To solve this problem, we will find the dimensions of the picture frame using the given information. Let’s break it into steps:

Step 1: Define variables

• Let the width of the picture inside the frame (not the entire frame) be $w$ inches.
• The length of the picture inside the frame is $2w$ inches, as given.

The area of the picture inside the frame is given as $40 \, \text{in}^2$. Hence:

$w \cdot 2w = 40$

Step 2: Solve for $w$

$2w^2 = 40$ $w^2 = 20$ $w = \sqrt{20} = 2\sqrt{5} \, \text{inches} \, (\text{approximately } 4.47 \, \text{inches}).$

The width of the picture is $w = 2\sqrt{5}$ inches, and the length of the picture is: $2w = 4\sqrt{5} \, \text{inches} \, (\text{approximately } 8.94 \, \text{inches}).$

Step 3: Add the frame thickness

The frame is 1 inch thick on all four sides. This means the total dimensions of the entire frame (picture + frame) are:

$\text{Frame width} = w + 2 \cdot 1 = 2\sqrt{5} + 2 \, \text{inches}.$ $\text{Frame length} = 2w + 2 \cdot 1 = 4\sqrt{5} + 2 \, \text{inches}.$

Final Answer:

The dimensions of the entire picture frame are:

• Width: $2\sqrt{5} + 2 \approx 6.47 \, \text{inches}$,
• Length: $4\sqrt{5} + 2 \approx 10.94 \, \text{inches}$.

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Would you like more details or a visual explanation of this solution? Here are 5 related questions to explore further:

1. What is the perimeter of the entire picture frame?
2. How does the area of the frame (excluding the picture) compare to the picture’s area?
3. If the frame’s thickness is increased to 2 inches, what would the new dimensions be?
4. What is the diagonal length of the entire picture frame?
5. How would the problem change if the frame was not uniform in thickness?

Tip: Always check units and dimensions carefully when solving real-world geometry problems.