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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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

$25.5$ $25.5$ inches by $35.5$ $35.5$ inches.

Explanation:

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

Then the area of the frame is given by:

$A_{frame} = A_{Picture with frame} - A_{Picture without frame}$ $A_"frame" = A_"Picture with frame" - A_"Picture without frame"$

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.

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

$402 = x^{2} + 6 x + 16 x + 96 - x^{2} - 10 x$ $402 = x^2 + 6x + 16x + 96 - x^2 - 10x$

$402 = 12 x + 96$ $402 = 12x + 96$

$306 = 12 x$ $306 = 12x$

$x = 25.5$ $x = 25.5$

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

Hopefully this helps!

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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?

1 Answer

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