2 cubes each of volume $64\text{ cm}^3$ are joined end to end. The surface area of the resulting cuboid is:
(a) $160\text{ cm}^2$
(b) $128\text{ cm}^2$
(c) $144\text{ cm}^2$
(d) $256\text{ cm}^2$
Step-by-Step Solution
Key Concept: Side of cube $a = \sqrt[3]{64} = 4\text{ cm}$. Cuboid dimensions: $l = 8\text{ cm}, b = 4\text{ cm}, h = 4\text{ cm}$.
Side of cube $a = 4\text{ cm}$. Cuboid $l = 8, b = 4, h = 4$. [0.5 Mark]
SA $= 2(lb + bh + hl) = 2(32 + 16 + 32) = 2(80) = 160\text{ cm}^2$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Finding side of cube $= 4\text{ cm}$ and cuboid dimensions: 0.5 Mark
Calculating surface area $= 160\text{ cm}^2$: 0.5 Mark
Correct Answer: $160\text{ cm}^2$