Question 1207821: An artist has 50 inches of oak trim to frame a painting. The frame is to have a border 3 inches wide surrounding the painting.
(a) If the painting is square, what are its dimensions? What are the dimensions of the frame?
(b) If the painting is rectangular with a length twice its width, what are the dimensions of the painting? What are the dimensions of the frame?
Found 3 solutions by greenestamps, ikleyn, mananth:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Technically, the statement of the problem is deficient, because it does not say that the artist uses the full 50 inches of trim.
Obviously the problem can be solved only if we assume that he does. So....
(a) If the painting is a square...
Let x be the side length of the painting; then the perimeter is 4x, and the perimeter of the painting plus frame is 4(x+3) = 4x+12.
4x+12 = 50
4x = 38
x = 38/4 = 19/2 = 9.5
ANSWERS:
The painting is a square 9.5 inches on a side.
The frame is a square 9.5+2(3) = 15.5 inches on a side.
(b) If the painting is a rectangle with a length twice its width...
Let x be the width; then the length is 2x; the perimeter is 6x, and the perimeter of the frame is 6x+12.
6x+12 = 50
6x = 38
x = 38/6 = 19/3 = 6 1/3
ANSWERS:
The painting is a rectangle 6 1/3 inches wide and 12 2/3 inches long.
The frame is a rectangle 6 1/3 + 2(3) = 12 1/3 inches wide and 12 2/3 + 2(3) = 18 2/3 inches long.
Answer by ikleyn(52852)  (Show Source):
You can put this solution on YOUR website! .
An artist has 50 inches of oak trim to frame a painting.
The frame is to have a border 3 inches wide surrounding the painting.
(a) If the painting is square, what are its dimensions?
What are the dimensions of the frame?
(b) If the painting is rectangular with a length twice its width, what are
the dimensions of the painting? What are the dimensions of the frame?
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(a) Imagine this painting surrounded by 3 inches wide frame.
Let x be size of the painting square, in inches.
Then from the problem, you have this equation for x
(x+3) + (x+3) + (x+3) + (x+3) = 50 inches,
or
4(x+3) = 50.
It gives
x+3 = 50/4 = 12.5 inches,
x = 12.5 - 3 = 9.5 inches.
Thus the dimensions of the painting are 9.5 by 9.5 inches. <<<---=== Answer to (a).
Outer dimensions of the frame are 9.5 + 3 + 3 = 15.5 inches. <<<---=== Answer to (a).
(b) Let w be the width of the painting, in inches.
Then the length of the painting is 2x inches.
From the problem, you have this equation for x
(2w+3) + (w+3) + (2w+3) + (w+3) = 50 inches,
or
6w+12 = 50.
It gives
w = (50-12)/6 = 38/6 = 6 inches (the width),
2x= 12 inches (the length).
Thus the dimensions of the painting are 6 inches by 12 inches. <<<---=== Answer to (b).
Outer dimensions of the frame are 6 + 3 + 3 = 12 and 12 + 3 + 3 = 18 inches. <<<---=== Answer to (b).
Solved in full.
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Be aware: both the solution method and the answers in the post by @mananth are incorrect.
They are incorrect, because his setup equations are incorrect, in both cases.
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! An artist has 50 inches of oak trim to frame a painting. The frame is to have a border 3 inches wide surrounding the painting.
(a) If the painting is square, what are its dimensions? What are the dimensions of the frame?
(b) If the painting is rectangular with a length twice its width, what are the dimensions of the painting? What are the dimensions of the frame?
An artist has 50 inches of oak trim to frame a painting. The frame is to have a border 3 inches wide surrounding the painting.
Let the side of the painting be x inches
There is a border of 3 inches all round.
length of the side of frame will be (x+6) ( 3 in on either side.)
Perimeter is 50 inches
4(x+6) =50 ( 4 sides)
4x+24 =50
4x= 26
x= 6.5 inches
Painting is a square with side 6.5 inches and frame with 12.5 inches side
(b) If the painting is rectangular with a length twice its width, what are the dimensions of the painting? What are the dimensions of the frame?
Let width of painting be x in. the length will be 2x
frame dimensions
x+6 and 2x+6
perimeter
2(x+6)+2(2x+6)=50
2x+12 +4x+12=50
6x = 26
x= 4.33 with
Length = 2x= 8.66 in
Add 6 to both sides to get the frame dimensions
The dimensions of the framed painting are 10.33 inches by 14.67 inches.
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