Roboguru

Perhatikan dengan baik bangun ruang di bawah ini, lalu hitunglah luas permukaannya!

Pertanyaan

Perhatikan dengan baik bangun ruang di bawah ini, lalu hitunglah luas permukaannya!


  

Pembahasan Soal:

Limas

Diketahui:

p equals 13 space cm l equals 6 space cm t equals 15 space cm

Untuk menentukan luas permukaan limas, tentukan tinggi sisi tegak limas tersebut.

Tinggi segitiga 1;

table attributes columnalign right center left columnspacing 0px end attributes row cell t subscript increment 1 end subscript end cell equals cell square root of open parentheses 1 half p close parentheses squared plus t squared end root end cell row blank equals cell square root of open parentheses 1 half cross times 13 close parentheses squared plus 15 squared end root end cell row blank equals cell square root of 6 comma 5 squared plus 15 squared end root end cell row blank equals cell square root of 42 comma 25 plus 225 end root end cell row blank equals cell square root of 267 comma 25 end root space cm end cell end table

Tinggi segitiga 2:

table attributes columnalign right center left columnspacing 0px end attributes row cell t subscript increment 2 end subscript end cell equals cell square root of open parentheses 1 half l close parentheses squared plus t squared end root end cell row blank equals cell square root of open parentheses 1 half cross times 6 close parentheses squared plus 15 squared end root end cell row blank equals cell square root of 3 squared plus 15 squared end root end cell row blank equals cell square root of 9 plus 225 end root end cell row blank equals cell square root of 234 end cell row blank equals cell 3 square root of 26 space cm end cell end table  

Luas permukaan limas segi empat:

table attributes columnalign right center left columnspacing 0px end attributes row cell L subscript limas end cell equals cell Luas space alas plus Luas space sisi space tegak end cell row blank equals cell p cross times l plus 2 open parentheses 1 half cross times p cross times t subscript increment 2 end subscript close parentheses plus 2 open parentheses 1 half cross times l cross times t subscript increment 1 end subscript close parentheses end cell row blank equals cell 13 cross times 6 plus open parentheses 13 cross times 3 square root of 26 close parentheses plus open parentheses 6 cross times square root of 267 comma 25 end root close parentheses end cell row blank equals cell 78 plus 39 square root of 26 plus 6 square root of 267 comma 25 end root end cell row blank equals cell 3 open parentheses 26 plus 13 square root of 26 plus 2 square root of 267 comma 25 end root close parentheses space cm squared end cell end table 

Jadi, luas permukaan limas segi empat tersebut adalah 3 open parentheses 26 plus 13 square root of 26 plus 2 square root of 267 comma 25 end root close parentheses space cm squared.

Balok

Diketahui:

p equals 28 space cm l equals 11 space cm t equals 13 space cm 

Luas permukaan balok:

table attributes columnalign right center left columnspacing 0px end attributes row cell L subscript balok end cell equals cell 2 open parentheses p cross times l plus p cross times t plus l cross times t close parentheses end cell row blank equals cell 2 open parentheses 28 cross times 11 plus 28 cross times 13 plus 11 cross times 13 close parentheses end cell row blank equals cell 2 open parentheses 308 plus 364 plus 143 close parentheses end cell row blank equals cell 2 open parentheses 815 close parentheses end cell row blank equals cell 1.630 space cm squared end cell end table

Jadi, luas permukaan balok tersebut adalah 1.630 space cm squared.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 20 Mei 2021

Ruangguru

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved