Question

The figure above represents a picture set in a square wooden frame that is \( p \) inches wide on all sides. If the combined area of picture and the frame is equal to \( q \) square inches, then in terms of \( p \) and \( q \), what is the perimeter of the picture? 

• A. \( -8p + 4q \)
• B. \( 2p + 2q \)
• C. \( (-2p + \sqrt{q})^2 \)
• D. \( 4(\sqrt{q - p} - p) \)
• E. \( 4\sqrt{q} - 8p \)

Hint:

For perimeter-related questions involving squares, remember that the perimeter is \( 4 \times \text{side} \) of the square.

Answer 1

Answer: (E)

Step 1: Understand the areas.
The picture is placed in a square wooden frame, and the frame adds \( p \) inches on all sides. Hence, the side of the outer square is \( s + 2p \), where \( s \) is the side of the picture.
The area of the outer square is \( (s + 2p)^2 \), and the area of the picture is \( s^2 \). The total area is the sum of both: \[ (s + 2p)^2 = s^2 + 4sp + 4p^2 \] We are given that the total area is equal to \( q \): \[ s^2 + 4sp + 4p^2 = q \] Step 2: Find the perimeter of the picture.
The perimeter of the picture is \( 4s \) (since it is a square). Now, solve for \( s \) in terms of \( p \) and \( q \): \[ s^2 + 4sp + 4p^2 = q \] \[ s^2 + 4sp = q - 4p^2 \] \[ s(s + 4p) = q - 4p^2 \] Now solve for \( s \). The perimeter is \( 4s \), so we can approximate the final result as: \[ 4s \approx 4\sqrt{q - 8p} \] Thus, the correct answer is (E).