Problem 29 · 2014 Math Kangaroo
Stretch
Geometry & Measurement
pythagorean-triplearea-decomposition
In triangle ABC, AB = 6 cm, AC = 8 cm and BC = 10 cm. M is the midpoint of side BC. AMDE is a square, and MD meets AC at point F. What is the area of the quadrilateral AFDE in cm²?
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Answer: B — 1258
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Hint 1 of 2
The 6–8–10 triangle is right-angled at A, which makes coordinates easy.
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Hint 2 of 2
Place A at the origin, build the square on AM, and find where MD meets AC.
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Approach: coordinates: right angle at A, square on AM, intersection F
- Since 6² + 8² = 10², angle A is right; set A = (0,0), B = (6,0), C = (0,8), so M = (3,4).
- Square AMDE has side AM = 5; building it and intersecting line MD with AC gives F = (0, 6.25).
- The quadrilateral AFDE then has area 125/8 cm².
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