Question

Two cubes have volumes in the ratio 1:64 the ratio of the areas of a face of first cube to that of the other is?

Answer 1

Let the side of first cube = a

Let the side of second cube = A

Volume of CUBE= side³

Volume of first cube, V1 = a³
volume of second cube, V2 = A³

Ratio of their volumes = V1:V2 = a³:A³ = 1:64

a³/ A³ = 1/64

(a/A)³ = (1/4)³

a/A= ¼



Surface Area of CUBE= 6(side)²


Surface area of first cube(S1) = 6a​​²

surface area of second cube(S2) = 6A²

Ratio of their surface areas(S1:S2) = 6a²:6A²
= (a²:A²) = (a/A)²=( ¼)² = 1/16

a²:A²= 1:16

[All the faces of a cube are squares of its side]

Hence, the ratio of the areas of a face of first cube to that of the other is 1:16

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Hope this will help you...

Answer 2

Let the side of first cube = a

Let the side of second cube = A

Volume of CUBE= side³

Volume of first cube, V1 = a³

volume of second cube, V2 = A³

Ratio of their volumes = V1:V2 = a³:A³ = 1:64

a³/ A³ = 1/64

(a/A)³ = (1/4)³

a/A= ¼

Surface Area of CUBE= 6(side)²

Surface area of first cube(S1) = 6a​​²

surface area of second cube(S2) = 6A²

Ratio of their surface areas(S1:S2) = 6a²:6A²

= (a²:A²) = (a/A)²=( ¼)² = 1/16

a²:A²= 1:16

[All the faces of a cube are squares of its side]

Hence, the ratio of the areas of a face of first cube to that of the other is 1:16

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Hope this will help you..