جواب تمرین صفحه 134 درس 8 ریاضی نهم (حجم و مساحت)

الف

\(\begin{array}{l}d = 1/28 \times {10^4}\,km\\\\R = 6400 = 6/4 \times {10^3}\,km\end{array}\)

ب

\(\begin{array}{l}R = 6/4 \times {10^3}\,km\\\\S = 4\pi {R^2} = \\\\4 \times 3/14 \times {(6/4 \times {10^3})^2} \simeq \\\\5/144 \times {10^8}\,k{m^2}\end{array}\)

ج

\(1/648/000\,k{m^2} = 1/648 \times {10^6}\,k{m^2}\)

\( = \frac{{1/648 \times {{10}^6}\,k{m^2}}}{{5/144 \times {{10}^8}\,k{m^2}}} \times 100 = 0/32\,\% \) نسبت

الف

\(h = 1 - 0/3 = 0/7\) استوانه

\(V = \pi {R^2}h = \) استوانه

\(\pi \times {(0/3)^2} \times 0/7 = 0/063\pi \)

\(V = \frac{1}{2} \times \frac{4}{3}\pi {R^3} = \frac{2}{3}\pi {R^3} = \) نیم کره

\(\frac{2}{3}\pi \times {(0/3)^3} = 0/018\pi \)

\(V = 0/063\pi + 0/018\pi = 0/081\pi \) کپسول

ب

\( = 2\pi Rh = \) مساحت جانبی استوانه

\(2\pi \times (0/3)(0/7) = 0/42\pi \)

\( = \pi {R^2} = \pi {(0/3)^2} = 0/09\pi \) مساحت قاعده استوانه

\( = 0/18\pi + 0/42\pi + 0/09\pi = \) مساحت کل

\( = 0/69\pi \simeq 0/69 \times 3/14 \simeq 2/17\,{m^2}\)

\( = 2/17\,{m^2} \times 100\frac{g}{{{m^2}}}\, \times \frac{{1\,kg}}{{1000\,g}}\, = 0/217\,kg\) رنگ لازم

\(V = \frac{1}{2} \times \frac{4}{3}\pi \times {(12)^3} = \frac{2}{3}\pi \times {(12)^3}\) نیم کره = حجم آب

\(V = \pi \times {(12)^2} \times x\) استوانه = حجم آب

\(\begin{array}{l} \Rightarrow \frac{2}{3}\pi \times {(12)^3} = \pi \times {(12)^2} \times x\\\\ \Rightarrow x = \frac{{\frac{2}{3}\pi \times {{(12)}^3}}}{{\pi \times {{(12)}^2}}} = \frac{2}{3} \times 12 = 8\,cm\end{array}\)

ارتفاع آب در استوانه برابر با 8 سانتی متر می شود.