SOLUTION: Please help me solve: A 5 in. by 7 in. photograph is surrounded by a frame of uniform width. The area of the frame equals the area of the photograph. Find the width of the frame.

Question 72895This question is from textbook Algebra and Trigonometry Structure and Method
: Please help me solve: A 5 in. by 7 in. photograph is surrounded by a frame of uniform width. The area of the frame equals the area of the photograph. Find the width of the frame. This question is from textbook Algebra and Trigonometry Structure and Method

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A 5 in. by 7 in. photograph is surrounded by a frame of uniform width.
The area of the frame equals the area of the photograph. Find the width of the frame.
:
Let x = the width of the frame.
:
Draw a rough sketch of this, label the picture dimensions, and width of the frame as x. Note that the dimensions of the frame will be: (7+2x) by (5+2x)
:
Area of the picture: 5*7 = 35 sq in
Area of the frame given as the same, frame = 35 sq in also
Therefore the total area (picture & frame) will be 70 sq in
:
A simple are equation:
(7+2x) * (5+2x) = 70
FOIL
35 + 14x + 10x + 4x^2 = 70
:
4x^2 + 24x + 35 - 70 = 0; subtract 70 from both sides:
:
4x^2 + 24x - 35 = 0
:
Unfortunately this will not factor easily, have to resort to the quad equation:
a = 4; b = 24; c = -35

:
; only worry about the positive solution here
:
x =
:
x = 1.213 inches is the width of the frame
:
:
Check our solution
(7 + 2(1.213)) * (5 + 2(1.213))
(7+2.426) * (5+2.426)
9.426 * 7.426 = 69.997 ~ 70
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