Area swept by the square?
A hard-level "Area" problem."I cannot find the length, but I can find length × length." That one move unlocks a surprising number of elementary geometry problems.
GivenA 6 cm square turned 90° about vertex A, in the direction of the arrow / π = 3.14
Hints
- As it turns, the point farthest from A is C, at the far end of the diagonal. So the outer boundary of the swept region is the arc traced by C.
- That arc is centred at A, radius AC, through 90°. It forms the roof of the swept region.
- Cut the swept region along the two diagonals into three pieces. The blue pieces at each end are half of the original square and half of the turned one. The orange middle is a sector.
- Two halves make exactly one square: 6 × 6 = 36 cm².
- The sector is "radius × radius × 3.14 ÷ 4". We cannot find AC itself with elementary arithmetic, but we can find AC × AC: a square's area is also "diagonal × diagonal ÷ 2", so 36 = AC × AC ÷ 2, i.e. AC × AC = 72.
- The sector is 72 × 3.14 ÷ 4 = 56.52 cm². Add the square: 36 + 56.52.
Answer 92.52cm²
More "Area" problems
- Area of the inner triangle?
- Area of triangle DOA?
- Area of quadrilateral ABCD?
- Area of the overlap?
- Total area of the two crescents?
- What area of the table is covered by paper?