Trisection points divide the base into three equal segments, and both triangles share the same vertex altitude. Area equals (1/2) x base x height, so their areas are identical. Straight dissection along altitudes enables rearrangement.
Discussion Forum Latest Questions
The intersection of the circles consists of two 120° sectors minus two equilateral triangles of side r, giving overlapping area (2π/3 – √3/2)r². The total union area equals 2πr² minus this overlap: (4π/3 + √3/2)r².
Radius OA is perpendicular to tangent BC, bisecting it into segments of length l/2. In right-angled triangle OAB, R² – r² = (l/2)² = l²/4. The ring area is pi(R² – r²) = (1/4)pi l².
Diagonals d1 and d2 cross perpendicularly. (i) Algebraically, the kite splits into two triangles on base d1 with heights summing to d2, giving area (1/2)d1d2. (ii) Geometrically, enclosing it in a d1 x d2 rectangle doubles its area.
The trapezium is split into a parallelogram of base a and a triangle of base (b – a), both sharing height h. Area = ah + (1/2)(b – a)h = (1/2)(2a + b – a)h = (1/2)(a + b)h.