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SMK Teknologi dan Rekayasa

Soal 61 Luas Permukaan Bangun Ruang


Sebuah\ tabung\ tertutup\ akan\ dibuat\ dari\ selembar\ seng\ dengan\\jari-jari\ 14\ cm\ dan\ tinggi\ 20\ cm,\ luas\ seng\ yang\ dibutuhkan\\adalah\ ....\\a.\ \ 2.376\ cm^{2}\ \ \ \ \ \ \ \ \ \ b.\ \ 2.773\ cm^{2}\ \ \ \ \ \ \ \ \ \ c.\ \ 2.992\ cm^{2}\ \ \ \ \ \ \ \ \ \ d.\ \ 3.192\ cm^{2}\ \ \ \ \ \ \ \ \ \ e.\ \ 3.216\ cm^{2}
Jawaban: c


\begin{array}{rcl}Luas\ seng&=&2\pi r^{2}+2\pi rt\\\\&=&2\pi r(r+t)\\\\&=&2\times \frac{22}{7}\times 14(14+20)\\\\&=&88+34\\\\&=&2.992\ cm^{2}\end{array}

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One thought on “Soal 61 Luas Permukaan Bangun Ruang

  1. Diperbanyak lagi soalnya :))

    Posted by Elenni Mi-chan | 24 Maret 2011, 10:40 AM

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