Roboguru

Volume can lukas permukaan bangun di camping adalah...

Pertanyaan

Volume can lukas permukaan bangun di camping adalah...

  1. straight V equals 1.232 space cm cubed comma space straight L equals 700 space cm squared

  2. straight V equals 1.232 space cm cubed comma space straight L equals 702 space cm squared

  3. straight V equals 1.232 space cm cubed comma space straight L equals 704 space cm squared

  4. straight V equals 1.232 space cm cubed comma space straight L equals 706 space cm squared

Pembahasan Soal:

Diketahui:

Jari minus jari space open parentheses straight r close parentheses equals 7 space cm

Tinggi space kerucut equals 24 space cm

Ditanyakan:

Volume space dan space luas space permukaan equals... ?

Penyelesaian:

pertama kita menentukan volume kerucut:

table attributes columnalign right center left columnspacing 0px end attributes row V equals cell 1 third cross times straight pi cross times straight r squared cross times straight t end cell row blank equals cell 1 third cross times 22 over 7 cross times 7 squared cross times 24 end cell row blank equals cell 1.232 space cm cubed end cell end table

Untuk mencari luas permukaan kerucut terlebih dahulu kita tentukan panjang garis pelukis (s), dengan menggunkan teorema Pythagoras:

table attributes columnalign right center left columnspacing 0px end attributes row straight s equals cell square root of straight t squared plus straight r squared end root end cell row blank equals cell square root of 24 squared plus 7 squared end root end cell row blank equals cell square root of 576 plus 49 end root end cell row blank equals cell square root of 625 end cell row blank equals cell 25 space cm end cell end table

Sehingga luas permukaan kerucut adalah:'

table attributes columnalign right center left columnspacing 0px end attributes row L equals cell πr open parentheses straight r plus straight s close parentheses end cell row blank equals cell 22 over 7 cross times 7 open parentheses 7 plus 25 close parentheses end cell row blank equals cell 22 cross times 32 end cell row blank equals cell 704 space cm squared end cell end table

Jadi, jawaban yang tepat adalah C

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Nasrullah

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 30 April 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Hitunglah luas permukaan dan volume bangun di bawah ini ! b.

Pembahasan Soal:

Ingat, 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell L subscript p end cell equals cell straight pi times r open parentheses r plus s close parentheses end cell row V equals cell 1 third times straight pi times r squared times t end cell end table end style 

sehingga, 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell L subscript p end cell equals cell straight pi times r open parentheses r plus s close parentheses end cell row blank equals cell 3 comma 14 times 6 open parentheses 6 plus 10 close parentheses end cell row blank equals cell 301 comma 44 space cm squared end cell row V equals cell 1 third times straight pi times r squared times t end cell row blank equals cell 1 third times 3 comma 14 times 6 squared times 8 end cell row blank equals cell 301 comma 44 space cm cubed end cell end table end style

Jadi, luas permukaan dan volume kerucut tersebut masing-masing adalah begin mathsize 14px style 301 comma 44 space cm squared space dan space 301 comma 44 space cm cubed end style.

0

Roboguru

Suatu kerucut memiliki jari-jari  dan tinggi . Jika luas permukaan kerucut adalah  dan volume kerucut adalah  maka tentukan: a. Nilai dari

Pembahasan Soal:

Diketaui kerucut dengan ukuran begin mathsize 14px style r equals 6 space cm end stylebegin mathsize 14px style t equals t space cm end style, begin mathsize 14px style L equals A space cm squared end style dan begin mathsize 14px style V equals A space cm cubed end style, maka:

begin mathsize 14px style s squared equals r squared plus t squared s equals plus-or-minus square root of r squared plus t squared end root end style

Karena panjang garis pelukis kerucut tidak mungkin negatif, maka yang memenuhi adalah begin mathsize 14px style s equals square root of r squared plus t squared end root end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell L equals πr left parenthesis straight r plus straight s right parenthesis A equals straight pi cross times 6 cross times open parentheses 6 plus square root of 6 squared plus straight t squared end root close parentheses end cell row A equals cell 36 straight pi plus 6 straight pi square root of 36 plus straight t squared end root end cell row blank blank blank row straight V equals cell 1 third πr squared straight t end cell row straight A equals cell 1 third straight pi cross times 6 squared cross times straight t end cell row straight A equals cell 12 πt end cell row cell 36 up diagonal strike straight pi plus 6 up diagonal strike straight pi square root of 36 plus straight t squared end root end cell equals cell 12 up diagonal strike straight pi straight t space left parenthesis kedua space ruas space divided by 6 right parenthesis end cell row cell 6 plus square root of 36 plus straight t squared end root end cell equals cell 2 straight t end cell row cell square root of 36 plus straight t squared end root end cell equals cell 2 straight t minus 6 space left parenthesis kedua space ruas space dikuadratkan right parenthesis end cell row cell 36 plus straight t squared end cell equals cell 4 straight t squared minus 24 straight t plus 36 end cell row cell 4 straight t squared minus 24 straight t plus 36 minus 36 minus straight t squared end cell equals 0 row cell 3 straight t squared minus 24 straight t end cell equals cell 0 space left parenthesis kedua space ruas space divided by 3 right parenthesis end cell row cell straight t squared minus 8 straight t end cell equals 0 row straight t equals cell 0 space atau space straight t equals 8 end cell end table end style 

Karena tinggi kerucut tidak mungkin bernilai 0, maka yang memenuhi adalah begin mathsize 14px style t equals 8 space cm end style.

Jadi tinggi kerucut tersebut adalah begin mathsize 14px style 8 space cm end style

0

Roboguru

Suatu kerucut memiliki jari-jari  dan tinggi . Jika luas permukaan kerucut adalah  dan volume kerucut adalah  maka tentukan: b. Nilai dari

Pembahasan Soal:

Diketaui kerucut dengan ukuranbegin mathsize 14px style r equals 6 space cm end style, begin mathsize 14px style t equals t space cm end style, begin mathsize 14px style L equals A space cm squared end style dan begin mathsize 14px style V equals A space cm cubed end style, maka:

begin mathsize 14px style s squared equals r squared plus t squared s equals plus-or-minus square root of r squared plus t squared end root end style

Karena panjang garis pelukis kerucut tidak mungkin negatif, maka yang memenuhi adalah begin mathsize 14px style s equals square root of r squared plus t squared end root end stylebegin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell L equals πr left parenthesis straight r plus straight s right parenthesis A equals straight pi cross times 6 cross times open parentheses 6 plus square root of 6 squared plus straight t squared end root close parentheses end cell row A equals cell 36 straight pi plus 6 straight pi square root of 36 plus straight t squared end root end cell row blank blank blank row straight V equals cell 1 third πr squared straight t end cell row straight A equals cell 1 third straight pi cross times 6 squared cross times straight t end cell row straight A equals cell 12 πt end cell row cell 36 up diagonal strike straight pi plus 6 up diagonal strike straight pi square root of 36 plus straight t squared end root end cell equals cell 12 up diagonal strike straight pi straight t space left parenthesis kedua space ruas space divided by 6 right parenthesis end cell row cell 6 plus square root of 36 plus straight t squared end root end cell equals cell 2 straight t end cell row cell square root of 36 plus straight t squared end root end cell equals cell 2 straight t minus 6 space left parenthesis kedua space ruas space dikuadratkan right parenthesis end cell row cell 36 plus straight t squared end cell equals cell 4 straight t squared minus 24 straight t plus 36 end cell row cell 4 straight t squared minus 24 straight t plus 36 minus 36 minus straight t squared end cell equals 0 row cell 3 straight t squared minus 24 straight t end cell equals cell 0 space left parenthesis kedua space ruas space divided by 3 right parenthesis end cell row cell straight t squared minus 8 straight t end cell equals 0 row straight t equals cell 0 space atau space straight t equals 8 end cell end table end style 

Karena tinggi kerucut tidak mungkin bernilai 0, maka yang memenuhi adalah begin mathsize 14px style t equals 8 space cm end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row V equals cell 1 third πr squared straight t end cell row cell straight A space cm cubed end cell equals cell 1 third cross times 3 comma 14 cross times open parentheses 6 space cm close parentheses squared cross times 8 space cm end cell row cell straight A up diagonal strike space cm cubed end strike end cell equals cell 301 comma 44 space up diagonal strike cm cubed end strike end cell row straight A equals cell 301 comma 44 end cell end table end style 

Jadi, nilai A adalah begin mathsize 14px style 301 comma 44 end style

0

Roboguru

Volume can lukas permukaan bangun di camping adalah...

