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Pertanyaan


 

Lp equals.... space cm squared 

straight V equals.... space cm cubed 

Pembahasan Soal:

Tentuka tinggi sisi tegak (segitiga) dengan Pythagoras.

table attributes columnalign right center left columnspacing 0px end attributes row cell t subscript 1 end cell equals cell square root of open parentheses AB over 2 close parentheses squared plus TO squared end root end cell row blank blank cell square root of open parentheses 36 over 2 close parentheses squared plus 24 squared end root end cell row blank equals cell square root of 18 squared plus 24 squared end root end cell row blank equals cell square root of 324 plus 576 end root end cell row blank equals cell square root of 900 end cell row blank equals cell 30 space cm end cell end table  

table attributes columnalign right center left columnspacing 0px end attributes row cell t subscript 2 end cell equals cell square root of open parentheses BC over 2 close parentheses squared plus TO squared end root end cell row blank equals cell square root of open parentheses 20 over 2 close parentheses squared plus 24 squared end root end cell row blank equals cell square root of 10 squared plus 24 squared end root end cell row blank equals cell square root of 100 plus 576 end root end cell row blank equals cell square root of 676 end cell row blank equals cell 26 space cm end cell end table  

Sehingga, luas permukaan limas tersebut:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row Lp equals cell straight L. space alas plus 4 times jumlah space sisi space tegak end cell row blank equals cell open parentheses p times l close parentheses plus open parentheses open parentheses 2 times 1 half times a subscript 1 times t subscript 1 close parentheses plus open parentheses 2 times 1 half times a subscript 2 times t subscript 2 close parentheses close parentheses end cell row blank equals cell open parentheses 36 times 20 close parentheses plus open parentheses up diagonal strike 2 times fraction numerator 1 over denominator up diagonal strike 2 end fraction times 20 times 30 close parentheses plus open parentheses up diagonal strike 2 times fraction numerator 1 over denominator up diagonal strike 2 end fraction times 36 times 26 close parentheses end cell row blank equals cell 720 plus 600 plus 936 end cell row blank equals cell 2.256 space cm squared end cell end table end style   

Sedangkan, volumenya:

table attributes columnalign right center left columnspacing 0px end attributes row straight V equals cell 1 third times straight L. space alas times straight t. space limas end cell row blank equals cell fraction numerator 1 over denominator up diagonal strike 3 end fraction times open parentheses 36 times 20 close parentheses times stack up diagonal strike 24 with 8 on top end cell row blank equals cell 720 times 8 end cell row blank equals cell 5.760 space cm cubed end cell end table 

Jadi, Lp equals 2.256 space cm squared dan straight V equals 5.760 space cm cubed.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Nur

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

Lengkapilah tabel berikut dengan menentukan luas permukaan dan volume bangun ruangnya.

Pembahasan Soal:

  • Pandang segitiga ABC

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell A B end cell equals 6 row cell A C end cell equals 8 row cell B C end cell equals cell square root of 6 squared plus 8 squared end root end cell row blank equals cell square root of 36 plus 64 end root end cell row blank equals cell square root of 100 end cell row blank equals 10 row blank blank blank row cell text Luas end text end cell equals cell 1 half times 6 times 8 end cell row blank equals 24 end table end style

Jadi, luas segitiga ABC adalah begin mathsize 14px style 24 space text cm end text squared end style 

  • Pandang segitiga ABT

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Jadi, luas segitiga ABT adalah begin mathsize 14px style 24 space text cm end text squared end style 

  • Pandang segitiga ACT

Asumsikan begin mathsize 14px style angle T A C end style siku-siku, sehingga diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell A T end cell equals 8 row cell A C end cell equals 8 row cell C T end cell equals cell square root of 8 squared plus 8 squared end root end cell row blank equals cell square root of 64 times 2 end root end cell row blank equals cell 8 square root of 2 end cell row blank blank blank row cell text Luas end text end cell equals cell 1 half times 8 times 8 end cell row blank equals 32 end table end style 

Jadi, luas segitiga ACT adalah begin mathsize 14px style 32 space text cm end text squared end style 

  • Pandang segitiga BCT

Segitiga BCT merupakan segitiga sama kaki, sehingga diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell B T end cell equals cell B C end cell row blank equals 10 row cell C T end cell equals cell 8 square root of 2 end cell row cell text Tinggi end text end cell equals cell square root of 10 squared minus open parentheses 4 square root of 2 close parentheses squared end root end cell row blank equals cell square root of 100 minus 32 end root end cell row blank equals cell square root of 68 end cell row blank equals cell 2 square root of 17 end cell row blank blank blank row cell text Luas end text end cell equals cell 1 half times 8 square root of 2 times 2 square root of 17 end cell row blank equals cell 8 square root of 34 end cell end table end style 

Jadi, luas segitiga BCT adalah begin mathsize 14px style 8 square root of 34 space text cm end text squared end style 

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell text Luas permukaan limas end text end cell equals cell text Luas ABT+Luas ABC+Luas ACT+Luas BCT end text end cell row blank equals cell 24 plus 24 plus 32 plus 8 square root of 34 end cell row blank equals cell 50 plus 8 square root of 34 end cell end table end style    

Jadi, luas permukaan limas adalah begin mathsize 14px style 50 plus 8 square root of 34 space text cm end text squared end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell text Volume end text end cell equals cell 1 third times text Luas alas·tinggi end text end cell row blank equals cell 1 third text ·Luas segitiga  end text A B C times text tinggi end text end cell row blank equals cell 1 third times 24 times 8 end cell row blank equals 64 end table end style

Jadi, volume limas adalah begin mathsize 14px style 64 space text cm end text cubed end style

 

