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Soal Ujian Mandiri-SPMB UIN-IAIN-STAIN Tes Potensi Akademik Aritmatika Tahun 2012 No. 51 – buku-on-line.com

Soal Ujian Mandiri-SPMB UIN-IAIN-STAIN Tes Potensi Akademik Aritmatika Tahun 2012 No. 51

SOAL

51. Seorang berjalan dari titik A ke Timur sejauh 4 m, lalu ke Selatan sejauh 3 m sampai ke titik B, lalu berbelok ke Timur sejauh 3 m terus ke selatan sejauh 4 m sampai ke titik C terus ke selatan sejauh 7 m sampai ke titik D. Berapa panjang ABCD?

(A)   19,4 m

(B)   18,8 m

(C)   18,5 m

(D)  17,0 m

(E)   17,8 m

JAWABAN DAN PEMBAHASAN

51. jawaban: D

       Panjang AB = √(4^2+3^2 )  = 5;  panjang BC = √(3^2+4^2 ) = 5,

       jadi panjang ABCD = AB + BC + CD = 5 + 5+ 7 = 17

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