[Risolto] Problemi, sistemi

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Base $\small =b;$

ciascun lato obliquo $\small =l;$

sistema (con sostituzione):

$\small \begin{Bmatrix} b+2l &=&64\\
2l-\dfrac{b}{2}&=&28\\
\end{Bmatrix}$

$\small \begin{Bmatrix} b&=&64-2l\\
4l-b&=&56\\
\end{Bmatrix}$

sostituisci la $\small b$ nella 2° equazione:

$\small \begin{Bmatrix} b&=&64-2l\\
4l-(64-2l)&=&56\\
\end{Bmatrix}$

$\small \begin{Bmatrix} b&=&64-2l\\
4l-64+2l&=&56\\
\end{Bmatrix}$

$\small \begin{Bmatrix} b&=&64-2l\\
6l&=&56+64\\
\end{Bmatrix}$

$\small \begin{Bmatrix} b&=&64-2l\\
6l&=&120\\
\end{Bmatrix}$

$\small \begin{Bmatrix} b&=&64-2l\\
\dfrac{\cancel6l}{\cancel6}&=&\dfrac{\cancel{120}^{20}}{\cancel6_1}\\
\end{Bmatrix}$

$\small \begin{Bmatrix} b&=&64-2l\\
l&=&20\\
\end{Bmatrix}$

sostituisci la $\small l$ nella 1°:

$\small \begin{Bmatrix} b&=&64-2·20\\
l&=&20\\
\end{Bmatrix}$

$\small \begin{Bmatrix} b&=&64-40\\
l&=&20\\
\end{Bmatrix}$

$\small \begin{Bmatrix} b&=&24\\
l&=&20\\
\end{Bmatrix}$

quindi risulta:

base $\small =b= 24\,cm;$

ciascun lato obliquo $\small =l = 20\,cm;$

altezza $\small h= \sqrt{l^2-\left(\dfrac{b}{2}\right)^2}= \sqrt{20^2-\left(\dfrac{24}{2}\right)^2} = \sqrt{20^2-12^2} =\sqrt{400-144} =\sqrt{256}= 16\,cm$ (teorema di Pitagora);

area $\small A= \dfrac{b·h}{2} = \dfrac{\cancel{24}^{12}·16}{\cancel2_1} = 12·16 = 192\,cm^2.$

4l-b = 56

2l+b = 64

6l = 120

l = 120/6 = 20 cm

base b = 64-40 = 24 cm 

semiperimetro p = 64/2 = 32 cm 

altezza h = √20^2-12^2 = 16,0 cm 

area A = b*h/2 = 24*8 = 192 cm 

area A = √32*(32-20)*(32-20)*(32-24) = √32*12^2*8 = 192 cm^2