27.9 Extra opgaven

$21$ ; $31$ ; $2 n + 1$

${(n + 1)}^{2} - n^{2}$ $= n^{2} + 2 n + 1 - n^{2} = 2 n + 1$

$10 + 5 \sqrt{2}$
$\frac{1}{2} \sqrt{2} + 1 \frac{2}{3} \sqrt{3} + 1 \frac{1}{6} \sqrt{6}$
$- \sqrt{2}$

$2 + 5 = 7$ cm , dus $70$ mm
$\sqrt{5} + \sqrt{20}$ cm, dus $57$ mm
$\sqrt{8} + \sqrt{17}$ cm, dus $70$ mm
$\sqrt{13} + 4$ cm, dus $76$ mm

$\sqrt{\frac{1}{100}} = \frac{1}{10}$	$\sqrt{\frac{144}{100}} = \frac{12}{10} = 1,2$
$2^{4} = 16$	$8$
$\sqrt{4} = 2$	$\sqrt{\frac{225}{4}} = \frac{15}{2} = 7 \frac{1}{2}$
$\sqrt{4} = 2$	$\frac{1}{\sqrt{25}} = \frac{1}{5}$

Piet meet: $\sqrt{3^{2} + {(0,2)}^{2}} \approx 3,0067$ meter. Het scheelt $0,67$ cm.

Linkerkolom:
$5 \sqrt{2} + 5 \sqrt{5} + 3 \sqrt{5} + 2 \sqrt{2} = 8 \sqrt{5} + 7 \sqrt{2}$
$\frac{1}{2} \sqrt{2} + \frac{1}{2} + \frac{1}{4} \sqrt{2} + \frac{1}{4} = \frac{3}{4} + \frac{3}{4} \sqrt{2}$

Rechterkolom:
$5 \sqrt{2} \cdot 5 \sqrt{5} \cdot 3 \sqrt{5} \cdot 2 \sqrt{2} = 5 \cdot 2 \cdot 5 \cdot 5 \cdot 3 \cdot 2 = 10 \cdot 10 \cdot 15 = 1500$
$\sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

$2 \sqrt{5}$

$\sqrt{25^{2} + 1^{2}} = \sqrt{626} \approx 25,02$ m, dus $2$ cm langer

Linkerkolom:
${(5 x)}^{2} = 50$
Manier 1:
$5 x = \sqrt{50} = 5 \sqrt{2} of 5 x = ‐ \sqrt{50} = ‐ 5 \sqrt{2}$
$x = \sqrt{2} of x = ‐ \sqrt{2}$
Manier 2:
$25 x^{2} = 50$
$x^{2} = 2$
$x = \sqrt{2} of x = ‐ \sqrt{2}$

${(\sqrt{2} x)}^{2} = 12$
Manier 1:
$\sqrt{2} x = \sqrt{12} of \sqrt{2} x = ‐ \sqrt{12}$
$x = \sqrt{6} of x = ‐ \sqrt{6}$
Manier 2:
$2 x^{2} = 12$
$x^{2} = 6$
$x = \sqrt{6} of x = ‐ \sqrt{6}$

Rechterkolom:
${(5 x)}^{2} = 20$
Manier 1:
$5 x = \sqrt{20} = 2 \sqrt{5} of 5 x = ‐ \sqrt{20} = ‐ 2 \sqrt{5}$
$x = \frac{2}{5} \sqrt{5} of x = ‐ \frac{2}{5} \sqrt{5}$
Manier 2:
$25 x^{2} = 20$
$x^{2} = \frac{20}{25}$
$x = \frac{2}{5} \sqrt{5} of x = ‐ \frac{2}{5} \sqrt{5}$

${(\sqrt{2} x)}^{2} = 10$
Manier 1:
$\sqrt{2} x = \sqrt{10} of \sqrt{2} x = ‐ \sqrt{10}$
$x = \sqrt{5} of x = ‐ \sqrt{5}$
Manier 2:
$2 x^{2} = 10$
$x^{2} = 5$
$x = \sqrt{5} of x = ‐ \sqrt{5}$

Twee zijden van $2 \sqrt{5}$ en een zijde van $2 \sqrt{2}$ .

$\sqrt{2} \cdot \sqrt{18} = 6$

$4 \cdot 5 = 20$	$6 \cdot 5 = 20$	$25 \sqrt{6}$
$2 \sqrt{3}$	$6$	$5 \sqrt{3}$
$2 \sqrt{6} + \sqrt{6} = 3 \sqrt{6}$	$2 \sqrt{6} + 3 \sqrt{6} = 5 \sqrt{6}$	$\sqrt{1000} = 10 \sqrt{10}$
$1 \frac{1}{2} + 2 \frac{1}{2} = 4$	$\sqrt{4} = 2$	$\frac{\sqrt{33}}{3} = \frac{1}{3} \sqrt{33}$
$2 \sqrt{2}$	$\frac{1}{2} \sqrt{6}$	$\frac{3}{5} \sqrt{5}$

$2 \sqrt{2}$

$2 + 4 \sqrt{2} \approx 7,7$ cm, dus $77$ mm

$2 \sqrt{2} \approx 2,83$ cm³, dus $283$ mm³

$2 \sqrt{3}$

$2 \sqrt{2}$

De verhouding is: $\sqrt{2} : \sqrt{6} : 2 \sqrt{2}$ . Als je door $\sqrt{2}$ deelt, krijg je: $1 : \sqrt{3} : 2$ .
$60 °$ , want $∠ A C D = 30 °$ , want driehoek $A C D$ is een $30 ‐ 60 ‐ 90$ -driehoek, blijkt uit de verhouding van de zijden van die driehoek.

De schuine zijde van de blauwe driehoek is $\frac{1}{2} \cdot 18,01 = 9,005$ meter
$x^{2} + 9^{2} = {9,005}^{2}$ geeft: $x \approx 0,30$ m, dus 30 cm.

$\frac{2 \sqrt{3}}{\sqrt{2}} = \sqrt{6}$

$6$

Zie plaatje.
$C S = D S = \frac{6}{\sqrt{2}} = 3 \sqrt{2}$
$A S = B S = \sqrt{3} \cdot 3 \sqrt{2} = 3 \sqrt{6}$
$A D = B C = 2 \cdot 3 \sqrt{2} = 6 \sqrt{2}$
$A B = 3 \sqrt{6} \cdot \sqrt{2} = 6 \sqrt{3}$

$\sqrt{80} = \sqrt{16} \cdot \sqrt{5} = 4 \sqrt{5}$

Per speler $\frac{60.000}{\sqrt{16}} = 15.000$ gulden, totaal: $16 \cdot 15.000 = 240.000$ gulden.

$\frac{60.000}{\sqrt{10}} \cdot 10 \approx 189.737$ gulden

$b = \frac{60.000}{\sqrt{n}} \cdot n = 60.000 \sqrt{n}$