STEREOMETRIA - rozwiązania

Zad.1

2r = a

d = a

$$r = \ \frac{1}{2}a$$

$$P_{c} = 4\pi{(\frac{1}{2}a)}^{2}$$

$$P_{c} = 4\pi\frac{1}{4}a^{2}$$

P_c =  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:

P_ABCD,  AB =  ?

Rozwiązanie:

24² + z² = 25²

576 + z² = 625

z² = 625 − 576

z² = 49

z = 7

|AB| = 14

P_ABCD = 14 * 16 = 224cm²

Zad.5

Dane:

h = 9cm

OM = 2cm

$$\alpha = \frac{\pi}{3}$$

Szukane:

P_ABCD,  |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

P_bw = P_bs

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%