SOLUTION: Hello I'm having trouble with this one, thanks for the help. The dimensions of a picture inside a frame of uniform width are 12 by 16 inches. If the whole area (picture and frame)

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Question 1201075: Hello I'm having trouble with this one, thanks for the help. The dimensions of a picture inside a frame of uniform width are 12 by 16
inches. If the whole area (picture and frame) is 288 in2, what is the width of
the frame?
Found 4 solutions by josgarithmetic, ikleyn, greenestamps, math_tutor2020:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!
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The dimensions of a picture inside a frame of uniform width are 12 by 16
inches. If the whole area (picture and frame) is 288 in2, what is the width of
the frame?
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Taking away the modifier in the first part,
=====
The dimensions of a picture -modifier- are 12 by 16
inches.
=====

Then the area of the picture alone, is square inches.

Make the drawing and label parts. If u is the uniform width of the frame, then the area of frame PLUS pucture is .

Second part of description,
=====
whole area (picture and frame) is 288 square inches
=====

You already know from description that the frame surrounds the picture, and is of a uniform width. Using the given total area,

And you can solve this equation with whatever skills you have or need. (You might want to first multiply both sides of equation by ).

Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website!
.
Hello I'm having trouble with this one, thanks for the help. T
he dimensions of a picture inside a frame of uniform width are 12 by 16 inches.
If the whole area (picture and frame) is 288 in2, what is the width of the frame?
~~~~~~~~~~~~~~~~~~~~~~

Let's the width of the frame be x inches. Then the dimensions of the frame are (12+2x) and (16+2x) inches, and we write the total area equation in the form (12+2x)*(16+2x) = 288 in^2. For simplicity, divide both sides by 4 (6+x)*(8+x) = 72 Simplify and find x 48 + 6x + 8x + x^2 = 72 x^2 + 14x - 24 = 0. It is not factorable (despite expectations); so, use the quadratic formula = = = . We deny the negative root and accept the positive one x = = 1.544 in (rounded). ANSWER. The width of the frame is = 1.544 in (rounded).

Solved.

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If you want to see many other similar solved problems, look into the lessons
    - Problems on the area and the dimensions of a rectangle surrounded by a strip
    - Cynthia Besch wants to buy a rug for a room
    - Problems on a circular pool and a walkway around it
in this site.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic
"Dimensions and the area of rectangles and circles and their elements".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

Just a comment to say that the problem is very poorly formulated.

The problem is clearly intended as an exercise for beginning algebra students. As such, the process of solving the problem should be the hardest part of the problem. The solution itself should be easy, which means the answer should be a "nice" number.

Ignoring the fact that the problem is probably for beginning algebra students, if the dimensions of the picture and the area whole numbers then the width of the frame should be a whole number. Having a width that is an irrational number to make the total area a whole number is not realistic.

Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!

Start with a 16 by 12 rectangle which represents the picture only.

Then extend each side out by some amount x (as indicated in red).
This represents the width of the frame. It's some positive real number.

Then form a larger rectangle like so

We go from a smaller 16 by 12 rectangle to a larger (16+2x) by (12+2x) rectangle.
We've added two copies of x to each original dimension.

Here's what it would look like if we erased the original rectangle and the red dashed lines

This represents the entire area of the picture plus the frame.

----------------------------------------

We've determined the larger rectangle is (16+2x) by (12+2x)
The area is:
A = L*W
A = (16+2x)(12+2x)
A = 16(12+2x)+2x(12+2x)
A = 192+32x+24x+4x^2
A = 192+56x+4x^2
A = 4x^2+56x+192

Set this equal to the stated area 288 since 4x^2+56x+192 represents the area of the picture itself plus the surrounding frame.

4x^2+56x+192 = 288
4x^2+56x+192-288 = 0
4x^2+56x-96 = 0
4(x^2+14x-24) = 0
x^2+14x-24 = 0

We could try to factor, but it turns out that there aren't two whole numbers that multiply to -24 and add to 14.
The factoring method won't work here.

Luckily the quadratic formula works.
Plug in: a = 1, b = 14, c = -24.

or

or

or

or
Ignore the negative x value.
It doesn't make sense to have a negative frame width.

Therefore, the frame is approximately 1.544 inches wide.
Round that however needed.

Check:
smaller length = 16
larger length = 16+2*x = 16+2*1.544 = 19.088
smaller width = 12
larger width = 12+2*x = 12+2*1.544 = 15.088
larger area = (larger length)*(larger width)
larger area = (19.088)*(15.088)
larger area = 287.999744
We don't land on 288 exactly, but we're fairly close.
This slight discrepancy is due to rounding error.
If you used a more accurate value of x, then you would get closer to 288.
(eg: use x = 1.5440037 to get an approximate area of 287.99999690245477 square inches)