0

Roboguru

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Luas space alas end cell equals cell P Q cross times Q R end cell row blank equals cell 30 cross times 30 end cell row blank equals cell 900 space cm squared end cell row blank blank blank row cell Luas space permukaan end cell equals cell luas space alas plus 4 cross times luas space sisi space tegak end cell row blank equals cell 900 plus 4 open parentheses 1 half cross times alas cross times tinggi space segitiga close parentheses end cell row blank equals cell 900 plus 4 open parentheses 1 half cross times 30 cross times 25 close parentheses end cell row blank equals cell 900 plus 1.500 end cell row blank equals cell 2.400 space cm squared end cell row blank blank blank row Volume equals cell 1 third cross times luas space alas cross times tinggi end cell row blank equals cell 1 third cross times 900 cross times 20 end cell row blank equals cell 6.000 space cm cubed end cell end table end style

0

Roboguru

\  Volume dan luas permukaan bangun di atas adalah ...

Pembahasan Soal:

Diasumsikan tinggi limas adalah begin mathsize 14px style 20 space cm end style dan panjang sisi yang saling tegak lurus adalah begin mathsize 14px style 8 space cm end style dan begin mathsize 14px style 15 space cm end style 
begin mathsize 14px style straight V subscript limas equals 1 third cross times 8 cross times 15 cross times 20 space space space space space space space space equals 800 space cm cubed end style 

dengan menggunakan tripel pytagoras didapatkan bahwa ukuran segitiga alas begin mathsize 14px style 8 space cm comma space 15 space cm space dan space 17 space cm end style,ukuran sisi tegaknya size 14px 6 size 14px space size 14px cm size 14px comma size 14px space size 14px 8 size 14px space size 14px cm size 14px comma size 14px 10 size 14px space size 14px cm  

begin mathsize 14px style L u a s space p e r m u k a a n equals 1 half cross times 8 cross times 15 plus 1 half cross times 8 cross times 6 plus 1 half cross times 10 cross times 15 plus 1 half cross times 17 cross times 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 60 plus 24 plus 75 plus 51 space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 210 space cm squared end style 

0

Roboguru

Sebuah limas segitiga memiliki tinggi  Alasnya berupa segitiga siku-siku dengan sisi siku-sikunya  dan  Luas sisi tegaknya masing-masing adalah  dan   Tentukan volume dan luas permukaan limas tersebut...

Pembahasan Soal:

Diketahui:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell t subscript l i m a s end subscript end cell equals cell 21 space cm end cell row a equals cell 5 space cm end cell row t equals cell 12 space cm end cell row cell L subscript 1 end cell equals cell 30 space cm squared end cell row cell L subscript 2 end cell equals cell 40 space cm squared end cell row cell L subscript 3 end cell equals cell 24 space cm squared end cell end table end style

Maka:

  begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row V equals cell 1 third times luas space alas times t subscript l i m a s end subscript end cell row blank equals cell 1 third times left parenthesis 1 half times a times t right parenthesis times t subscript l i m a s end subscript end cell row blank equals cell 1 third times left parenthesis 1 half times 5 times 12 right parenthesis times 21 end cell row blank equals cell 1 third times 30 times 21 end cell row blank equals cell 210 space cm cubed end cell row blank blank blank row cell L subscript p end cell equals cell luas space alas plus luas space sisi space tegak end cell row blank equals cell left parenthesis 1 half times a times t right parenthesis plus left parenthesis L subscript 1 plus L subscript 2 plus L subscript 3 right parenthesis end cell row blank equals cell left parenthesis 1 half times 5 times 12 right parenthesis plus left parenthesis 30 plus 40 plus 24 right parenthesis end cell row blank equals cell 30 plus 94 end cell row blank equals cell 124 space cm squared end cell end table end style

Jadi, volume dan luas permukaan limas tersebut adalah begin mathsize 14px style 210 space cm cubed end style dan begin mathsize 14px style 124 space cm squared. end style 

0

Roboguru

Sebuah limas dengan alas berbentuk persegi mempunyai volume dan panjang rusuk alas . Tentukan luas permukaan limas tersebut!

Pembahasan Soal:

  • Menentukan tinggi limas:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight V. limas end cell equals cell 1.296 space cm cubed end cell row cell 1 third cross times straight L. alas cross times Tinggi space limas end cell equals cell 1.296 space cm cubed end cell row cell 1 third cross times open parentheses 18 space cm cross times 18 space cm close parentheses cross times tinggi space limas end cell equals cell 1.296 space cm cubed end cell row cell 108 space cm squared cross times tinggi space limas space end cell equals cell space 1.296 space cm cubed end cell row cell tinggi space limas space end cell equals cell fraction numerator 1.296 space cm cubed over denominator 108 space cm squared end fraction end cell row cell tinggi space limas end cell equals cell 12 space cm end cell end table

  • Menentukan tinggi sisi tegak:

table attributes columnalign right center left columnspacing 0px end attributes row cell tinggi space sisi space tegak end cell equals cell square root of 12 squared plus 9 squared end root end cell row blank equals cell square root of 144 plus 81 end root end cell row blank equals cell square root of 225 end cell row blank equals 15 end table

  • Menentukan luas permukaan limas:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight L. permukaan end cell equals cell straight L. alas plus 4 space straight L. sisi space tegak end cell row blank equals cell open parentheses 18 space cm close parentheses squared plus 4 open parentheses 1 half cross times space straight a. sisi space tegak cross times straight t. sisi space tegak close parentheses end cell row blank equals cell 324 space cm squared plus 4 open parentheses 1 half cross times 18 space cm cross times 15 space cm close parentheses end cell row blank equals cell 324 space cm squared plus 540 space cm squared end cell row blank equals cell 864 space cm squared end cell end table end style

Jadi, luas permukaan limas adalah 864 space cm squared.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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