Pertanyaan

Cermati Gambar berikut! Gambar 4.35 .Kombinasi segitiga siku-siku Dengan menemukan hubungan antarsudut-sudut dan panjang sisi-sisi pada segitiga siku-siku yang ada pada gambar, hitunglah panjang AD, EC, BC, BD, AB, FB, AE, dan DE!

Cermati Gambar berikut!

Gambar 4.35. Kombinasi segitiga siku-siku

Dengan menemukan hubungan antarsudut-sudut dan panjang sisi-sisi pada segitiga siku-siku yang ada pada gambar, hitunglah panjang AD, EC, BC, BD, AB, FB, AE, dan DE!

F. Ayudhita

Master Teacher

Jawaban terverifikasi

Pembahasan

Perhatikan segitiga siku-siku ABD, berlaku: Jadi, dan . Dengan Teorema Pythagoras, diperoleh Perhatikan! Pada segitiga ADE, Sehingga . Perhatikan segitiga BDF, berlaku: Diperoleh . Dengan Terorema Pythagoras, diperoleh Perhatikan segitiga ABC, berlaku: Dengan Teorema Pythagoras, diperoleh Selanjutnya, karena , dengan , maka Perhatikan . Diperoleh . Jadi, berdasarkan perhitungan tersebut dapat diperoleh .

Perhatikan segitiga siku-siku ABD, berlaku:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin 30 degree end cell equals cell fraction numerator B D over denominator A D end fraction end cell row cell 1 half end cell equals cell fraction numerator B D over denominator A D end fraction end cell row blank blank blank end table end style$

Jadi, dan .

Dengan Teorema Pythagoras, diperoleh

Perhatikan!

Pada segitiga ADE,

Sehingga .

Perhatikan segitiga BDF, berlaku:

$table attributes columnalign right center left columnspacing 0px end attributes row cell size 14px sin size 14px angle size 14px B size 14px D size 14px E end cell size 14px equals cell size 14px sin size 14px 45 size 14px degree size 14px equals fraction numerator size 14px B size 14px F over denominator size 14px B size 14px D end fraction end cell row cell fraction numerator square root of size 14px 2 over denominator size 14px 2 end fraction end cell size 14px equals cell fraction numerator size 14px B size 14px F over denominator size 14px x end fraction end cell end table$

Diperoleh $begin mathsize 14px style B F equals fraction numerator x square root of 2 over denominator 2 end fraction end style$ .

Dengan Terorema Pythagoras, diperoleh

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell D F end cell equals cell square root of B D squared minus B F squared end root end cell row blank equals cell square root of x squared minus open parentheses fraction numerator x square root of 2 over denominator 2 end fraction close parentheses squared end root end cell row blank equals cell square root of x squared minus fraction numerator 2 x squared over denominator 4 end fraction end root end cell row blank equals cell square root of x squared over 2 end root end cell row blank equals cell fraction numerator x over denominator square root of 2 end fraction end cell row blank equals cell fraction numerator x square root of 2 over denominator 2 end fraction end cell row blank blank blank end table end style$

Perhatikan segitiga ABC, berlaku:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin angle B A C end cell equals cell sin 45 degree equals fraction numerator B C over denominator A B end fraction end cell row cell fraction numerator square root of 2 over denominator 2 end fraction end cell equals cell fraction numerator B C over denominator x square root of 3 end fraction end cell row cell B C end cell equals cell fraction numerator x square root of 6 over denominator 2 end fraction end cell row blank blank blank row blank blank blank end table end style$

Dengan Teorema Pythagoras, diperoleh

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell A C end cell equals cell square root of A B squared minus B C squared end root end cell row blank equals cell square root of open parentheses x square root of 3 close parentheses squared minus open parentheses fraction numerator x square root of 6 over denominator 2 end fraction close parentheses squared end root end cell row blank equals cell square root of 3 x squared minus fraction numerator 6 x squared over denominator 4 end fraction end root end cell row blank equals cell square root of fraction numerator 6 x squared over denominator 4 end fraction end root end cell row blank equals cell fraction numerator x square root of 6 over denominator 2 end fraction end cell row blank blank blank row blank blank blank end table end style$

Selanjutnya, karena , dengan $begin mathsize 14px style C E equals B F equals fraction numerator x square root of 2 over denominator 2 end fraction end style$ , maka

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell A E end cell equals cell A C minus C E end cell row blank equals cell fraction numerator x square root of 6 over denominator 2 end fraction minus fraction numerator x square root of 2 over denominator 2 end fraction end cell row blank equals cell fraction numerator x open parentheses square root of 6 minus square root of 2 close parentheses over denominator 2 end fraction end cell end table end style$

Perhatikan $begin mathsize 14px style E F equals B C equals fraction numerator x square root of 6 over denominator 2 end fraction end style$ .

Diperoleh $begin mathsize 14px style D E equals D F plus E F equals fraction numerator x square root of 2 over denominator 2 end fraction plus fraction numerator x square root of 6 over denominator 2 end fraction equals fraction numerator x open parentheses square root of 6 plus square root of 2 close parentheses over denominator 2 end fraction end style$ .

Jadi, berdasarkan perhitungan tersebut dapat diperoleh $begin mathsize 14px style A D equals 2 x comma blank E C equals fraction numerator x square root of 2 over denominator 2 end fraction comma blank B C equals fraction numerator x square root of 6 over denominator 2 end fraction comma blank B D equals x comma blank A B equals x square root of 3 comma blank F B equals fraction numerator x square root of 2 over denominator 2 end fraction comma blank A E equals fraction numerator x open parentheses square root of 6 minus square root of 2 close parentheses over denominator 2 end fraction comma blank dan blank D E equals fraction numerator x open parentheses square root of 6 plus square root of 2 close parentheses over denominator 2 end fraction end style$ .