Question

Two rectangular picture frames have the same area of 45 square inches but have different side lengths. Frame A has a length of 634

inches, and Frame B has a length of 712
inches.

Answer 1

The widths of the two frames are approximately 0.0709 inches for Frame A and 0.0632 inches for Frame B.

Let's denote the length and width of Frame A as
`\(L_A\) and \(W_A\)`respectively, and similarly
`, \(L_B\) and \(W_B\)` for Frame B.

The area
`(\(A\))` of a rectangle is given by the formula
`\(A = \text{length} * \text{width}\).`

For Frame A:

`\[ A_A = L_A * W_A \]`

For Frame B:

`\[ A_B = L_B * W_B \]`

Given that both frames have the same area of 45 square inches:

`\[ A_A = A_B = 45 \]`

You mentioned the length of Frame A
`(\(L_A\))` is 634 inches, and the length of Frame B
`(\(L_B\))` is 712 inches.

We can set up a system of equations to solve for the widths
`(\(W_A\) and \(W_B\))`:

`\[ L_A * W_A = 45 \]`

`\[ L_B * W_B = 45 \]`

Substitute the given lengths:

`\[ 634 * W_A = 45 \]`

`\[ 712 * W_B = 45 \]`

Now, solve for
`\(W_A\) and \(W_B\):`

`\[ W_A = (45)/(634) \]`

`\[ W_B = (45)/(712) \]`

You can then calculate the values:

`\[ W_A \approx 0.0709 \, \text{inches} \]`

`\[ W_B \approx 0.0632 \, \text{inches} \]`

So, the widths of the two frames are approximately 0.0709 inches for Frame A and 0.0632 inches for Frame B.