An oil painting is 10in. longer than it is wide and is bordered on all sides by a 3in. wide frame. If the area of the frame alone is 402in^2, what are the dimensions of the painting?

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Answer 1 · 2016-12-21T05:17:43

#25.5# inches by #35.5# inches.

We let the width be #x# and the length #x + 10#. Then the frame has a width of #x + 3 + 3 = x + 6# and a length of #x + 10 + 3 + 3= x + 16#.

Then the area of the frame is given by:

#A_"frame" = A_"Picture with frame" - A_"Picture without frame"#

We know the area of the frame. Since the frame with the picture and without the picture is a rectangle, the area is given by length times width.

#402 = (x + 6)(x + 16) - x(x + 10)#

#402 = x^2 + 6x + 16x + 96 - x^2 - 10x#

#402 = 12x + 96#

#306 = 12x#

#x = 25.5#

The picture therefore has dimensions of #25.5# inches by #35.5# inches.

Hopefully this helps!