A cuboid of dimensions 18 cm × 12 cm × 8 cm is melted and recast into smaller cubes each having an edge of 4 cm.
A cuboid of dimensions 18 cm × 12 cm × 8 cm is melted and recast into smaller cubes each having an edge of 4 cm. Find the ratio of the total surface area of one small cube to that of the original cuboid.
18 सेमी × 12 सेमी × 8 सेमी विमाओं वाले एक घनाभ को पिघलाकर 4 सेमी किनारे वाले छोटे घनों में ढाला जाता है। एक छोटे घन के कुल पृष्ठीय क्षेत्रफल का मूल घनाभ के कुल पृष्ठीय क्षेत्रफल से अनुपात ज्ञात कीजिए।
Detailed Solution & Logic
2 : 19
Step 1: Surface area of the original cuboid
Surface area of a cuboid = $2(lw + lh + wh)$
$l = 18$, $w = 12$, $h = 8$
$2(18 \cdot 12 + 18 \cdot 8 + 12 \cdot 8) = 2(216 + 144 + 96) = 2 \cdot 456 = 912$ cm²
So, original cuboid’s surface area = 912 cm²
Step 2: Surface area of one small cube
Edge of small cube = 4 cm
Surface area of a cube = $6 \cdot (\text{edge})^2 = 6 \cdot 4^2 = 6 \cdot 16 = 96$ cm²
Step 3: Ratio of one small cube’s SA to cuboid’s SA
Ratio = 96 : 912 = 2 : 19
Answer: 2 : 19
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