ROZWIĄZANIA
Zad.1
2r = a
d = a
$$r = \ \frac{1}{2}a$$
$$P_{c} = 4\pi{(\frac{1}{2}a)}^{2}$$
$$P_{c} = 4\pi\frac{1}{4}a^{2}$$
Pc = a2π
$$V = \frac{4}{3}\pi\frac{a^{3}}{8} = \frac{a^{3}}{6}$$
Zad.2
$$2r = a\sqrt{3}$$
$$r = \frac{a\sqrt{3}}{2}$$
$$P_{c} = 4\pi{(\frac{a\sqrt{3}}{a})}^{2}$$
$$P_{c} = 4\pi\frac{a^{2}3}{4} = 3a^{2}\pi$$
$$V = \frac{4}{3}\pi{(\frac{a\sqrt{3}}{a})}^{3} = \sqrt{\frac{3}{2}}\pi a^{3}$$
Zad.3
Dane:
r = 17cm
h = 16cm
Szukane:
x = ?
Kwadrat
|AB| = |CB| = 16
|AB| = 16
|AM| = 8
172 = x2 + 82
289 − 64 = x2
x = 15cm
Zad.4
Dane:
r = 25cm
h = 16cm
x = OM = 24 cm
Szukane:
PABCD, AB = ?
Rozwiązanie:
242 + z2 = 252
576 + z2 = 625
z2 = 625 − 576
z2 = 49
z = 7
|AB| = 14
PABCD = 14 * 16 = 224cm2
Zad.5
Dane:
h = 9cm
OM = 2cm
$$\alpha = \frac{\pi}{3}$$
Szukane:
PABCD, |AB| = ?
AO1B jest rownoramiennym o ∢ miedzy ramionami 60, czyli AO1B jest rownobocznym.
|O1M| jest wysokoscia.
O1M = 2
$$O_{1}M = \frac{|AB|\sqrt{3}}{2}$$
$$2 = \frac{|AB|\sqrt{3}}{2}$$
$$\left| \text{AB} \right| = \frac{4\sqrt{3}}{3}\text{cm}$$
$$P_{\text{ABCD}} = \frac{4\sqrt{3}}{3}*9 = {12\sqrt{3}\text{cm}}^{2}$$
Zad.6
Dane:
h
Pbw = Pbs
2πrh = πr
l = 2h
Szukane:
V, r, α = ?
Rozwiązanie:
h2 + r2 = l2
h2 + r2 = 4h2
r2 = 3h2
$$r = \sqrt{3}h$$
$$V = \frac{1}{3}\pi{(\sqrt{3}h)}^{2}h$$
$$V = \frac{1}{3}\pi 3h^{3} = \pi h^{3}$$
$$\sin\frac{\alpha}{2} = \frac{h\sqrt{3}}{2h}$$
$$\sin\frac{\alpha}{2} = \frac{\sqrt{3}}{2}$$
$$\frac{\alpha}{2} = 60$$
α = 120
Zad.7
$$\frac{V_{1}}{V} = \ ?$$
$$V_{1} = \frac{1}{3}\pi x^{2}*\frac{1}{3}h$$
V = V1 + V2
$$\frac{\frac{1}{3}h}{x} = \frac{h}{r}$$
$$\frac{1}{3}hr = hx$$
$$\frac{1}{3}r = x$$
$$x = \frac{1}{3}r$$
$$V_{1} = \frac{1}{3}\pi{(\frac{1}{3}r)}^{2}*\frac{1}{3}h$$
$$V_{1} = \frac{1}{3}*\pi*\frac{1}{9}r^{2}*\frac{1}{3}h = \frac{1}{81}\text{πh}r^{2}$$
$$V = \frac{1}{3}*\pi r^{2}*h$$
$$\frac{V_{1}}{V} = \frac{\pi r^{2}h}{81}*\frac{3}{\pi r^{2}h}$$
$$\frac{V_{1}}{V} = \frac{1}{27}$$
$$V_{1} = \frac{1}{27}V$$
V = 27V1
27 − 100%
1 − x
$$x = \frac{100\%}{27}$$
x = 3, 7%