An expert question on perimeter and area for P6, with a step-by-step solution, diagrams and the answer. Common in the Hua Xia Cup's P4 to P6 papers; harder ones work out areas from the ratios between triangles' areas.
The question
In triangle ABC, F is on AB and E is on AC;
BE and CF cross at O, cutting the triangle into four pieces as shown.
Triangle BOF has area 11 cm², triangle BOC 12 cm² and triangle COE 11 cm².
What is the area of quadrilateral AFOE (shaded), in cm²? (Not drawn to scale.)
Show the hints (3)
💡Hint 1: Join AO, splitting the quadrilateral into triangles AFO and AOE; call their areas x and y.
💡Hint 2: Triangles BOF and BOC share a height, so FO : OC = 11 : 12. Triangles AFO and AOC, on bases FO and OC with one height, are in the same ratio.
💡Hint 3: Likewise EO : OB = 11 : 12, so triangle AOE : triangle AOB = 11 : 12. Write both as equations in x and y and solve them together.
Worked solution
Have a go first, then read the solution step by step.
1.Join AO (dashed). Let triangle AFO have area x and triangle AOE area y.
2.BOF and BOC share a height (both from B), so their bases FO : OC = 11 : 12. AFO and AOC (= y + 11) stand on FO and OC too, with one height (from A), so x : (y + 11) = 11 : 12, i.e. 12 × x = 11 × (y + 11).
3.Likewise COE and BOC share a height (both from C), so EO : OB = 11 : 12. AOE and AOB (= x + 11) share one, so y : (x + 11) = 11 : 12, i.e. 12 × y = 11 × (x + 11).
4.The second says 12 × y = 11 × x + 121. Multiply both sides of the first by 12: 144 × x = 11 × (12 × y) + 1452 = 11 × (11 × x + 121) + 1452 = 121 × x + 2783.
5.Take 121 × x from both sides: 23 × x = 2783, so x = 121. Into the second: 12 × y = 11 × 121 + 121 = 1452, so y = 121.
6.Quadrilateral AFOE = 121 + 121 = 242 cm².
⚠️Common mistake: assuming AFO and AOE are equal, or answering from one ratio. Two unknowns need both ratios.
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