Mathematics
19 Online
OpenStudy (shaleiah):
Area
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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:
\(perimeter= 5 \times side=5 \times 9=...?\)
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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:
\[\Large \begin{gathered}
area = \frac{{perimeter \times apothem}}{2} = \hfill \\
\hfill \\
= \frac{{45 \times 6.192}}{2} = ...? \hfill \\
\end{gathered} \]
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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
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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):
\[\frac{ 13 }{ 9 }\]
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OpenStudy (michele_laino):
correct! So the requested area, is:
\[A = 711 \cdot {\left( {\frac{{13}}{9}} \right)^2} = ...?\]
OpenStudy (shaleiah):
1483.4