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Andreas93 [3]
3 years ago
5

A sculpture of a clothespin is 20 feet high. A normal clothespin of this shape is five inches high. Using this scale, how many f

eet tall would a sculpture of a 5-foot, 7-inch woman be?
Mathematics
2 answers:
zepelin [54]3 years ago
8 0

Answer:

<h2>The sculpture of the woman must be 268 inches tall.</h2>

Step-by-step explanation:

This problem is about proportions and scales factors.

According to the problem, a sculpture of a clothespin is 20 feet, and a normal clothespin is 5 inches high.

So, the ratio is \frac{20}{5}=4

If a woman is 5 feet and 7 inches, the height of her sculpture must be in the given ratio.

First, we need to use only inches units. We know that 1 foot is 12 inches.

5ft\frac{12in}{1ft} + 7=60+7=67in

So, the woman height is 67 inches.

Using the given scale factor, we multiply it with the height of the woman

67 \times 4 = 268 in

Therefore, the sculpture of the woman must be 268 inches tall.

kodGreya [7K]3 years ago
7 0
This is the concept of scale factor. The linear scale factor is given by:
(height of sculpture)/(height of clothespin)
=20/(5/12)
=48
Given that a woman is 5ft 7 in, the sculpture of this woman would be:
height of sculpture=(scale factor)*(height of the woman)
height of the woman=5 ft 7 inches=5 7/12 ft=67/12 ft
hence the height of the sculpture will be:
67/12×48
=268 ft

Answer: 268 ft

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