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Dari segitiga diketahui panjang sisi  dan besar , maka luas daerah segitiga .

Pertanyaan

Dari segitiga A B C diketahui panjang sisi A C italic equals italic 6 italic space c m italic comma italic space A B italic equals italic space italic 8 italic space c m dan besar italic angle B A C italic equals italic 60 italic degree, maka luas daerah segitiga A B C italic equals italic. italic. italic. italic space cm squared.

  1. begin mathsize 14px style 48 end style 

  2. begin mathsize 14px style 24 square root of 3 end style 

  3. size 14px 16 square root of size 14px 3 

  4. size 14px 24 

  5. size 14px 12 square root of size 14px 3 

Pembahasan Video:

Pembahasan Soal:

Segitiga A B C dapat diilustrasikan:

Luas segitiga dapat dicari menggunakan aturan sinus sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row Luas italic equals cell italic 1 over italic 2 italic cross times A C italic cross times C T end cell row blank italic equals cell italic 1 over italic 2 italic cross times A C italic cross times A B italic cross times s i n italic space italic angle B A C end cell row blank italic equals cell italic 1 over italic 2 italic cross times italic 6 italic cross times italic 8 italic cross times s i n italic space italic 60 italic degree end cell row blank italic equals cell italic 24 italic cross times italic 1 over italic 2 square root of italic 3 end cell row blank italic equals cell italic 12 square root of italic 3 end cell end table

Jadi, jawaban yang tepat adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Entry

Terakhir diupdate 07 Juni 2021

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Pertanyaan yang serupa

2. Diberikan segitiga sama kaki ABC dengan  Luas segitiga ABC adalah  Hitunglah panjang sisi BC.

Pembahasan Soal:

Dengan menggunakan aturan sinus, perhatikan gambar dan penghitungan berikut!

  

  Luas space ABC equals 1 half cross times AB cross times AC cross times sin space straight A 600 equals 1 half cross times 42 cross times 42 cross times sin space straight A 600 equals 882 cross times sin space straight A 600 over 882 equals sin space straight A 100 over 147 equals sin space straight A  identitas space trigonometri cos space straight A equals square root of 1 minus sin squared space straight A end root cos space straight A equals square root of 1 minus open parentheses 100 over 147 close parentheses squared end root cos space straight A equals square root of 1 minus fraction numerator 10.000 over denominator 21.609 end fraction end root cos space straight A equals square root of fraction numerator 21.609 over denominator 21.609 end fraction minus fraction numerator 10.000 over denominator 21.609 end fraction end root cos space straight A equals square root of fraction numerator 11.609 over denominator 21.609 end fraction end root cos space straight A equals 0 comma 73  sehingga comma space dengan space aturan space cosinus space diperoleh colon BC equals square root of AB squared plus AC squared minus 2 AB times AC space cos space straight A space end root BC equals square root of 42 squared plus 42 squared minus 2 open parentheses 42 close parentheses times open parentheses 42 close parentheses open parentheses 0 comma 73 close parentheses space end root BC equals square root of 3.528 minus 2.575 comma 44 end root BC equals square root of 952 comma 56 end root BC equals 30 comma 86 space cm  

Jadi, panjang sisi BC adalah 30 comma 86 space cm. 

0

Roboguru

Sebuah limas tegak alasnya berbentuk segidelapan dengan panjang sisinya 10 cm dan tinggi limas tersebut 15 cm. Tentukan volume limas tersebut!

Pembahasan Soal:

Perhatikan gambar alas limas tersebut berikut 


 


Menentukan luas ABO 

table attributes columnalign right center left columnspacing 0px end attributes row cell angle AOB end cell equals cell fraction numerator 360 degree over denominator 8 end fraction end cell row blank equals cell 45 degree end cell end table 

