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Pertanyaan

Pada gambar berikut, OABC adalah bangun geometri segi empat. Titik-titik straight P dan straight Q adalah titik-titik tengah ruas garis OB dan ruas garis AC.

Tunjukkan bahwa OA with rightwards arrow on top plus OC with rightwards arrow on top plus BA with rightwards arrow on top plus BC with rightwards arrow on top equals 4 PQ with rightwards arrow on top.

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

pembuktian bahwa table attributes columnalign right center left columnspacing 0px end attributes row cell OA with rightwards arrow on top plus OC with rightwards arrow on top plus BA with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell 4 PQ with rightwards arrow on top end cell end table telah diuraikan di atas.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut ada pada uraian di bawah bahwa Error converting from MathML to accessible text..

Misalkan vektor-vektor posisi titik text A end text dan titik text B end text berturut-turut adalah a with rightwards arrow on top dan b with rightwards arrow on top. Titik text C end text terletak pada ruas garis text AB end text dengan perbandingan text AC end text colon text CB end text equals m colon n, maka vektor posisi text C end text adalah c with rightwards arrow on top ditentukan dengan rumus:

c with rightwards arrow on top equals fraction numerator m b with rightwards arrow on top plus n a with rightwards arrow on top over denominator m plus n end fraction

Penyelesaian:

Misalkan vektor-vektor posisi titik text A end text, titik text B end text, dan titik text C end text berturut-turut adalah a with rightwards arrow on topb with rightwards arrow on top, dan c with rightwards arrow on top. Diperoleh persamaan (1) berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell OA with rightwards arrow on top plus OC with rightwards arrow on top plus BA with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell a with rightwards arrow on top plus c with rightwards arrow on top plus open parentheses a with rightwards arrow on top minus b with rightwards arrow on top close parentheses plus open parentheses c with rightwards arrow on top minus b with rightwards arrow on top close parentheses end cell row cell OA with rightwards arrow on top plus OC with rightwards arrow on top plus BA with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell 2 open parentheses a with rightwards arrow on top minus b with rightwards arrow on top plus c with rightwards arrow on top close parentheses end cell end table

Vektor posisi titik text P end text diwakili oleh ruas garis berarah stack text OP end text with rightwards arrow on top.

OP with rightwards arrow on top equals 1 half OB with rightwards arrow on top equals 1 half b with rightwards arrow on top

sebab titik text P end text merupakan titik tengah OB with rightwards arrow on top.

Vektor posisi titik text Q end text diwakili oleh ruas garis berarah OQ with rightwards arrow on top

OQ with rightwards arrow on top equals 1 half open parentheses a with rightwards arrow on top plus c with rightwards arrow on top close parentheses

sebab titik text Q end text merupakan titik tengah text AC end text.

Dengan menggunakan segitiga text OPQ end text pada gambar, diperoleh hubungan persamaan (2) berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell OP with rightwards arrow on top plus PQ with rightwards arrow on top end cell equals cell OQ with rightwards arrow on top end cell row cell PQ with rightwards arrow on top end cell equals cell OQ with rightwards arrow on top minus OP with rightwards arrow on top end cell row cell PQ with rightwards arrow on top end cell equals cell 1 half open parentheses a with rightwards arrow on top plus c with rightwards arrow on top close parentheses minus 1 half b with rightwards arrow on top end cell row cell PQ with rightwards arrow on top end cell equals cell 1 half open parentheses a with rightwards arrow on top minus b with rightwards arrow on top plus c with rightwards arrow on top close parentheses end cell row cell 2 open parentheses a with rightwards arrow on top minus b with rightwards arrow on top plus c with rightwards arrow on top close parentheses end cell equals cell 4 PQ with rightwards arrow on top end cell end table

Substitusi persamaan (2) ke persamaan (1), diperoleh hubungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell OA with rightwards arrow on top plus OC with rightwards arrow on top plus BA with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell 2 open parentheses a with rightwards arrow on top minus b with rightwards arrow on top plus c with rightwards arrow on top close parentheses end cell row blank equals cell 4 PQ with rightwards arrow on top end cell end table

Jadi, terbukti bahwa table attributes columnalign right center left columnspacing 0px end attributes row cell OA with rightwards arrow on top plus OC with rightwards arrow on top plus BA with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell 4 PQ with rightwards arrow on top end cell end table 

Dengan demikian, pembuktian bahwa table attributes columnalign right center left columnspacing 0px end attributes row cell OA with rightwards arrow on top plus OC with rightwards arrow on top plus BA with rightwards arrow on top plus BC with rightwards arrow on top end cell equals cell 4 PQ with rightwards arrow on top end cell end table telah diuraikan di atas.

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