A picture frame measures $14$ cm by $20$ cm. $160$ cm$^2$ of the picture shows inside the frame. Find the width of the frame.

Question

A picture frame measures $14$cm by $20$cm. $160$cm$^2$ of the picture shows inside the frame. Find the width of the frame.

This is the question I was given, word-for-word. Is it asking for the width of the picture? Becuase the way I see it, the width is simply $14$...

EDIT: Is this the correct interpretation? (Sorry did a really quick drawing in MS paint, red represents what I am supposed to find)

Answer 1

You know what a picture frame looks like, right? Two concentric rectangles.

You are given the dimensions of the outer rectangle. You are given the area of the inner rectangle.

You are asked to find the thickness of the border between them.

$$20\textrm{cm}\left\{ \vphantom{\bbox[green, 2ex, border:solid 1pt]{\bbox[white, 1ex, border:solid 1pt]{\begin{array}{l}\qquad\\160\textrm{cm}^2\\~\\~\end{array}}}} \right. % \overbrace{\bbox[green, 2ex, border:solid 1pt]{\bbox[white, 1ex, border:solid 1pt]{\begin{array}{l}\qquad\\160\textrm{cm}^2\\~\\~\end{array}}}}^{14\textrm{cm}}$$

Answer 2

The question pretty ambiguous. My interpretation would be that you have a rectangular picture frame whose exterior dimensions are $14\times 20$. The frame has uniform width, so all sides are the same width. If you place a picture in the frame then $160$ square cm of the picture will be visible. What is the width of a side of the frame?

Answer 3

You are expected to assume that the picture frame has a constant width. That width creates a border around the picture. Note that the area of the frame is $280 \text{ cm}^2$, which is larger than the picture. You are supposed to find th width of the frame.

Answer 4

Yes, I think that's what you have to solve. So, if your picture frame's constant width is x cm, you can get this;

14*20=160+2(14-x)x+2(20-x)x

280=160+28x-2x^2+40x-2x^2

4x^2-68x-120=0

x^2-17x+30=0

(x-15)(x-2)=0

x=2 or 15

But, x<14

Therefore, x=2

2cm

Answer 5

$$(14-2w)(20-2w)=160.$$

This quadratic equations has the two solutions $w=2$ and $w=15$; the latter must naturally be rejected.

Hence the width is $2$ cm.