How do i use composition and decomposition to determine the area of a polygon?

Mathematics

1 answer:

wolverine [178]3 years ago

3 0

Answer: Decompose means the opposite, to take apart

Step-by-step explanation: In this lesson, the words composition and decomposition are used to describe how irregular figures can be separated into triangles and other polygons. The area of these parts can then be added together to calculate the area of the whole figure.

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Nesterboy [21]

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Ilia_Sergeevich [38]

I'll go out on a limb and guess the system is

$\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x$

with initial condition $\mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top$ . The coefficient matrix has eigenvalues $\lambda$ such that

$\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12$

The corresponding eigenvectors $\eta$ are such that

$\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}$

$\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}$

So the characteristic solution to the ODE system is

$\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

When $t=0$ , we have

$\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}$

from which it follows that $C_1=2$ and $C_2=5$ , making the particular solution to the IVP

$\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

$\mathbf x(t)=\begin{bmatrix}2e^{t/2}\\2e^{t/2}+5e^{-t/2}\end{bmatrix}$

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bezimeni [28]