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Mathematics 19 Online
OpenStudy (shaleiah):

Area

OpenStudy (shaleiah):

@Michele_Laino

OpenStudy (shaleiah):

We should draw a pentagon first.

OpenStudy (michele_laino):

here we have to apply the subsequent formula: \[area = \frac{{perimeter \times apothem}}{2}\]

OpenStudy (shaleiah):

The instructor says the formula of a regular polygon.

OpenStudy (michele_laino):

hint: perimeter is: \(perimeter= 5 \times side=5 \times 9=...?\)

OpenStudy (shaleiah):

45

OpenStudy (michele_laino):

correct!

OpenStudy (michele_laino):

and apothem is: \(apothem = side \times 0.688= 9 \times 0.688=...?\) here \(0.688\) is a fixed nuumber

OpenStudy (shaleiah):

6.192

OpenStudy (michele_laino):

correct! so the requested area, is: \[\Large \begin{gathered} area = \frac{{perimeter \times apothem}}{2} = \hfill \\ \hfill \\ = \frac{{45 \times 6.192}}{2} = ...? \hfill \\ \end{gathered} \]

OpenStudy (shaleiah):

139.32

OpenStudy (michele_laino):

that's right! Now we have to round such result as requested from the problem

OpenStudy (michele_laino):

hint: we have to round to the nearest tenth

OpenStudy (shaleiah):

140

OpenStudy (michele_laino):

hint: if I write 139.32 I have rounded to the nearest hundredth

OpenStudy (michele_laino):

we have to take only one decimal figure

OpenStudy (shaleiah):

139.3

OpenStudy (michele_laino):

that's right! :)

OpenStudy (michele_laino):

the similitude ratio is: \[\frac{{39}}{{27}} = ...?\] please simplify such ratio

OpenStudy (shaleiah):

\[\frac{ 13 }{ 9 }\]

OpenStudy (michele_laino):

correct! So the requested area, is: \[A = 711 \cdot {\left( {\frac{{13}}{9}} \right)^2} = ...?\]

OpenStudy (shaleiah):

1483.4

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