Pembahasan Soal:

Diketahui:

Jari minus jari space open parentheses straight r close parentheses equals 7 space cm

Tinggi space kerucut equals 24 space cm

Ditanyakan:

Volume space dan space luas space permukaan equals... ?

Penyelesaian:

pertama kita menentukan volume kerucut:

table attributes columnalign right center left columnspacing 0px end attributes row V equals cell 1 third cross times straight pi cross times straight r squared cross times straight t end cell row blank equals cell 1 third cross times 22 over 7 cross times 7 squared cross times 24 end cell row blank equals cell 1.232 space cm cubed end cell end table

Untuk mencari luas permukaan kerucut terlebih dahulu kita tentukan panjang garis pelukis (s), dengan menggunkan teorema Pythagoras:

table attributes columnalign right center left columnspacing 0px end attributes row straight s equals cell square root of straight t squared plus straight r squared end root end cell row blank equals cell square root of 24 squared plus 7 squared end root end cell row blank equals cell square root of 576 plus 49 end root end cell row blank equals cell square root of 625 end cell row blank equals cell 25 space cm end cell end table

Sehingga luas permukaan kerucut adalah:'

table attributes columnalign right center left columnspacing 0px end attributes row L equals cell πr open parentheses straight r plus straight s close parentheses end cell row blank equals cell 22 over 7 cross times 7 open parentheses 7 plus 25 close parentheses end cell row blank equals cell 22 cross times 32 end cell row blank equals cell 704 space cm squared end cell end table

Jadi, jawaban yang tepat adalah C

0

Roboguru

Hitunglah Luas permukaan dan volume kerucut dibawah ini!

Pembahasan Soal:

Diketahui:

Jari-jari alas equals r equals 6 space dm

Tinggi kerucut equals t equals 9 space dm

Ditanyakan:

Luas permukaan dan volume kerucut adalah...?

Penyelesaian:

  • Luas permukaan:

Untuk menentukan luas permukaan kerucut di atas, dapat menggunakan rumus:

L p equals πr open parentheses straight r plus straight s close parentheses

Pertama kita menentukan garis pelukis (s) dari kerucut tersebut:

table attributes columnalign right center left columnspacing 0px end attributes row straight s equals cell square root of straight r squared plus straight t squared end root end cell row blank equals cell square root of 6 squared plus 9 squared end root end cell row blank equals cell square root of 36 plus 81 end root end cell row blank equals cell square root of 117 end cell row blank equals cell 3 square root of 13 space dm end cell end table

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell L p end cell equals cell πr open parentheses straight r plus straight s close parentheses end cell row blank equals cell 3 comma 14 cross times 6 open parentheses 6 plus 3 square root of 13 close parentheses end cell row blank blank cell 18 comma 84 open parentheses 16 comma 82 close parentheses space end cell row blank equals cell 316 comma 89 space dm squared end cell end table

Volume kerucut

table attributes columnalign right center left columnspacing 0px end attributes row V equals cell 1 third πr squared straight t end cell row blank equals cell 1 third cross times 3 comma 14 cross times 6 squared cross times 9 end cell row blank equals cell 1 third cross times 3 comma 14 cross times 36 cross times 9 end cell row blank equals cell 339 comma 12 space dm cubed end cell end table

Jadi, Luas permukaan dan volume kerucut masing-masing adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 316 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 89 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell dm squared end cell end table dan table attributes columnalign right center left columnspacing 0px end attributes row blank blank 339 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 12 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell dm cubed end cell end table

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

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