15.
Pole równoległoboku:
|AB| = a₁ = 5 cm
|DE| = h = 2 cm
P₁ = a₁ × h = 5 cm × 2 cm = 10 cm²
Pole zacieniowanego prostokąta:
h = 2 cm
|BE| = a₂ = |AD| - 2 cm = 5 cm - 2 cm = 3 cm
[tex]\frac{P_2}{P_1} = \frac{6 \ cm^{2}}{10 \ cm^{2}} = \frac{3}{5}\\\\Odp. \ B.[/tex]
16.
70 ÷ 6 = 11 r. 4
4 cm × (6 cm)² = 4 cm × 36 cm² = 144 cm³
Odp. B.
