GP - edge lengths

The block edge lengths are made up of three consecutive GP members. The sum of the lengths of all edges is 84 cm, and the volume block is 64 cm³. Determine the surface of the block.

Correct answer:

S = 672 cm²

Step-by-step explanation:

a + b + c = 84 c m V = 64 cm^{3} b = q a c = q^{2} a V = a b c = a^{3} \cdot q^{3} = (a \cdot q)^{3} = b^{3} b = 3 V = 364 = 4 cm a + c = 84 - b a c = V / b a (84 - b - a) = V / b a (84 - b - a) = V / b a (84 - 4 - a) = 64 / 4 - a^{2} + 80 a - 16 = 0 a^{2} - 80 a + 16 = 0 p = 1; q = - 80; r = 16 D = q^{2} - 4 p r = 8 0^{2} - 4 \cdot 1 \cdot 16 = 6336 D > 0 a_{1, 2} = \frac{- q \pm D}{2 p} = \frac{8 0 \pm 6 3 3 6}{2} = \frac{8 0 \pm 2 4 1 1}{2} a_{1, 2} = 40 \pm 39.799497 a_{1} = 79.799497484 a_{2} = 0.200502516 a = a_{2} = 0.2005 ≐ 0.2005 c = V / (b \cdot a) = 64 / (4 \cdot 0.2005) ≐ 79.7995 cm S = 2 \cdot (a \cdot b + b \cdot c + a \cdot c) = 2 \cdot (0.2005 \cdot 4 + 4 \cdot 79.7995 + 0.2005 \cdot 79.7995) = 672 = 672 cm^{2} Verifying Solution: V_{2} = a \cdot b \cdot c = 0.2005 \cdot 4 \cdot 79.7995 = 64 cm^{3} s_{2} = a + b + c = 0.2005 + 4 + 79.7995 = 84 cm q_{1} = b / a = 4 / 0.2005 ≐ 19.9499 q_{2} = c / b = 79.7995 / 4 ≐ 19.9499

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