the outside of a picture frame has a length which is 4 cm more than width. The area enclosed by the outside of the picture frame is 165 square cm. Find the width of the outside of the picture frame.
L = w + 4
L w = 165
(w+4)w = 165
w^2 + 4 w - 165 = 0
(w+15)(w-11) = 0
w = 11
Well, isn't this a puzzling frame! Let's solve it together.
Let's say the width of the outside of the picture frame is "x" cm. According to the problem, we know that the length is 4 cm more than the width, which means the length is "x + 4" cm.
The area of the frame can be calculated by subtracting the area of the picture from the area of the outside frame. Since the area of the frame is given as 165 square cm, we have:
(x + 4) * x - x * (x - 4) = 165
Simplifying that equation, we have:
x^2 + 4x - (x^2 - 4x) = 165
Canceling out the x^2 terms and combining like terms, we get:
8x = 165
Dividing both sides by 8, we find:
x = 20.625
So, the width of the outside of the picture frame is approximately 20.625 cm. Although, I must say, that's quite an unusual width for a frame!
Let's say that the width of the outside of the picture frame is x cm.
According to the given information, the length of the outside of the picture frame is 4 cm more than the width. So the length would be x + 4 cm.
The formula for the area of a rectangle is A = length * width. In this case, the area is given as 165 square cm.
So we can write the equation as:
165 = (x + 4) * x
Expanding the equation:
165 = x^2 + 4x
Rearranging the equation to get it in standard form:
x^2 + 4x - 165 = 0
Now, we can solve this equation either by factoring, completing the square, or using the quadratic formula. Let's solve it using factoring:
(x + 15)(x - 11) = 0
So, x + 15 = 0 or x - 11 = 0
If we solve for x in each case:
x = -15 or x = 11
Since the width cannot be negative, the width of the outside of the picture frame is 11 cm.
Therefore, the width of the outside of the picture frame is 11 cm.
To find the width of the outside of the picture frame, we can set up the problem as a quadratic equation. Let's assume the width of the frame is x cm.
According to the given information, the length of the frame is 4 cm more than the width. So, the length would be x + 4 cm.
The area enclosed by the outside of the picture frame is given as 165 square cm. The formula for the area of a rectangle is length multiplied by width. Therefore, we have:
Area = Length * Width
165 = (x + 4) * x
Now, let's solve this quadratic equation for x by expanding and rearranging the equation:
165 = x^2 + 4x
Rearranging the equation, we get:
x^2 + 4x - 165 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 4, and c = -165. Plugging these values into the quadratic formula:
x = (-4 ± √(4^2 - 4 * 1 * -165)) / (2 * 1)
x = (-4 ± √(16 + 660)) / 2
x = (-4 ± √676) / 2
x = (-4 ± 26) / 2
Simplifying further, we have two possible solutions:
x1 = (-4 + 26) / 2 = 11
x2 = (-4 - 26) / 2 = -15
Since the width cannot be negative, we take the positive value as the width of the outside of the picture frame. Therefore, the width of the outside of the picture frame is 11 cm.