Matemaatika:Gümnaasium/Kordamine eksamiks/Geomeetria tasandil

Valemid

Ruut

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Ruut

$\begin{array}{c}d=a\sqrt{2}\end{array}$

$\begin{array}{c}S=a^2\end{array}$

Kolmnurk

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Kolmnurk

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

$S=\sqrt{p(p-a)(p-b)(p-c)},$ kus $p=\frac{1}{2}(a+b+c)$

$\begin{array}{c}{a}^2=b^2+c^2-2bc\ast cos\alpha \\ b^2=a^2+c^2-2ac\ast cos\beta \\ c^2=a^2+b^2-2ab\ast cos\gamma \end{array}\}$Koosinusteoreem

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

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

Täisnurkne kolmnurk

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Täisnurkne kolmnurk

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

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

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

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

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

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

Ringjoon, ring, sektor

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Ring

$\begin{array}{c}C=2\ast \pi \ast r=\pi \ast d\end{array}$, C on ümberringjoone pikkus

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

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

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

Ristkülik

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Ristkülik

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

$\begin{array}{c}S=a\ast b\end{array}$

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

Romb

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Romb

$\begin{array}{c}S=a\ast h\end{array}$

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

$\begin{array}{c}S=\frac{1}{2}{d}_1{d}_2\end{array}$

Näited

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

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

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

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

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

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

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

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

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

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