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Question

Part 1
The volumes of two similar figures are given. The surface area of the larger figure is given. Find the surface area of the smaller figure.
Upper Vequals2240 m cubed
Upper Vequals4375 m cubed
S.A.equals1600 m squared
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Part 1
The surface area of the smaller figure is
 

Answer 1

S.A = 1600 m²
The volumes of the two similar figures are:
V₁ = 2240 m³
V₂ = 4375 m³

Since the figures are similar, their volumes are related by the ratio of their sides cubed:
(V₂/V₁) = (S₂/S₁)³

Solving for the ratio of their sides:
(4375/2240) = (S₂/S₁)³
1.9513 = (S₂/S₁)³
Taking the cube root of both sides:
1.2684 = S₂/S₁

Since surface area is proportional to the square of the sides, we have:
(S₂/S₁)² = S.A₂/S.A₁
(1.2684)² = S.A₂/1600
1.607 = S.A₂/1600
Solving for the surface area of the smaller figure:
S.A₂ = 1.607 * 1600
S.A₂ = 2571.2 m²

Therefore, the surface area of the smaller figure is 2571.2 m².

Answer 2

wrong

Answer 3

I apologize for the mistake. Let's try again:

Given:
V₁ = 2240 m³
V₂ = 4375 m³
S.A = 1600 m²

Since the figures are similar, their volumes are related by the ratio of their sides cubed:
(V₂/V₁) = (S₂/S₁)³

Solving for the ratio of their sides:
(4375/2240) = (S₂/S₁)³
1.9513 = (S₂/S₁)³
Taking the cube root of both sides:
1.2684 = S₂/S₁

Since surface area is proportional to the square of the sides, we have:
(S₂/S₁)² = S.A₂/S.A₁
(1.2684)² = S.A₂/1600
1.607 = S.A₂/1600
Solving for the surface area of the smaller figure:
S.A₂ = 1.607 * 1600
S.A₂ = 2571.2 m²

Therefore, the surface area of the smaller figure is 2571.2 m².