Question

If the total area of a picture measuring 10 inches by 12 inches plus a meeting of a uniform width surrounding the picture is 224 square inches, the width of the matting is...

Answer 1

To find the width of the matting, you can solve a quadratic equation using the information given.

We know that the new area of 224 square inches will be found by multiplying the width and the height.  If the original dimensions are 10 in by 12 in, the new dimensions would be 10 +x and 12 + x.

Use these in the formula for area of a rectangle.

A = (10 + x) (12 + x)
224 = x^2 + 22x + 120
-120                  -120
104 = x^2 + 22x 
-104       -104
0 = x^2 + 22x -104
0 = (x + 26) (x - 4)
The only answer for x that would solve this problem would be the positive 4.  

The width of the matting is 4 inches.

Check your answer:
(10 +4) (12 + 4)
14 x 16
224 square inches.

Answer 2

Answer:

The width of the matting is 4 inches.

Step-by-step explanation:

Since we are given the area of the picture including the frame we can use the formula for the Area of a rectangle to solve this problem.

$A = L * W$

since the picture frame has a uniform width we can use the variable x and plug in the values in the formula above. We then solve for x to find out the width of the matting.

$224 = (10+x) * (12+x)$ .....use distributed multiplication

$224 = 120+10x+12x+x^{2}$

$224 = 120+22x+x^{2}$ .... subtract 224 on both sides

$0 = x^{2} +22x-104$

Now we have a quadratic equation which we can use the quadratic formula to solve.

$x = \frac{-b (+ or -)\sqrt{b^{2}-4ac } }{2a}$

$x = \frac{-22 (+ or -)\sqrt{22^{2}-4(1)(-104) } }{2(1)}$

$x = \frac{-22 (+ or -)\sqrt{484+416} }{2}$

$\frac{-22 + 30}{2} = 4$

$\frac{-22 - 30}{2} = -26$

After Solving the Quadratic Equation we are given two values, but since our answer cannot be a negative, then the only valid answer is 4 in . So the width of the matting is 4 inches.

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