Kerjakan nomor 13 pakai cara

Kelas            : 12
Mapel           : Matematika
Kategori      : Baris dan Deret
Kata kunci  : Deret tak hingga
Kode            : 12.2.7 (Kelas 12 Matematika Bab 7-Baris dan Deret)

[tex]AB=6 \\ BC=10 \\ AC= \sqrt{10^2-6^2}=8 \\ L\triangle ABC= \frac{1}{2}\times AB\times AC= \frac{1}{2}\times 6\times 8=24 \\ L\triangle ABC= \frac{1}{2}\times BC\times AB_1 \\ 24= \frac{1}{2}\times 10\times AB_1 \\ 5AB_1=24 \\ AB_1=4,8 \\ B_1C= \sqrt{8^2-(4,8)^2}=6,4 \\ L\triangle B_1AC= \frac{1}{2}\times B_1C\times AB_1= \frac{1}{2}\times 6,4\times 4,8=15,36 \\ deret\; tak\; hingga\; luas\; segitiga: \\ 24+15,36+... \\ a=24 \\ r= \frac{15,36}{24}=0,64= \frac{64}{100}= \frac{16}{25} [/tex]
[tex]\\ S\infty= \frac{a}{1-r} \\ S\infty= \frac{24}{1- \frac{16}{25} } \\ S\infty= \frac{24}{ \frac{25}{25}- \frac{16}{25} } \\ S\infty= \frac{24}{ \frac{9}{25} } \\ S\infty=24\times \frac{25}{9} \\ S\infty= \frac{600}{9} [/tex]

Jawaban: B

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