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1992 AJHSME

Problem 10

Problem 10 · 1992 AJHSME Hard
Geometry & Measurement count-congruent-pieces
Figure for AJHSME 1992 Problem 10
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Answer: B — 20.
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Hint 1 of 3
The pieces are all congruent — identical. So instead of measuring the odd-shaped shaded region directly, what one easy number unlocks the whole figure?
Still stuck? Show hint 2 →
Hint 2 of 3
When a shape is cut into equal pieces, find ONE piece's area (total ÷ number of pieces), then shaded area is just "piece area × pieces shaded." Counting beats measuring.
Still stuck? Show hint 3 →
Hint 3 of 3
Get the big triangle's area from its legs first; the small pieces each get an equal share of it.
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Approach: one equal piece's area × the number shaded
  1. The whole triangle is a right triangle with legs 8, so its area is ½ · 8 · 8 = 32. It's split into 16 congruent pieces, so each piece has area 32 ÷ 16 = 2.
  2. Counting the shaded little triangles in the picture gives 10 of them.
  3. Shaded area = 10 × 2 = 20.
  4. Why this transfers: "equal pieces" is your friend — once every piece is the same size, area becomes pure counting. You never have to compute the strange shaded outline itself, only how many unit-pieces it contains.
  5. Sanity check: 10 of 16 pieces are shaded, a bit over half, and 20 is a bit over half of 32. Consistent.
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