To determine the area of the picture when the projection unit is moved to a distance of 30 feet from the wall, we can use the concept of similar triangles, since the distance from the projection unit to the wall and the size of the picture are proportional.
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Calculate the height of the picture at 25 ft:
The 25 ft distance produces a 100 ft picture. This means that the ratio of the distance to height is \( \frac{25 \text{ ft}}{100 \text{ ft}} = \frac{1}{4} \). -
Apply the ratio to the new distance:
When the projection unit is moved to a distance of 30 ft, we can set up the ratio to find the new height \( h \) of the picture using the similar triangles concept: \[ \frac{30 \text{ ft}}{h} = \frac{25 \text{ ft}}{100 \text{ ft}} \] -
Cross-multiply and solve for \( h \): \[ 30 \text{ ft} \times 100 \text{ ft} = 25 \text{ ft} \times h \] \[ 3000 = 25h \] \[ h = \frac{3000}{25} = 120 \text{ ft} \]
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Calculate the area of the picture:
The area \( A \) of a rectangle is given by the formula \( A = \text{length} \times \text{width} \). Assuming the picture is rectangular and based on the earlier projection, we can consider this height as either the length or width. If we just have the height, to find the area we need to consider if the width changes as well. Using the height of 120 ft, and maintaining the proportionality, we need to confirm the original measurement adjustments if any regarding the width at changed projections.
However, since we don't specify a width, let's calculate under the assumption that the initial dimensions were calculated similarly for any square measure:
Assuming the original picture was a square where both length and width are equal. If 100 ft was taken with a fixed height corresponding 100 ft initially, and it matches both dimensions at that scale: Thus, area can be inferred as:
\[ \text{Area} = height \times height = 120 \text{ ft} \times 120 \text{ ft} = 14400 \text{ ft}^2 \]
Therefore, if moving to 30 ft results in a marked increase in relative size consistently applied you're concluding the area of the picture is \( \mathbf{14400 \text{ ft}^2} \).