Scaling linear dimensions by factor k increases area by k²: 2² = 4 and 3² = 9. Yes, in each case exactly 4 or 9 congruent copies tile and perfectly fit into the scaled figure without gaps.
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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.
Inverting a second congruent trapezium alongside the first forms a parallelogram of base (a + b) and height h. Its area is (a + b)h, so one trapezium has area (1/2)(a + b)h.
Drawing a diagonal splits the trapezium into two triangles having bases a and b, each sharing perpendicular height h. Summing their areas gives (1/2)ah + (1/2)bh = (1/2)(a + b)h.
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.