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Maths olympiad question · P5–P6★★★ Hard

Perimeter and area: olympiad question 7

A hard question on perimeter and area for P5–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 trapezium ABCD, AD is parallel to BC, and the diagonals AC and BD cross at O, as shown.

Triangle AOD has an area of 27 cm² and triangle AOB 36 cm².

What is the area of triangle BOC (shaded), in cm²?

Show the hints (3)
  1. Hint 1: How are triangles AOB and COD related? Compare triangles ABD and ACD.
  2. Hint 2: AOD and AOB share the corner A, and their bases OD and OB lie on one line. What does their area ratio tell you?
  3. Hint 3: OD : OB = 3 : 4. Triangles COD and COB stand on OD and OB too.

Worked solution

Have a go first, then read the solution step by step.

  1. 1.Triangles ABD and ACD share the base AD, and since AD is parallel to BC they have the same height (dashed), so their areas are equal.
  2. 2.Taking triangle AOD from both: triangle COD = triangle AOB = 36 cm².
  3. 3.Triangles AOD and AOB have the same height (from A to BD, dashed), so their bases OD : OB = 27 : 36 = 3 : 4.
  4. 4.Triangles COD and COB also share a height (from C to BD), on bases OD : OB = 3 : 4 again, so triangle BOC = 36 ÷ 3 × 4 = 48 cm².
  5. ⚠️Common mistake: taking AOD : BOC = 3 : 4. In fact AOD : COD = 3 : 4 and COD : BOC = 3 : 4 again, so the ratio is used twice.

Answer: 48 cm²

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