osirus0830 wrote:18

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sorry for my horrible attempt to draw the figure above. It should look like a rectangle inside another rectangle. The shaded region is the area between the two rectangles.


The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?



Two things:

1) Can someone show me how to solve this

2) Can someone explain the approach to solving problems that relate the quantites in terms of ratios?

Thanks in advance

Hey Osirus,

This is a total shot in the dark because I'm not totally sure from your picture and explanation what the question is asking but hopefully my approach can help. I'm not sure what the shaded region has anything to do with things but maybe I am not visualizing properly.

I look at this as an algebra problem mixed with geometry. It seems like you are struggling so I'll try to make more in depth explanations (and because I'm not sure I'm getting the problem)

Aframe = area of the frame
Apicture = area of the picture
Aoutter = area of the rectangle represented by the frame
Lframe = Loutter = length of the frame, which is the same as the length of the outter rectangle
Wframe = Woutter = width of the frame, which is the same as the width of the outter rectangle
Lpicture = length of the picture
Wpicture = width of the picture

Thus for the first equation we have, from the fact that the area of the picture equals the area of the frame...

Apicture = Aoutter - Apicture
2*(Apicture) = Aoutter
2*(Lpicture)*(Wpicture) = (Lframe) * (Wframe)

For the second part we are given a ratio. This is how I think of things. In any case, whether they give you the fact that ratios are equal or not, you can always represent ratios as follows.

(Lpicture) / (Wpicture) = ALPHA
(Lframe) / (Wframe) = BETA

I have no idea what ALPHA and BETA are. But in this case it doesnt matter. It says that the length width ratio of the frame is equal to the length width ratio of the picture. What the problem tells you is that the ratios are the same, and thus

ALPHA = BETA

So

(Lpicture) / (Wpicture) = (Lframe) / (Wframe)

The words "same ratio" should scream to you "a fractional equality" and thus an equation like the above.

Since you know the Length and Width of the frame to be 18 and 15 respectively, you now have 2 equations and 2 unknowns.

(1) (Lpicture) * (Wpicture) = 18*15 / 2
(2) (Lpicture) / (Wpicture) = 18/15

Now its just plug and chug equation solving. I get that the length of the picture is 18*root(2).

Helpful?

osirus0830 wrote:18

The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?

is the length=15 inches and width=9 inhes

pls confirm the ans

thephoenix wrote:

osirus0830 wrote:18

The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?

is the length=15 inches and width=9 inhes

pls confirm the ans

That doesn't satisfy the condition that the ratio of the length of the frame to the width of the frame is equal to the ratio of the length of the picture to the width of the picture.

Lpicture / Wpicture = 15/9 (from your answer)
Lframe / Wframe = 18/15 (from the question)

15/9 != 18/15

Again, unless I am reading the question wrong...

Thanks guys for the replies. The length is 18 and the width is 15.

osirus0830 wrote:Thanks guys for the replies. The length is 18 and the width is 15.

Whats the length of the picture?

it looks like this

osirus0830 wrote:18

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| | | | 15
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sorry for my horrible attempt to draw the figure above. It should look like a rectangle inside another rectangle. The shaded region is the area between the two rectangles.


The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?



Two things:

1) Can someone show me how to solve this

2) Can someone explain the approach to solving problems that relate the quantites in terms of ratios?

Thanks in advance

Osirus I am totally confused what u r trying to show in figure.. Well if u can draw figure in some paint tool and repost it that would be of great help to all of us here.

osirus0830 wrote:it looks like this

Okay then I think that I understood you correctly. I stand by my explanation!

ldoolitt wrote:

osirus0830 wrote:it looks like this

Okay then I think that I understood you correctly. I stand by my explanation!

Your explanation was very helpful, the only thing is the answer is 9 sqrt(2)

osirus0830 wrote:

ldoolitt wrote:

osirus0830 wrote:it looks like this

Okay then I think that I understood you correctly. I stand by my explanation!

Your explanation was very helpful, the only thing is the answer is 9 sqrt(2)

OOOO, super apologies Osirus. When I solved the last two equations on a sheet of paper i copied them down wrong from the screen and misplaced a 2. If you solve (1) and (2) you'll note that you get Lpainting = 9*root(2). The explanation is exactly the same. My math sucks.