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January 31, 2015

Posted by **Muffy** on Friday, October 16, 2009 at 10:01pm.

So I know the width would be x and the length would be x+3

if I do 12x=208, I get 17.3333

I'm really not sure where to go with this can you please help?

- PRE-CALC -
**Reiny**, Friday, October 16, 2009 at 10:11pmthe width of the painting would be x-2 feet (12 inches = 1 foot)

and the length would be x+1

(I have no idea what your equation 12x=208 is supposed to say)

area of whole picture = x(x+3)

area of painting = (x-2)(x+1)

the difference in those areas is 208

x(x+3) - (x-2)(x+1) = 208

solve

- PRE-CALC -
**Muffy**, Friday, October 16, 2009 at 10:19pmCan you please explain the x-2 and x+1, I'm not sure how you got that.

- PRE-CALC -
**Reiny**, Friday, October 16, 2009 at 10:31pmdid you make a diagram?

if the outside width is x feet and you are coming in 12 inches or 1 foot on each side for the frame, isn't the inside width of the picture x - 2 feet

same for the length, which is x+3 on the outside.

come in 1 foot on each side and you have to subtract 2 feet from the original length.

Now what is (x+3) - 2 ?

- PRE-CALC -
**Muffy**, Friday, October 16, 2009 at 10:45pmOh, these word problems make me feel so dense sometimes. Okay, so I got 54.5 and 51.5 as my demensions. Is that correct?

- PRE-CALC -
**Reiny**, Friday, October 16, 2009 at 11:01pmI got the same answer, and I verified it by

51.5x54.5 = 2806.75

49.5x52.5 = 2598.75

and 2806.75-2598.75 = 208

- PRE-CALC -
**Muffy**, Friday, October 16, 2009 at 11:06pmThanks, that was a big help :)

- PRE-CALC -
**Ash**, Friday, October 14, 2011 at 3:48amUmmm the answer in the teacher's book says 11 ft by 14 ft. No idea where you guys got 54.5 and 51...

- PRE-CALC -
**Aerin**, Friday, October 14, 2011 at 9:32pmI do believe the x = 12. Because the area of the painting and the frame is 208 FT^2 and the thickness of the frame is 1 foot. The dimensions of the painting are x by x+3, so they are x^2+3x. The dimensions of the painting and the frame therefore must be x^2+5x+4 from (x+4)(x+1). When you subtract x^2+3x from x^2+5x+4, you get 2x+4. If you look at the various factors of 208, you will notice that the only ones that have a difference of 3 are 13 and 16. So the dimensions of the painting and frame together would be 13 ft by 16 ft thus equaling 208 ft. If 13= x+1, x must be 12 and if 16=x+4, again, x must equal 12. So the dimensions of the painting would be 12 ft by 15 ft as (x)(x+3) equals the dimensions. And to prove my point, if you go back to the 2x+4 and plug in 12 for x, you will get the answer 28. 12 times 15= 180. 208-180=28 which equals the difference between the dimensions of the entire frame and painting and of the painting alone. And there you have your answer. :D

- PRE-CALC -
**Aerin**, Friday, October 14, 2011 at 9:51pmOkay. I just realized something. X will equal 11 and the dimensions would be 11 by 14 because the width of the frame would have to be considered twice. So the dimensions of the painting and the frame altogether would be (x+2)(x+5). The dimensions of the painting would still be (x)(x+3). The difference would change to 4x+10. And if you plug in 11 for x and solve, the difference between the dimensions would equal what the difference between 208 ft^2 and 11 by 14 [154 ft^2]. So in the end, the dimensions of the painting are 11 by 14. PLEASE DISREGARD THE ANSWER ABOVE THIS ONE EXCEPT FOR THE STRATEGY USED AND THEN MAKE SURE YOU READ THIS ANSWER. =]

- PRE-CALC -
**Elle**, Tuesday, November 22, 2011 at 4:36pmI got a different answer than all of you guys :P

HINT: To actually visualize the problem, draw it out! :)

Obviously the dimensions of the painting are: w and 3+w

We know that the frame is 12 in wide, aka 1 ft wide. And we also are given that the total area of the painting and the frame is 208ft^2.

We know that in order to get 208ft^2, we must multiply the length of the painting plus the width of the frame by the width of the painting plus the width of the frame. So therefore:

(3+w+2)(2w)

*Confused where we got the 2 from? We got 2 because you need to add the width of the frame to both sides of the width and both sides of the length.

Next we must distribute, then we will get: 6w+2w^2+4w or 2w^2+10w

Now we know that: 208=2w^2+10w

Our next step is to transfer this equation into Standard Form (ax^2+bx+c): 2w^2+10w-208= 0

Then use the Quadratic Formula to solve for x! (I can't type out the Quadratic Formula, so hopefully you already know what it is, if not just look it up)

You will get x= 8 or x=-13

Obviously, you cannot have a negative dimension so x=8 will be your answer.

The dimensions of the painting will be 8ft by 11ft.

Hope this helps!

- PRE-CALC -
**Adrian**, Monday, April 15, 2013 at 8:22pm**********Correct easy answer*************

See this last answer would've been right up to one point

I got a different answer than all of you guys :P

HINT: To actually visualize the problem, draw it out! :)

Obviously the dimensions of the painting are: w and 3+w

We know that the frame is 12 in wide, aka 1 ft wide. And we also are given that the total area of the painting and the frame is 208ft^2.

We know that in order to get 208ft^2, we must multiply the length of the painting plus the width of the frame by the width of the painting plus the width of the frame. So therefore:

(3+w+2)(2+w)

*Confused where we got the 2 from? We got 2 because you need to add the width of the frame to both sides of the width and both sides of the length.

Then we get (w+5)*(w+2)(dimensions of frame which also incldes the picture)=280 =

w^2+2w+5w+10=280 =

w^2+7w+10=280

Our next step is to transfer this equation into Standard Form (ax^2+bx+c):

w^2+7w-198=0

factor

(w+18)(w-11)=0

w can't be a negative dimension so x=11

the length*width of the picture frame works

(11+5)(11+2)=208

inside the picture works too

280-(11+3)(11)(picture)=54 (the area of the picture frame alone)

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