Dany jest walec o promieniu podstawy...- Zadanie 1: MATeMAtyka 3. Zakres podstawowy i rozszerzony. Po gimnazjum

Rozwiązanie

$V = π r^{2} \cdot h$

$P_{b} = 2 π r h$

$P_{c} = 2 π r (r + h)$

a)

$r = 5$

$h = 7$

$V = π 5^{2} \cdot 7 = π \cdot 25 \cdot 7 = 175 π$

$P_{b} = 2 π \cdot 5 \cdot 7 = 2 π \cdot 35 = 70 π$

$P_{c} = 2 π \cdot 5 \cdot (5 + 7) = 10 π \cdot 12 = 120 π$

b)

$r = 4$

$d = 17$

$(2 r)^{2} + h^{2} = d^{2}$

$8^{2} + h^{2} = 17^{2}$

$64 + h^{2} = 289$

$h^{2} = 225$

$h = 15$

$V = π 4^{2} \cdot 15 = π \cdot 16 \cdot 15 = 240 π$

$P_{b} = 2 π \cdot 4 \cdot 15 = 2 π \cdot 60 = 120 π$

$P_{c} = 2 π \cdot 4 \cdot (4 + 15) = 8 π \cdot 19 = 152 π$

c)

$h = 6$

$α = 30^{o}$

$\frac{h}{2 r} = t g α$

$\frac{6}{2 r} = t g 30^{o}$

$\frac{3}{r} = \frac{3}{3}$

$3 \cdot 3 = r \cdot 3 ∣ : 3$

$\frac{9}{3} = r$

$\frac{9 3}{3} = r$

$33 = r$

$V = π (33)^{2} \cdot 6 = π \cdot 27 \cdot 6 = 162 π$

$P_{b} = 2 π \cdot 33 \cdot 6 = 363 π$

$P_{c} = 2 π \cdot 33 (33 + 6) = 63 π (33 + 6) = 54 π + 363 π$

d)

$d = 8$

$α = 60^{o}$

$\frac{h}{d} = sin α$

$\frac{h}{8} = sin 60^{o}$

$\frac{h}{8} = \frac{3}{2}$

$2 h = 83 ∣ : 2$

$h = 43$

$h^{2} + (2 r)^{2} = d^{2}$

$(43)^{2} + 4 r^{2} = 8^{2}$

$48 + 4 r^{2} = 64 ∣ - 48$

$4 r^{2} = 16 ∣ : 4$

$r^{2} = 4$

$r = 2$

$V = π 2^{2} \cdot 43 = π \cdot 4 \cdot 43 = 163 π$

$P_{b} = 2 π \cdot 2 \cdot 43 = 163 π$

$P_{c} = 2 π \cdot 2 \cdot (2 + 43) = 4 π (2 + 43) = 8 π + 163 π$

Magdalena Matusik

Nauczycielka matematyki