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(D) Let the dimensions of the cuboid be $x, y,$ and $z$. The faces have dimensions $(x, y), (y, z),$ and $(x, z)$.The sum of the perimeter and area fo

Vedclass · Accepted Answer

$28$ and $35$

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(D) Let the dimensions of the cuboid be $x, y,$ and $z$. The faces have dimensions $(x, y), (y, z),$ and $(x, z)$.The sum of the perimeter and area for each face is given by:$2(x+y) + xy = 16 \quad (i)$$2(y+z) + yz = 24 \quad (ii)$$2(x+z) + xz = 31 \quad (iii)$Adding $4$ to both sides of each equation to factorize:$xy + 2x + 2y + 4 = 16 + 4 \implies (x+2)(y+2) = 20 \quad (iv)$$yz + 2y + 2z + 4 = 24 + 4 \implies (y+2)(z+2) = 28 \quad (v)$$xz + 2x + 2z + 4 = 31 + 4 \implies (x+2)(z+2) = 35 \quad (vi)$Let $X = x+2, Y = y+2, Z = z+2$. Then $XY=20, YZ=28, XZ=35$.Multiplying these: $(XYZ)^2 = 20 	imes 28 	imes 35 = 19600 \implies XYZ = 140$.$Z = \frac{XYZ}{XY} = \frac{140}{20} = 7 \implies z+2 = 7 \implies z = 5$.$X = \frac{XYZ}{YZ} = \frac{140}{28} = 5 \implies x+2 = 5 \implies x = 3$.$Y = \frac{XYZ}{XZ} = \frac{140}{35} = 4 \implies y+2 = 4 \implies y = 2$.The volume $V = xyz = 3 	imes 2 	imes 5 = 30$.Since $30$ lies between $28$ and $35$,option $(d)$ is correct.

On each face of a cuboid,the sum of its perimeter and its area is written. Among the six numbers so written,there are three distinct numbers and they are $16, 24$ and $31$. The volume of the cuboid lies between

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