Matemaatika:Kordamine eksamiks/Gümnaasium/Geomeetria tasandil

Valemid

Ruut

Ruut

$\begin{matrix} d=a\sqrt2 \end{matrix}$

$\begin{matrix} S=a^2 \end{matrix}$

Kolmnurk

Kolmnurk

$S=\frac{b*h}{2}=\frac{a*b*sin\gamma}{2}$

$S=\sqrt{p(p-a)(p-b)(p-c)},$ kus $p=\frac1 2(a+b+c)$

$\left.\begin{matrix} a^2=b^2+c^2-2bc*cos\alpha \\ b^2=a^2+c^2-2ac*cos\beta\\ c^2=a^2+b^2-2ab*cos\gamma \end{matrix}\right\}$Koosinusteoreem

$\frac{a}{sin\alpha}=\frac{b}{sin\beta}=\frac{c}{sin\gamma}=2R$Siinusteoreem

$\begin{matrix} \alpha+\beta+\gamma=180^0 \\ \mbox{R - } \ddot{u} \mbox{mberringjoone raadius} \end{matrix}$

Täisnurkne kolmnurk

Täisnurkne kolmnurk

$a^2+b^2=c^2\Rightarrow c=\sqrt{a^2+b^2}$ Pythagorase teoreem

$S=\frac{ab}{2}=\frac{ch}{2}$

$sin\alpha=\frac{a}{c}=cos\beta$

$cos\alpha=\frac{b}{c}=sin\beta$

$tan\alpha=\frac{a}{b}$

$\begin{matrix} \alpha+\beta=90^0 \\a^2=f*c; & sin(90^0-\alpha)=cos\alpha \\ b^2=g*c; & cos(90^0-\alpha)=sin\alpha \\ h^2=f*g; & tan(90^0-\alpha)=\frac{1}{tan\alpha}\end{matrix}$

Ringjoon, ring, sektor

Ring

$\begin{matrix} C=2*\pi*r=\pi*d \end{matrix}$, C on ümberringjoone pikkus

$S=\pi*r^2=\frac{\pi*d^2}{4}$, S on täispindala

$S_s=\frac{\pi*r^2}{2\pi}*x=\frac{x*r^2}{2}=\frac{l*r}{2}$, S_s on sektori pinda

$l=\frac{2*\pi*r}{2\pi}*x=x*r$, l on sektori kaare pikkus

Ristkülik

Ristkülik

$\begin{matrix} d=\sqrt{a^2+b^2} \end{matrix}$

$\begin{matrix} S=a*b \end{matrix}$

$\begin{matrix} P=2(a+b) \end{matrix}$

Romb

Romb

$\begin{matrix} S = a*h \end{matrix}$

$\begin{matrix} S=a^2sin\alpha \end{matrix}$

$\begin{matrix} S=\frac{1}{2}d_1d_2 \end{matrix}$

Näited

1. Rombi ümber asetseb minimaalse suurusega ring. Leia mitu korda on romb ringist väiksem, kui antud on rombi lühem diagonaal ja alus.

$h=\sqrt{d_2^2-\frac{a^2}{4}}$

$S_{romb}=a*\sqrt{d_2^2-\frac{a^2}{4}}$

$d_1=\frac{a*\sqrt{d_2^2-\frac{a^2}{4}}}{d_2}$

$r=\frac{\frac{a*\sqrt{d_2^2-\frac{a^2}{4}}}{d_2}}{2}$

$S_{ring}=\pi*{\frac{\frac{a*\sqrt{d_2^2-\frac{a^2}{4}}}{d_2}}{2}}^2$

$x=\frac{S_{ring}}{S_{romb}}\Rightarrow x=\frac{\pi*{\frac{\frac{a*\sqrt{d_2^2-\frac{a^2}{4}}}{d_2}}{2}}^2}{a*\sqrt{d_2^2-\frac{a^2}{4}}}\Rightarrow x= \pi*\frac{4a^2*\frac{d_2^2-\frac{a^2}{4}}{d_2^2}}{a*\sqrt{d_2^2-\frac{a^2}{4}}} \Rightarrow$

$\Rightarrow x=4\pi a\frac{(d_2^2-\frac{a^2}{4})*\sqrt{d_2^2-\frac{a^2}{4}}}{d_2^{2}} \Rightarrow$

$\Rightarrow x=(4\pi a-\frac{\pi a^3}{d_2^2})\sqrt{d_2^2+\frac{a^2}{4}}$

Vastus: Romb on ringist $(4\pi a-\frac{\pi a^3}{d_2^2})\sqrt{d_2^2+\frac{a^2}{4}}$ korda väiksem.