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SMK Teknologi dan Rekayasa

Soal 68 Volume Bangun Ruang


Sebuah\ tempat\ air\ berbentuk\ kerucut\ berdiameter\ 18\ cm.\\Kerucut\ dapat\ menampung\ air\ 1188\ cm^{3}.\ Tinggi\ kerucut\ adalah\ ....\\a.\ \ 28\ cm\ \ \ \ \ \ \ \ \ \ b.\ \ 21\ cm\ \ \ \ \ \ \ \ \ \ c.\ \ 14\ cm\ \ \ \ \ \ \ \ \ \ d.\ \ 7\ cm\ \ \ \ \ \ \ \ \ \ e.\ \ 3,5\ cm
Jawaban: c

\begin{array}{rcl}Volume\ kerucut&=&\frac{1}{3}\pi r^{2}t\\\\\frac{1}{3}\pi r^{2}t&=&1188\\\\\frac{1}{3}\times \pi \times 9\times 9\times t&=&1188\\\\27\pi t&=&1188\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (dibagi\ dengan\ 27)\\\\\pi t&=&44\\\\\frac{22}{7}\times t&=&44\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (dibagi\ dengan\ 22)\\\\\frac{1}{7}\times t&=&2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (dikali\ dengan\ 7)\\\\t&=&14\end{array}
Tinggi\ kerucut\ adalah\ 14\ cm.

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