To find the area of the picture when the projection unit is moved to a distance of 30 feet from the wall, we can use the principle of similar triangles.
Given:
- At a distance of 25 ft, the area of the picture is 100 ft².
- We need to find the area of the picture when the distance is increased to 30 ft.
- Relationship based on distance: The area of the projected image is inversely proportional to the square of the distance from the projection unit to the wall. That is, if \(A_1\) is the area at distance \(d_1\) and \(A_2\) is the area at distance \(d_2\), then:
\[ \frac{A_1}{A_2} = \left(\frac{d_2}{d_1}\right)^2 \]
- Substituting the known values: Here, \(d_1 = 25\) ft, \(A_1 = 100\) ft², and \(d_2 = 30\) ft. We need to find \(A_2\).
\[ \frac{100}{A_2} = \left(\frac{30}{25}\right)^2 \]
- Calculating the ratio:
\[ \frac{30}{25} = 1.2 \] \[ \left(1.2\right)^2 = 1.44 \]
- Setting up the equation:
\[ \frac{100}{A_2} = 1.44 \]
- Solving for \(A_2\):
\[ A_2 = \frac{100}{1.44} \approx 69.44 \text{ ft²} \]
Therefore, the area of the picture when the projection unit is moved to a distance of 30 feet from the wall is approximately 69.44 ft².