A photograph print measuring 24cm by 15 cm is enclosed in a frame.
A uniform space of width x cm is left in between the edges of the photograph and the frame. If the area of the space is 270cm', find the value of x.
new dimension: 24+2x by 15+2x
entire area = (24+2x)(15+2x)
(24+2x)(15+2x) - (24)(15) = 270
solve the resulting quadratic for x
Hint: the simplified quadratic factors nicely, make sure to reject the negative
To find the value of x, we can start by calculating the area of the photograph.
The area of the photograph is given by the formula:
Area = Length x Width
Substituting the given values:
Area = 24 cm x 15 cm
Area = 360 cm²
Next, we can calculate the total area of the space including the gaps on all four sides of the photograph.
The total area of the space is given by the formula:
Total Space Area = (2x + 24) x (2x + 15)
Substituting the given value for the total area of the space:
270 = (2x + 24) x (2x + 15)
Expanding this equation, we have:
270 = 4x² + 78x + 360
Rearranging the equation to have all terms on one side:
4x² + 78x + 360 - 270 = 0
4x² + 78x + 90 = 0
Next, we can solve this quadratic equation. However, this equation doesn't factor easily, so we will use the quadratic formula:
x = (-b ± √(b² - 4ac)) / 2a
For our equation:
a = 4, b = 78, and c = 90
Substituting these values into the quadratic formula:
x = (-78 ± √(78² - 4 * 4 * 90)) / (2 * 4)
Simplifying:
x = (-78 ± √(6084 - 1440)) / 8
x = (-78 ± √4644) / 8
Finding the square root:
x = (-78 ± 68.14) / 8
Now, we can calculate the two possible values of x:
1. x = (-78 + 68.14) / 8 = -9.86 / 8 = -1.23
2. x = (-78 - 68.14) / 8 = -146.14 / 8 = -18.27
Since we are working with a width, the value of x cannot be negative. Therefore, the solution for x is:
x = -1.23 cm
To find the value of x, we need to consider the area of the photograph, the area of the frame, and the area of the space between them.
The area of the photograph is given by multiplying its length by its width: 24 cm × 15 cm = 360 cm².
Let's assume the width of the space between the photograph and the frame is x cm. Then the dimensions of the frame would be (24 + 2x) cm by (15 + 2x) cm.
The area of the frame can be calculated by subtracting the area of the photograph from the area of the frame and space combined. This can be represented by the equation: (24 + 2x)(15 + 2x) - 360 = 270.
Simplifying the equation, we have: 360 + 48x + 30x + 4x² - 360 = 270.
Combining like terms, we get: 78x + 4x² = 270.
Rearranging the equation, we have: 4x² + 78x - 270 = 0.
We can now solve this quadratic equation using factoring, completing the square, or the quadratic formula. For simplicity, let's solve it using the quadratic formula.
The quadratic formula states that for an equation of the form ax² + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b² - 4ac)) / (2a).
In our equation, a = 4, b = 78, and c = -270. Substituting these values into the formula, we get:
x = (-78 ± √(78² - 4(4)(-270))) / (2(4)).
Simplifying further, we have:
x = (-78 ± √(6084 + 4320)) / 8.
x = (-78 ± √(10404)) / 8.
x = (-78 ± 102) / 8.
Now, we have two possible values for x:
1. x = (-78 + 102) / 8 = 24 / 8 = 3.
2. x = (-78 - 102) / 8 = -180 / 8 = -22.5.
Since we are looking for a width value, x cannot be negative. Therefore, the value of x is 3 cm.
So, the width of the space left in between the edges of the photograph and the frame is 3 cm.