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### Fitting a Picture to a Frame

```
Date: 11/14/96 at 23:00:27
From: Anonymous
Subject: test

I have a problem that I worked out in a drawing but do not know how to
solve with an algebra formula: A picture frame is 10 by 12 inches
with the shorter side horizontal. A photographer wants to put a photo
in the frame which will have twice as much border at the sides as at
the top and bottom. If the photo is to take up half the area of the
frame, how wide should the border be at the top and bottom?

I figured the area of the frame was 120 sq in so the area of the
picture was 60 sq in. I played around with height and width and found
6 x 10 to allow a 1" border at the top and bottom and a 2" frame on
each side. But what is the algebraic formula?
```

```
Date: 11/20/96 at 11:45:19
From: Doctor Lynn
Subject: Re: test

I'm hoping you understand quadratic equations, because you can't
answer questions like this without them.  If not, the best thing to do

Assuming the photo is central (you could work it out if it wasn't, but
it would be a bit more complicated), the frame looks like this:

+-----------------+----  (Excuse my bad diagram, it's just
|                 |   h   meant to show that the photo is in
|     +-----+----------   the middle of a border of height h
|     |  P  |     |       top and bottom and a border of width
|<-w->|  H  |<-w->|       w left and right and that the total
12"|     |  O  |     |       width is 10" and the total height is
|     |  T  |     |       12")
|     |  O  |     |
|     +-----+----------
|                 |   h
+-----------------+----
10"

The width of the border is w, the height of the border is h and the
area of the frame is 120 square inches, so half the area is 60 square
inches.

The width of the photo is 10-2w (Total width - width of border on
either side).  The height of the photo is 12-2h (Total height - border
height on top & bottom).

So (10-2w)(12-2h) = 60
(width of photo * height of photo = area of photo)

But we know that there is twice as much border at the sides as at the
top, so w = 2h.

So    (10-2w)(12-w) = 60
120-10w-24w+2w^2 = 60
2w^2-34w+60 = 0
w^2-17w+30 = 0
(w-15)(w-2) = 0

So the width of the border is either 15 or 2, because these are the
only values of w for which the equation is true.  But we know the
width can't be 15", because the width of the frame is only 10 inches.
This means the width must be 2 inches and the height must be 1 inch.

-Doctor Lynn,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Euclidean/Plane Geometry
High School Geometry
High School Practical Geometry
High School Triangles and Other Polygons
Middle School Algebra
Middle School Geometry
Middle School Triangles and Other Polygons
Middle School Two-Dimensional Geometry

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