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EC² = EG² + CG², EG merupakan diagonal sisi EFGH, sehingga...
(√a²38)² = (EF² + FG²) + CG²
a²38 = EF²(panjang) + FG²(lebar) + CG²(tinggi), misal panjang = 5. lebar = 3, dan tinggi = 2, maka :
a²38 = 5² + 3² + 2²
a²38 = 25 + 9 + 4
a²38 = 38
a² = 38÷38
a² = 1
a = 1, jika a = 1 maka memenuhi persamaan a > 0, maka p=5, l = 3, dan t = 2, akhirnya LP balok adalah....
2 (pl + pt + lt)
2 (5*3 + 5*2 + 3*2)
2(15 + 10 + 6)
2(31)
62 cm², semoga terpecahkan