Question

Perbandingan panjang,lebar,dan tinggi balok ABCD.EFGH = 3:2:1 . luas permukaan balok itu = 88 cm. hitunglah panjang,lebar dan tinggi balok !

Answer 1

Panjang = 3x
lebar = 2x
tinggi = x
Lp. Balok =  88 cm²

Lp. Balok = 2 (pl + pt + lt)
88 cm² = 2 (3x * 2x + 3x * x + 2x * x)
88 cm² = 2 ( 6x² + 3x² + 2x²)
88 cm² = 2 * 11x²
88 cm² = 22x²
88 cm² / 22 = x²
4 cm² = x²
√4 cm² = x
x = 2

panjang = 3x = 3 * 2 = 6 cm
lebar = 2x = 2 * 2 = 4 cm
tinggi = x = 2 cm

Answer 2

DIT = PANJANG = 3
LEBAR = 2
TINGGI = 1
DIK = LUAS PERMUKAAN BALOK=
JADI DIK = LUAS PERMUKAAN BALOK PERBANDINGAN 
=(3X2)+(2X1)+(3X1) X 2
= (6 +2 + 3) X 2
= 11 X 2
= 22
SOL = LUAS PERMUKAAN ASLINYA = 88

PANJANG = DIT/DIK X SOL
= 3/22 X 88
= 12 : 2 = 6
LEBAR = DIT/DIK X SOL
= 2/22 X 88
= 8 : 2 = 4
TINGGI = DIT/DIK X SOL
= 1/22 X 88
4 : 2 = 2
KET = DIBAGI 2 KARENA = √88 : 22 =√4 = 2

SEMOGA MEMBANTU.........