# 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?

Dec 21, 2016

$25.5$ inches by $35.5$ inches.

#### Explanation:

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:

${A}_{\text{frame" = A_"Picture with frame" - A_"Picture without frame}}$

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.

$402 = \left(x + 6\right) \left(x + 16\right) - x \left(x + 10\right)$

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

$402 = 12 x + 96$

$306 = 12 x$

$x = 25.5$

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

Hopefully this helps!