Misal AO equals BO equals x, dengan menggunakan aturan cosinus diperoleh 

table attributes columnalign right center left columnspacing 0px end attributes row cell A B squared end cell equals cell A O squared plus B O squared minus 2 times A O times B O times cos A O B end cell row cell 10 squared end cell equals cell x squared plus x squared minus 2 times x times x times cos space 45 degree end cell row 100 equals cell 2 x squared minus 2 x squared times 1 half square root of 2 end cell row 100 equals cell x squared open parentheses 2 minus square root of 2 close parentheses end cell row cell x squared end cell equals cell fraction numerator 100 over denominator 2 minus square root of 2 end fraction end cell row cell x squared end cell equals cell fraction numerator 100 over denominator 2 minus square root of 2 end fraction times fraction numerator 2 plus square root of 2 over denominator 2 plus square root of 2 end fraction end cell row cell x squared end cell equals cell 50 open parentheses 2 plus square root of 2 close parentheses end cell end table 

Dengan menggunakan aturan sinus luas segitiga diperoleh 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight L subscript ABO end cell equals cell 1 half times AO times BO times cos space AOB end cell row blank equals cell 1 half times x times x times cos space 45 degree end cell row blank equals cell 1 half times x squared times 1 half square root of 2 end cell row blank equals cell 1 fourth square root of 2 x squared end cell row blank equals cell 1 fourth square root of 2 open parentheses 50 open parentheses 2 plus square root of 2 close parentheses close parentheses end cell row blank equals cell 1 fourth open parentheses 100 square root of 2 plus 100 close parentheses end cell row blank equals cell 25 open parentheses square root of 2 plus 1 close parentheses end cell end table 

Sehingga luas alas limas yaitu 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight L subscript alas end cell equals cell 8 cross times straight L subscript ABO end cell row blank equals cell 8 cross times 25 open parentheses square root of 2 plus 1 close parentheses end cell row blank equals cell 200 open parentheses square root of 2 plus 1 close parentheses end cell end table 

Maka volume limas adalah 

table attributes columnalign right center left columnspacing 0px end attributes row straight V equals cell 1 third cross times straight L subscript alas cross times tinggi end cell row blank equals cell 1 third cross times 200 open parentheses square root of 2 plus 1 close parentheses cross times 15 end cell row blank equals cell 1.000 open parentheses square root of 2 plus 1 close parentheses space cm cubed end cell end table 

Jadi, volume limas tersebut dalah 1.000 open parentheses square root of 2 plus 1 close parentheses space cm cubed.  

0

Roboguru

Diketahui AB = AD, DC = BC = 4 cm. Tentukan luas segiempat ABCD!

Pembahasan Soal:

Dengan aturan cosinus maka akan ditentukan panjang DB sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell DB squared end cell equals cell CD squared plus CB squared minus 2 times CD times CB times cos space 120 degree end cell row blank equals cell 4 squared plus 4 squared minus up diagonal strike 2 times 4 times 4 times negative fraction numerator 1 over denominator up diagonal strike 2 end fraction end cell row blank equals cell 16 plus 16 plus 16 end cell row cell DB squared end cell equals 48 end table  

Selanjutnya akan ditentukan panjang AB dengan aturan cosinus berikut. 

misal AB equals straight x, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell DB squared end cell equals cell AB squared plus AD squared minus 2 times AB times AD times cos space 60 degree end cell row 48 equals cell x squared plus x squared minus up diagonal strike 2 times x times x fraction numerator 1 over denominator up diagonal strike 2 end fraction end cell row 48 equals cell 2 x squared minus x squared end cell row 48 equals cell x squared end cell row x equals cell plus-or-minus square root of 48 end cell row x equals cell plus-or-minus 4 square root of 3 end cell end table 

Panjang AD tidak mungkin negative 4 square root of 3 space cm jadi panjang AD adalah 4 square root of 3 space cm 

Luas segiempat ABCD dapat ditentukan dengan cara berikut. 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight L subscript ABCD end cell equals cell straight L subscript DCB plus straight L subscript DAB end cell row blank equals cell 1 half times DC times BC times sin space 120 degree plus 1 half times AB times AD times sin space 60 degree end cell row blank equals cell fraction numerator 1 over denominator up diagonal strike 2 end fraction times up diagonal strike 4 times 4 times fraction numerator 1 over denominator up diagonal strike 2 end fraction square root of 3 plus fraction numerator 1 over denominator up diagonal strike 2 end fraction times up diagonal strike 4 square root of 3 times 4 square root of 3 times fraction numerator 1 over denominator up diagonal strike 2 end fraction square root of 3 end cell row blank equals cell 4 square root of 3 plus 12 square root of 3 end cell row blank equals cell 16 square root of 3 space cm squared end cell end table  

Jadi, luas segiempat ABCD tersebut adalah 16 square root of 3 space end root space cm squared

0

Roboguru

Perhatikan gambar! Luas segitiga PSR di atas adalah ...

