Mathematics • Year 10 • Unit 3 • Lesson 17
Area of Non-Right Triangles — Skill Drill
Build fluency with the two key area formulas from Lesson 17: A = ½ ab sin C when you know two sides and the included angle (SAS), and Heron's A = √[s(s−a)(s−b)(s−c)] with s = (a+b+c)/2 when you know all three sides (SSS). Pick the right formula and substitute carefully.
1. I do — fully worked example
Read every step. Each one has a short reason on the right so you can see why, not just what.
Problem. Find the area of a triangle with sides a = 8 cm, b = 10 cm and included angle C = 50°.
Step 1 — Identify the formula.
SAS (two sides + included angle) → A = ½ ab sin C
Reason: the angle C is between the two known sides a and b, so the area formula applies directly.
Step 2 — Substitute.
A = ½ × 8 × 10 × sin 50°
A = 40 × sin 50°
Step 3 — Calculate.
A = 40 × 0.766 ≈ 30.6 cm²
Reason: sin 50° ≈ 0.766. Units are cm × cm = cm².
Step 4 — Check reasonableness.
If C = 90° → A = ½ × 8 × 10 = 40 cm² (the maximum).
Since C = 50° < 90°, A should be less than 40. 30.6 < 40 ✓
Answer: A ≈ 30.6 cm².
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Use Heron's formula this time. 5 marks
Problem. A triangle has sides 7 cm, 8 cm and 9 cm. Find its area using Heron's formula.
Step 1 — Identify the pattern. Three sides known → ______ → use ______'s formula.
Step 2 — Find the semi-perimeter:
s = (7 + 8 + 9) / ______ = ______ / ______ = ______
Step 3 — Find s − a, s − b, s − c:
s − a = ______ − 7 = ______
s − b = ______ − 8 = ______
s − c = ______ − 9 = ______
Step 4 — Apply Heron's formula:
A = √[ ______ × ______ × ______ × ______ ] = √______
Step 5 — Final answer with units:
A ≈ ______ cm²
3. You do — independent practice
Show your working in the space under each problem. Round to 1 dp where needed. The first four are foundation. The middle two are standard. The last two are extension.
Foundation — substitute into A = ½ ab sin C
3.1 Find the area of a triangle with a = 6 cm, b = 8 cm, ∠C = 30°. (sin 30° = ½ exactly.) 2 marks
3.2 Find the area of a triangle with a = 12 cm, b = 15 cm, ∠C = 40°. 2 marks
3.3 A triangle has two equal sides of 10 cm and an included angle of 90°. Find its area. 1 mark
3.4 A triangle has a = 9, b = 9, ∠C = 60°. Find its area exactly using sin 60° = √3 / 2. 2 marks
Standard — Heron's formula
3.5 A triangle has sides 5, 6, 7. Find s, then use Heron's formula to find the area. Leave your answer in surd form if possible. 3 marks
3.6 A triangle has sides 13, 14, 15. Use Heron's formula to find the area. (This is the famous "13-14-15 triangle" with integer area.) 2 marks
Extension — push your thinking
3.7 Two sides of a triangle are 8 cm and 12 cm. What included angle gives the maximum area, and what is that maximum area? Justify briefly. 3 marks
3.8 A triangle has area 36 cm² with sides a = 9 and b = 12. Find the two possible values of the included angle C. (Hint: sin C may give an acute or an obtuse answer.) 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (Heron's formula on 7-8-9)
SSS → Heron's formula.
s = (7 + 8 + 9) / 2 = 24 / 2 = 12.
s − a = 12 − 7 = 5; s − b = 12 − 8 = 4; s − c = 12 − 9 = 3.
A = √(12 × 5 × 4 × 3) = √720 = 12√5 ≈ 26.8 cm².
3.1 — a = 6, b = 8, C = 30°
A = ½ × 6 × 8 × sin 30° = ½ × 48 × ½ = 12 cm².
3.2 — a = 12, b = 15, C = 40°
A = ½ × 12 × 15 × sin 40° = 90 × 0.6428 ≈ 57.9 cm².
3.3 — two sides 10, C = 90°
A = ½ × 10 × 10 × sin 90° = ½ × 100 × 1 = 50 cm². (Right triangle with legs 10 and 10.)
3.4 — a = 9, b = 9, C = 60°
A = ½ × 9 × 9 × (√3 / 2) = 81 × √3 / 4 = 81√3 / 4 cm² ≈ 35.1 cm². (This is the area of an equilateral triangle of side 9.)
3.5 — Heron's on 5-6-7
s = (5 + 6 + 7) / 2 = 9.
A = √(9 × 4 × 3 × 2) = √216 = 6√6 ≈ 14.7 cm².
3.6 — Heron's on 13-14-15
s = (13 + 14 + 15) / 2 = 21.
A = √(21 × 8 × 7 × 6) = √7056 = 84 cm² (exact integer area).
3.7 — sides 8 and 12; max area
A = ½ × 8 × 12 × sin C = 48 sin C. Maximum when sin C = 1, i.e. C = 90°. Max area = 48 cm². (Any other angle produces sin C < 1 and hence a smaller area.)
3.8 — area 36, sides 9 and 12; find C
36 = ½ × 9 × 12 × sin C = 54 sin C → sin C = 36 / 54 = 2/3 ≈ 0.6667.
C₁ = sin⁻¹(2/3) ≈ 41.8° (acute).
C₂ = 180° − 41.8° = 138.2° (obtuse).
Both give the same area because sin C = sin(180° − C).