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Dari gambar di bawah ini, jika , , maka tentukan: ...

Dari gambar di bawah ini, jika begin mathsize 14px style BD equals 4 space cm end stylebegin mathsize 14px style DC equals 12 space cm end style, maka tentukan:

a. panjang AB

b. panjang AC

c. panjang AD

Jawaban:

Perhatikan bahwa begin mathsize 14px style increment ABC end style dan begin mathsize 14px style increment ABD end style sebangun.

Digambarkan sebagai berikut:

Bangun datar dikatakan sebangun apabila memenuhi syarat:

  1. Sudut-sudut yang bersesuaian sama besar.
  2. Sisi-sisi yang bersesuaian memiliki perbandingan yang sama.

Sisi-sisi yang bersesuaian pada segitiga memiliki perbandingan yang sama, yaitu:

begin mathsize 14px style AB over BD equals BC over AB equals AC over AD end style

Sehingga diperoleh:

  • Panjang AB

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell AB over BD end cell equals cell BC over AB end cell row cell AB squared end cell equals cell BC cross times BD end cell row cell AB squared end cell equals cell 12 cross times 4 end cell row AB equals cell square root of 12 cross times 4 end root end cell row AB equals cell square root of 48 end cell row AB equals cell square root of 16 cross times 3 end root end cell row AB equals cell 4 square root of 3 space cm end cell end table end style

begin mathsize 14px style increment ABC end style merupakan segitiga siku-siku. Maka gunakan Teorema Pythagoras untuk mengetahui panjang AC.

  • Panjang AC

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell AC squared end cell equals cell BC squared minus AB squared end cell row AC equals cell square root of BC squared minus AB squared end root end cell row AC equals cell square root of 16 squared minus open parentheses 4 square root of 3 close parentheses squared end root end cell row AC equals cell square root of 256 minus 48 end root end cell row AC equals cell square root of 208 end cell row AC equals cell square root of 16 cross times 13 end root end cell row AC equals cell 4 square root of 13 space cm end cell end table end style

Gunakan perbandingan segitiga sebangun untuk mengetahui panjang AD.

  • Panjang AD

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell AB over BD end cell equals cell AC over AD end cell row cell fraction numerator 4 square root of 3 over denominator 4 end fraction end cell equals cell fraction numerator 4 square root of 13 over denominator AD end fraction end cell row cell AD open parentheses down diagonal strike 4 square root of 3 close parentheses end cell equals cell down diagonal strike 4 cross times 4 square root of 13 end cell row cell AD open parentheses square root of 3 close parentheses end cell equals cell 4 square root of 13 end cell row AD equals cell fraction numerator 4 square root of 13 over denominator square root of 3 end fraction end cell row AD equals cell fraction numerator 4 square root of 13 over denominator square root of 3 end fraction cross times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row AD equals cell fraction numerator 4 square root of 39 over denominator 3 end fraction end cell end table end style

Maka, diperoleh panjang begin mathsize 14px style AB equals 4 square root of 3 space cm end stylebegin mathsize 14px style AC equals 4 square root of 13 space cm end style, dan begin mathsize 14px style AD equals 4 over 3 square root of 39 space cm end style.

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