Pembahasan Soal:

Perhatikan perhitungan berikut.

Ingat, luas segitiga dengan aturan sinus:

Luas equals 1 half times a times b times sin space straight C

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row Luas equals cell 1 half times 3 times 4 times sin space 60 degree end cell row blank equals cell 1 half times 3 times 4 times 1 half square root of 3 end cell row blank equals cell 3 square root of 3 space cm squared end cell end table

Jadi, luas segitiga PSR adalah 3 square root of 3 space cm squared.

0

Roboguru

Segitiga  merupakan segitiga sama sisi,  dan  tegak lurus dengan , berapakah luas ?

Pembahasan Soal:

Informasi di atas dapat digambar seperti berikut ini:



Misalkan panjang begin mathsize 14px style C D equals x end style.

Segitiga begin mathsize 14px style B D F end style sebangun dengan segitiga begin mathsize 14px style A E F end style.

Tentukan panjang begin mathsize 14px style A E end style, dimana sudut-sudut segitiga sama sisi-sisi masing- masing adalah begin mathsize 14px style 60 degree end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 60 degree end cell equals cell fraction numerator A F over denominator A E end fraction end cell row cell 1 half end cell equals cell fraction numerator b over denominator A E end fraction end cell row cell A E end cell equals cell 2 b end cell end table end style 

Sedangkan panjang begin mathsize 14px style E F end style adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 60 degree end cell equals cell fraction numerator E F over denominator A E end fraction end cell row cell 1 half square root of 3 end cell equals cell fraction numerator E F over denominator 2 b end fraction end cell row cell 2 E F end cell equals cell 2 b square root of 3 end cell row cell E F end cell equals cell b square root of 3 end cell end table end style 

Maka dengan menggunakan konsep kesebangunan, panjang begin mathsize 14px style B F end style adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator A F over denominator B F end fraction end cell equals cell fraction numerator A E over denominator B D end fraction end cell row cell fraction numerator b over denominator B F end fraction end cell equals cell fraction numerator 2 b over denominator 3 x end fraction end cell row cell B F times 2 b end cell equals cell 3 x b end cell row cell B F end cell equals cell fraction numerator 3 x b over denominator 2 b end fraction end cell row cell B F end cell equals cell 3 over 2 x end cell end table end style 

Sehingga didapat panjang begin mathsize 14px style B F equals 3 over 2 x end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell A B end cell equals cell B C end cell row cell b plus 3 over 2 x end cell equals cell 2 x end cell row b equals cell 2 x minus 3 over 2 x end cell row b equals cell 1 half x end cell row x equals cell 2 b end cell end table end style

Karena begin mathsize 14px style x equals 2 b end style, maka:

begin mathsize 14px style A B equals B C equals A C equals 2 x space space space space space space space space space space space space space space space space space space space space space space space equals 2 left parenthesis 2 b right parenthesis space space space space space space space space space space space space space space space space space space space space space space space equals 4 b end style 

Sehingga, luas begin mathsize 14px style F B C E end style adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell L subscript F B C E end subscript end cell equals cell L subscript A B C end subscript minus L subscript A F E end subscript end cell row blank equals cell 1 half times A B times A C times sin space A minus 1 half times A F times E F end cell row blank equals cell 1 half times 4 b times 4 b times sin space 60 degree minus 1 half times b times b square root of 3 end cell row blank equals cell 8 b squared times 1 half square root of 3 minus 1 half b squared square root of 3 end cell row blank equals cell 4 b squared square root of 3 minus 1 half b squared square root of 3 end cell row blank equals cell 7 over 2 b squared square root of 3 end cell end table end style

Jadi, luas begin mathsize 14px style F B C E end style adalah begin mathsize 14px style 7 over 2 b squared square root of 3 end style

1

Roboguru

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