1) A rectangle has perimeter 38 feet and an area of 78 square feet. Find the dimensions of the rectangle assuming the width is smaller than the length.
L= W=
2) A rectangular canvas picture measures 9 inches by 7 inches. The canvas is mounted inside a frame of uniform width, increasing the total area covered by both canvas and frame to 120 square inches. Find the uniform width of the frame.
LW = 78, so L = 78/W
L+W = 19
78/W + W = 19
W^2 - 19W + 78 = 0
W=6,L=13
for the picture, if the frame has width w, then
(7+2w)(9+2w) = 120
now just solve for w.
Just by looking, I'd say that 10x12 = 120, so w=1.5
1) Well, the length and width of the rectangle seem to be a bit shy, always standing behind the perimeter and area! Let's help them unveil themselves:
Given that the perimeter is 38 feet, we know that 2(length + width) = 38. From this, we can infer that length + width = 19.
Now, we also know that the area is 78 square feet. So length * width = 78.
Let's solve this mystery! We have two clues:
l + w = 19,
lw = 78.
By substituting 19 - l for w in the second equation, we get:
l * (19 - l) = 78.
Expanding and rearranging this equation, we get:
l^2 - 19l + 78 = 0.
Solving this quadratic equation, we find two possible solutions for l, which are 6 and 13.
So, the dimensions of the rectangle are:
Length (L) = 13 feet,
Width (W) = 6 feet.
2) Ah, a picture framed by a frame! It's like a work of art getting dressed up for a fancy party.
Let's solve this puzzle step-by-step, shall we? Given that the rectangular canvas measures 9 inches by 7 inches, we need to find the width of the frame (let's call it F) so that the total area covered by both canvas and frame is 120 square inches.
So, the area of the canvas alone is 9 * 7 = 63 square inches.
Now, let's add the width of the frame to each dimension. The new dimensions are (9 + 2F) inches by (7 + 2F) inches.
The total area, which is the canvas area plus the frame area, is given as 120 square inches. So, we have the equation:
(9 + 2F)(7 + 2F) = 120.
Expanding this equation, we get:
63 + 18F + 14F + 4F^2 = 120.
Combining like terms, we have:
4F^2 + 32F - 57 = 0.
Solving this quadratic equation, we find two possible solutions for F, which are approximately -3.57 (negative width is not possible in this context) and 3.07.
Since the width of the frame cannot be negative, the uniform width of the frame is approximately 3.07 inches.
So, it seems our framed picture will have a frame width of 3.07 inches, adding a touch of elegance and enhancing its total area to 120 square inches!
1) To find the dimensions of the rectangle, let's assume the length of the rectangle is L and the width is W.
Since the perimeter of a rectangle is given by the formula P = 2L + 2W, we can set up the equation:
2L + 2W = 38
Since the area of a rectangle is given by the formula A = L * W, we can set up the equation:
L * W = 78
From the first equation, we can solve it for L:
2L = 38 - 2W
L = (38 - 2W) / 2
Substitute this value of L into the second equation:
(38 - 2W) / 2 * W = 78
Multiply both sides by 2 to eliminate the fraction:
38 - 2W = 156 - 4W
2W - 4W = 156 - 38
-2W = 118
W = 118 / -2
W = -59
Since the width cannot be negative, we can say that it is not possible to have a rectangle with the given dimensions.
Therefore, there are no valid dimensions for this rectangle.
2) Let's assume the uniform width of the frame is x inches.
The length of the overall picture (including the frame) can be calculated by adding twice the frame width to the original length:
Length including frame = 9 + 2x
The width of the overall picture (including the frame) can be calculated by adding twice the frame width to the original width:
Width including frame = 7 + 2x
The area of the overall picture (including the frame) is calculated by multiplying the length and width:
Area including frame = (9 + 2x) * (7 + 2x)
Given that the total area covered by both canvas and frame is 120 square inches, we can set up the equation:
(9 + 2x) * (7 + 2x) = 120
Expanding the equation, we get:
63 + 18x + 14x + 4x^2 = 120
4x^2 + 32x + 63 = 120
Rearranging the equation by subtracting 120 from both sides:
4x^2 + 32x + 63 - 120 = 0
4x^2 + 32x - 57 = 0
Now, we can solve this quadratic equation. We can either factor it or use the quadratic formula. In this case, factoring is not possible, so we will use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
By substituting the values a = 4, b = 32, and c = -57 into the quadratic formula, we get:
x = (-32 ± sqrt(32^2 - 4 * 4 * (-57))) / (2 * 4)
Simplifying further:
x = (-32 ± sqrt(1024 + 912)) / 8
x = (-32 ± sqrt(1936)) / 8
x = (-32 ± 44) / 8
This results in two possible values for x:
x1 = (-32 + 44) / 8 = 12 / 8 = 1.5 inches
x2 = (-32 - 44) / 8 = -76 / 8 = -9.5 inches
Since the width of the frame cannot be negative, the uniform width of the frame is 1.5 inches.
To find the dimensions of a rectangle given its perimeter and area, and assuming the width is smaller than the length, we can use the following steps:
1) Let's start with the first question:
Given: Perimeter = 38 feet, Area = 78 square feet
Let L represent the length and W represent the width of the rectangle.
We know that the perimeter of a rectangle is given by the formula: Perimeter = 2(L + W).
So, we can set up the equation: 38 = 2(L + W).
Next, we know that the area of a rectangle is given by the formula: Area = L * W.
So, we can set up the equation: 78 = L * W.
Now, we have a system of equations:
Equation 1: 38 = 2(L + W)
Equation 2: 78 = L * W
To solve this system, we can use substitution.
From Equation 1, we can isolate L:
L + W = 19
L = 19 - W
Substitute this value of L into Equation 2:
78 = (19 - W) * W
78 = 19W - W^2
Rearrange this equation to form a quadratic equation:
0 = W^2 - 19W + 78
Now, you can solve this quadratic equation using factoring, completing the square, or using the quadratic formula. Once you find the values of W, you can substitute them back into the equation L = 19 - W to find the corresponding lengths.
2) Now let's move on to the second question:
Given: Canvas dimensions = 9 inches by 7 inches, Total area = 120 square inches
Let W represent the width of the frame.
The inner area covered by the canvas is given by the dimensions 9 inches by 7 inches, so its area is 9 * 7 = 63 square inches.
The total area covered by both the canvas and frame is 120 square inches.
To find the width of the frame, we can subtract the inner area from the total area:
Frame area = Total area - Inner area
Frame area = 120 - 63 = 57 square inches
Since the frame has uniform width on all sides, we can represent the width as W on each side.
The total area of the frame can be calculated by multiplying the width by the length of the frame.
Total area = 2W * (9 + 2W) + 2W * (7 + 2W)
Setting this equal to the frame area:
57 = 2W * (9 + 2W) + 2W * (7 + 2W)
Simplify and rearrange the equation:
57 = 4W^2 + 32W
Now, you can solve this quadratic equation to find the value of W, which will be the uniform width of the frame.
Once you find the value of W, you can add it to the dimensions of the canvas (9 inches by 7 inches) to find the dimensions of the entire framed picture.