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The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is

CAT · 2024 · Quant Slot 1
Question:
The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is

Options

750π
1125π√2
1125π
750π√2
Answer: 1125π√2

Explanation:
Surface area = 2(ab + bc + ca) = 846 → ab + bc + ca = 423 Sum of all edges = 4(a + b + c) = 144 → a + b + c = 36 Now use the identity: (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) So, a^2 + b^2 + c^2 = (a + b + c)^2 − 2(ab + bc + ca) = 36^2 − 2 × 423 = 1296 − 846 = 450 Thus, the space diagonal of the box is: d = sqrt(a^2 + b^2 + c^2) = sqrt(450) = 15√2 Since the box is inscribed in the sphere, the diagonal = diameter of sphere: Diameter = 15√2 Radius r = (15√2) / 2 Volume of sphere = (4/3)πr^3 Compute r^3: r^3 = (15√2 / 2)^3 = (15^3 × (√2)^3) / 2^3 = (3375 × 2√2) / 8 = (3375√2) / 4 Now volume: Volume = (4/3)π × (3375√2 / 4) = (3375√2 / 3)π = 1125√2 π

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