Mathematics • Year 10 • Unit 3 • Lesson 17
Triangle Area — Mixed Challenge
Stretch every idea from Lesson 17: choose between A = ½ ab sin C and Heron's formula, reverse-engineer a side or angle from a given area, verify one method against the other, and design two triangles with the same area.
1. Mixed problems — choose the right idea
Decide which formula fits each piece of information before you start. Show all working. Round to 1 dp / 0.1°. 2-3 marks each
1.1 Find the area of a triangle with sides 11 cm and 14 cm and included angle 65°. 2 marks
1.2 A triangle has sides 9, 10, 17. Find its area using Heron's formula, to 1 dp. 2 marks
1.3 A triangle has area 50 cm² and two sides 10 cm and 12 cm. Find the included angle (give both possible values). 3 marks
1.4 A triangle has area 60 cm², one side 15 cm, and the angle between the 15 cm side and an unknown side is 30°. Find the unknown side. 2 marks
1.5 A triangle has sides 7, 8, 9. Find the area using Heron's formula, then verify it using A = ½ ab sin C (you will need to first find an angle with the cosine rule). The two methods must agree. 3 marks
1.6 An equilateral triangle has side length 12 cm. Find its area exactly (in surd form), using A = ½ ab sin 60°. 2 marks
2. Find the mistake
A Year 10 student has tried to find the area of a triangle with sides 6, 7, 8 using Heron's formula. Their working is shown. Exactly one line contains a mistake. Spot it, explain why it is wrong, then re-do the working correctly. 3 marks
Student's working:
Line 1: s = 6 + 7 + 8 = 21
Line 2: s − a = 21 − 6 = 15
Line 3: s − b = 21 − 7 = 14
Line 4: s − c = 21 − 8 = 13
Line 5: A = √(21 × 15 × 14 × 13) = √57 330 ≈ 239.4 cm²
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write the corrected working from that line onward, and give the final area.
Stuck? Sanity check: a triangle with sides 6, 7, 8 cannot possibly have area 239 cm² — that is way more than the bounding 8 × 8 = 64 cm² square. So the calculation has gone wrong early.3. Open-ended challenge — design your own triangles
Be creative but show every number. 4 marks
3.1 Design two different non-similar triangles, each with area exactly 30 cm². The two triangles must have different side lengths (they cannot be congruent or scaled copies).
For each triangle you design:
(i) State the side(s) and angle (or all three sides) you are using.
(ii) Use the appropriate area formula to verify the area is 30 cm².
(iii) State which formula you used (½ ab sin C or Heron's) and why.
Bonus: Use the fact that sin C = sin(180° − C) to design a third triangle with area 30 cm² using the same two sides as one of your triangles but a different included angle.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — a = 11, b = 14, C = 65°
A = ½ × 11 × 14 × sin 65° = 77 × 0.9063 ≈ 69.8 cm².
1.2 — sides 9, 10, 17
s = (9 + 10 + 17) / 2 = 18.
A = √(18 × 9 × 8 × 1) = √1 296 = 36.0 cm². (Note: 9 + 10 = 19 > 17, so the triangle just exists; it is very flat.)
1.3 — area 50, sides 10 and 12
50 = ½ × 10 × 12 × sin C = 60 sin C → sin C = 50 / 60 = 5/6 ≈ 0.8333.
C₁ = sin⁻¹(5/6) ≈ 56.4°; C₂ = 180° − 56.4° = 123.6°.
1.4 — area 60, one side 15, included angle 30°
60 = ½ × 15 × b × sin 30° = ½ × 15 × b × ½ = (15/4) × b → b = 60 × 4 / 15 = 16.
1.5 — Heron's vs ½ ab sin C on 7-8-9
Heron's: s = 12, A = √(12 × 5 × 4 × 3) = √720 = 12√5 ≈ 26.8 cm².
Find ∠C (opposite side 9): cos C = (49 + 64 − 81)/(2·7·8) = 32/112 ≈ 0.2857. C ≈ 73.4°, so sin C ≈ 0.9583.
Verify: A = ½ × 7 × 8 × 0.9583 ≈ 26.8 cm² ✓ Methods agree.
1.6 — equilateral side 12
A = ½ × 12 × 12 × sin 60° = ½ × 144 × (√3 / 2) = 144√3 / 4 = 36√3 cm² ≈ 62.4 cm².
2 — Find the mistake
(a) The mistake is on Line 1.
(b) The student used perimeter (6 + 7 + 8 = 21) instead of semi-perimeter. The formula needs s = (a + b + c) / 2, so s should be 21 / 2 = 10.5. All later lines are knock-on errors from this missing ÷ 2.
(c) Corrected:
s = (6 + 7 + 8) / 2 = 10.5.
s − a = 4.5; s − b = 3.5; s − c = 2.5.
A = √(10.5 × 4.5 × 3.5 × 2.5) = √413.4375 ≈ 20.3 cm². (Sanity check: this fits inside an 8 × 8 box, so it is reasonable.)
3 — Open-ended (sample triangles with area 30 cm²)
Triangle A — SAS: a = 10, b = 12, C = sin⁻¹(½) = 30°. Area = ½ × 10 × 12 × sin 30° = 60 × ½ = 30 cm². ✓ (Used ½ ab sin C because SAS was supplied directly.)
Triangle B — SSS: sides 5, 12, 13 (Pythagorean triple, right-angled). s = 15; A = √(15 × 10 × 3 × 2) = √900 = 30 cm². ✓ (Used Heron's because three sides were given and no angle.) Note this is also = ½ × 5 × 12 = 30 since it is a right triangle.
Bonus — same two sides as A, different angle: Keep a = 10, b = 12. Need ½ × 10 × 12 × sin C = 30 → sin C = ½, so C = 30° (Triangle A) OR C = 150° (Triangle C). Triangle C has the same two sides but obtuse 150° between them; the area is identical because sin 150° = sin 30° = ½.
Marking: 1 mark for each correct triangle with verification (× 2 = 2 marks), 1 mark for naming the formula used and justifying, 1 mark for the bonus that exploits sin C = sin(180° − C).