A square mirror has sides measuring 2 ft. less than the sides of a square painting. If the difference between their areas is 32 ft^2, find the lengths of the sides of the mirror and the painting?
Square mirror= x-2
Painting=x
I don't know where to go from there...
The dimensions of a HP flat panel monitor are such that its length is 3in more than its width. If the length were doubled and if the width were decreased by 1in. the area would be increased by 150 in^2. What are the length and the width?
I don't know where to start....
first one:
translate the English into math.
<A square mirror has sides measuring 2 ft. less than the sides of a square painting>
let the sides of the square painting be x ft.
Then the sides of the mirror are each x-2 ft.
area of mirror = (x-2)(x-2)
area of painting = (x)(x)
<the difference in their areas is 32 ft^2
----> x^2 - (x-2)^2 = 32
x^2 - (x^2 - 4x + 4 = 32
x^2 - x^2 + 4x - 4 = 32
4x = 36
x = 9
Number two:
old width : x
old length : x+3
old area : x(x+3)
new length : 2(x+3)
new width : x+3 - 1 or x+2
new area : 2(x+3)(x+2)
<area is increased by 150 in^s> or
the difference in their areas is 150
2(x+3)(x+2) - x(x+3) = 150
Can you take it from there?
Let me know what you got.
new width should be x-1
new area should be 2(x+3)(x-1)
equation should be
2(x+3)(x-1) - x(x+3) = 150
To solve the first problem, you can set up an equation to represent the given information.
Let x be the length of the sides of the square painting.
According to the problem, the length of the sides of the square mirror is 2 feet less than the sides of the square painting. Therefore, the length of the sides of the square mirror is x - 2.
The difference between their areas is given as 32 ft^2. The area of the square painting is x^2, and the area of the square mirror is (x - 2)^2.
So, the equation becomes:
x^2 - (x - 2)^2 = 32
You can expand the right side of the equation and simplify it:
x^2 - (x^2 - 4x + 4) = 32
x^2 - x^2 + 4x - 4 = 32
4x - 4 = 32
4x = 36
x = 9
Therefore, the length of the sides of the square painting is 9 feet, and the length of the sides of the square mirror is 9 - 2 = 7 feet.
For the second problem, let's denote the width of the HP flat panel monitor as w.
According to the problem, the length of the monitor is 3 inches more than its width, so the length can be represented as w + 3.
If we double the length and decrease the width by 1 inch, the new dimensions become 2(w + 3) for the length and w - 1 for the width.
We are given that this change increases the area by 150 in^2. The original area is w(w + 3), and the new area is 2(w + 3)(w - 1).
So, the equation becomes:
2(w + 3)(w - 1) - w(w + 3) = 150
You can expand and simplify the equation:
2(w^2 + 2w - 3) - (w^2 + 3w) = 150
2w^2 + 4w - 6 - w^2 - 3w = 150
w^2 + w - 6 = 150
w^2 + w - 156 = 0
Now, you can factor or solve the quadratic equation using the quadratic formula to find the values of w.
To find the lengths of the sides of the mirror and the painting, we can set up an equation based on the given information.
Let's denote the length of the painting as x. According to the problem, the length of the mirror is 2 ft less than the length of the painting, so it would be x - 2.
The area of a square can be found by squaring its side length. Therefore, the area of the painting is x^2, and the area of the mirror is (x - 2)^2.
Now, we are given that the difference between their areas is 32 ft^2. We can set up the following equation:
x^2 - (x - 2)^2 = 32
Expanding (x - 2)^2, we get:
x^2 - (x^2 - 4x + 4) = 32
Simplifying the equation, we have:
x^2 - x^2 + 4x - 4 = 32
4x - 4 = 32
Adding 4 to both sides:
4x = 36
Dividing both sides by 4:
x = 9
So, the length of the painting is 9 ft.
Using the given information that the mirror has sides 2 ft. less than the sides of the painting, the length of the mirror would be 9 - 2 = 7 ft.
Therefore, the lengths of the sides of the mirror and the painting are 7 ft and 9 ft, respectively.
Now let's move on to the second question.
To find the length and width of the monitor, we need to set up an equation based on the given information.
Let's denote the width of the monitor as x. According to the problem, the length is 3 inches more than the width, so it would be x + 3.
The area of a rectangle can be found by multiplying its length by its width. Therefore, the area of the original monitor is x(x + 3), and the area of the modified monitor is 2(x + 3)(x - 1).
According to the problem, the area increases by 150 in^2 when the length is doubled and the width is decreased by 1 inch. So, we can set up the following equation:
2(x + 3)(x - 1) - x(x + 3) = 150
Expanding and simplifying the equation, we have:
2x^2 + 4x - 2x - 6 - x^2 - 3x = 150
x^2 - x - 6 - 3x = 150
Combining like terms, we get:
x^2 - 4x - 156 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula.
Factoring:
(x - 13)(x + 12) = 0
Setting each factor equal to zero:
x - 13 = 0 or x + 12 = 0
Solving each equation:
x = 13 or x = -12
Since we are dealing with dimensions, the width cannot be negative. Therefore, the width of the monitor is 13 inches.
Using the given information that the length is 3 inches more than the width, the length of the monitor would be 13 + 3 = 16 inches.
So, the length and width of the monitor are 16 inches and 13 inches, respectively.