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 parallelogram ABCD, E is on AD with AE :
ED = 1 :
2, and F is on DC with DF :
FC = 1 :
2.
BE, EF and FB are joined, as shown.
Triangle DEF has an area of 2 cm².
What is the area of triangle BEF (shaded), in cm²?
Show the hints (3)
💡Hint 1: Triangle BEF is hard to find directly. Which three triangles, taken from the whole parallelogram, leave BEF?
💡Hint 2: First get the whole parallelogram from triangle DEF: DEF and DAF share a height, and so do DAF and DAC.
💡Hint 3: The parallelogram's area is 18 cm². Now find triangles ABE and BCF.
Worked solution
Have a go first, then read the solution step by step.
1.Triangles DEF and DAF share a height (from F to AD, dashed), so their areas compare as their bases ED : AD = 2 : 3: triangle DAF = 2 ÷ 2 × 3 = 3 cm².
2.Triangles DAF and DAC share a height (from A to DC, dashed), so they compare as DF : DC = 1 : 3: triangle DAC = 3 × 3 = 9 cm².
3.The diagonal AC cuts the parallelogram into two equal halves, so ABCD = 9 × 2 = 18 cm².
4.Triangles ABE and ABD share a height (from B to AD), and ABD is half the parallelogram (9); AE : AD = 1 : 3, so triangle ABE = 9 ÷ 3 = 3 cm².
5.Likewise triangles BCF and BCD (half again) share a height (from B to DC), and CF : CD = 2 : 3, so triangle BCF = 9 ÷ 3 × 2 = 6 cm².
6.The whole parallelogram less the three corner triangles: triangle BEF = 18 − 3 − 6 − 2 = 7 cm².
⚠️Common mistake: hunting for BEF's own base and height. None of its sides lies along the parallelogram, so take the three corner triangles from the whole instead.
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