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a rectangular painting measures 10 inches by 17 inches and is surrounde…

Question

a rectangular painting measures 10 inches by 17 inches and is surrounded by a frame of uniform width around the four edges. the perimeter of the rectangle formed by the painting and its frame is 78 inches. determine the width of the frame.

Explanation:

Step1: Define total width/height with frame

Let the frame width be $x$.
New width: $10 + 2x$, New length: $17 + 2x$

Step2: Set up perimeter equation

Perimeter formula: $P=2(l+w)$
$78 = 2[(17+2x)+(10+2x)]$

Step3: Simplify the equation

Simplify inside the brackets:
$78 = 2(27 + 4x)$
Divide both sides by 2:
$39 = 27 + 4x$

Step4: Solve for x

Subtract 27 from both sides:
$4x = 39 - 27 = 12$
Divide by 4:
$x = \frac{12}{4} = 3$

Answer:

3 inches