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SMK Teknologi dan Rekayasa

Soal 53 Keliling Bangun Datar


Gambar\ di\ bawah\ ini\ adalah\ trapesium\ sama\ kaki\ ABCD.

Jika\ AC=15\ cm,\ BF=3\ cm,\ dan\ DE=9\ cm,\ maka\\keliling\ trapesium\ adalah\ ....\\a.\ \ 12+\sqrt{10}\ \ \ \ \ \ \ \ \ \ b.\ \ 18+3\sqrt{10}\ \ \ \ \ \ \ \ \ \ c.\ \ 24+6\sqrt{10}\\d.\ \ 29+6\sqrt{10}\ \ \ \ \ \ \ \ \ e.\ \ 57+6\sqrt{10}
Jawaban: c

Keliling=AB+BC+CD+DA
Dari\ \Delta ACF\ siku-siku\ di\ F\ diperoleh:
AC=15=3.5,\ CF=9=3.3,\ maka\ AF=3.4=12\\sehingga\ AB=AF+FB=12+3=15
Karena\ samakaki,\ maka\ AE=FB=3,\ sehingga:\\EF=9=CD
Dari\ \Delta CFB\ siku-siku\ di\ F\ diperoleh:
BC^{2}=9^{2}+3^{2}=81+9=90\\BC=\sqrt{90}=\sqrt{9\times 10}=\sqrt{9}\times \sqrt{10}=3\sqrt{10}=DA
Keliling=15+3\sqrt{10}+9+3\sqrt{10}=24+6\sqrt{10}

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