = 0.6667. angle = 41.8 degrees or 180 - 41.8 = 138.2 degrees. Both values are valid since both give the same sine value.
⚠️ Common Errors
Watch Out!
1. Wrong: Using the perpendicular height formula when only two sides and the included angle are given Correct: If you know two sides and the included angle, use Area = 1/2 x a x b x sinC. The perpendicular height may not be given.
2. Wrong: Forgetting the 1/2 in the formula: Area = a x b x sinC Correct: Area = 1/2 x a x b x sinC. Without the 1/2 you will get double the correct area.
3. Wrong: Using an angle that is NOT between the two given sides Correct: The angle in the formula must be the INCLUDED angle (the angle between the two sides). Using a different angle gives the wrong area.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A triangular plot of land has sides 15 m and 12 m with included angle 110 degrees. (a) Find the area of the plot. (b) Find the length of the third side. (c) A second plot has the same two side lengths but included angle 70 degrees. Which plot has the larger area?
(a) Area = 1/2 x 15 x 12 x sin(110) = 90 x 0.9397 = 84.6 m squared.
(b) Cosine rule: c squared = 225 + 144 - 2(15)(12)cos(110) = 369 + 123.1 = 492.1. c = 22.2 m.
(c) Second plot: Area = 1/2 x 15 x 12 x sin(70) = 90 x 0.9397 = 84.6 m squared. Both plots have the SAME area because sin(110) = sin(70) since sin(180-x) = sin(x).
Mark scheme: M1 area formula, A1 84.6, M1 cosine rule, A1 22.2m, M1 second area, A1 same area with sin reasoning
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Two triangles share two sides of equal length but have different included angles.
(a) For what included angle does the triangle have the maximum area?
(b) If one triangle has included angle 30 degrees and the other has 150 degrees, compare their areas.
(c) A student says "A triangle with a larger angle always has a larger area." Is this true? Explain.
Answers: (a) Maximum area when sin(angle) = 1, i.e. when the included angle is 90 degrees. (b) sin(30) = 0.5 and sin(150) = 0.5. They have the SAME area because sin(x) = sin(180-x). (c) Not true — area depends on sin(angle), not the angle directly. Angles of 30 and 150 give the same area, and 91 degrees gives slightly LESS area than 90 degrees. The relationship is through sine, which peaks at 90 degrees.
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