SOLUTION: A frame of uniform width is to be placed around a painting so that the area of the frame is twice the area of the picture. If the outside dimensions of the frame of 40cm and 60cm,

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A frame of uniform width is to be placed around a painting so that the area of the frame is twice the area of the picture. If the outside dimensions of the frame of 40cm and 60cm,       Log On


   



Question 1155671: A frame of uniform width is to be placed around a painting so that the area of the frame is twice the area of the picture. If the outside dimensions of the frame of 40cm and 60cm, then find the width of the frame.
Found 2 solutions by greenestamps, josmiceli:
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The dimensions including the frame are 40 by 60.

With a frame of uniform width x, the dimensions of the painting alone are (40-2x) by (60-2x).

The area of the frame is twice the area of the painting; that means the area with the frame is 3 times the area of the painting alone.

60%2A40+=+3%2860-2x%29%2840-2x%29

Solve using basic algebra....

If a formal algebraic solution is not required, the problem can be solved mentally in a few seconds.

With the frame, the area is 40*60 = 2400. That is the product of two numbers whose difference is 20 with a product of 2400.

The area of the painting alone is one-third of the total area, which is 800.

With a frame of uniform width, the dimensions of the painting are two numbers whose difference is also 20.

A couple of moments of mental arithmetic find the dimensions of the painting to be 20 by 40.


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The area of picture and frame is:
+40%2A60+=+2400+ cm2
Let +A+ = the area of the picture
+2400+-+A+ = the area of the frame
--------------------------------------------
+2400+-+A+=+2A+
+3A+=+2400+
+A+=+800+ cm2
-------------------------
Let +W+ = the width of the frame
+%28+40+-+2W+%29%2A%28+60+-+2W+%29+=+800+
+2400+-+120W+-+80W+%2B+4W%5E2+=+800+
+4W%5E2+-+200W+%2B+1600+=+0+
+W%5E2+-+50W+%2B+400+=+0+
+%28+W+-+10+%29%2A%28+W+-+40+%29+=+0+
+W+ can't be +40+ since that is one of
the outside dimensions, so
+W+=+10+
The width is 10 cm
--------------------------
check:
+%28+40+-+2W+%29%2A%28+60+-+2W+%29+=+800+
+%28+40+-+2%2A10+%29%2A%28+60+-+2%2A10+%29+=+800+
+%28+40+-+20+%29%2A%28+60+-+20+%29+=+800+
+20%2A40+=+800+
+800+=+800+
OK