The length of a rectangle is 2 inches more than the width. If the diagonal is 20−−√ inches, find the area of the rectangle.
Responses
Let's call the width of the rectangle x inches.
According to the problem, the length of the rectangle is 2 inches more than the width. Therefore, the length of the rectangle is x + 2 inches.
We know that the diagonal of the rectangle forms a right triangle with the width and length of the rectangle. Using the Pythagorean theorem, we can set up the following equation:
x^2 + (x + 2)^2 = (20 - √2)^2
Expanding and simplifying the equation, we get:
x^2 + x^2 + 4x + 4 = 400 - 40√2 + 2
Combining like terms, we have:
2x^2 + 4x + 2 = 400 - 40√2
Rearranging the equation, we get:
2x^2 + 4x - 398 = 40√2
Dividing both sides of the equation by 2, we have:
x^2 + 2x - 199 = 20√2
At this point, we can solve for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values from our equation, we get:
x = (-2 ± √(2^2 - 4*1*(-199))) / (2*1)
Simplifying the expression under the square root, we have:
x = (-2 ± √(4 + 796)) / 2
x = (-2 ± √800) / 2
x = (-2 ± 20√2) / 2
Simplifying further, we have:
x = -1 ± 10√2
Since the width of a rectangle cannot be negative, we discard the negative solution and only consider the positive solution:
x = -1 + 10√2
Now that we have the width of the rectangle, we can calculate the length by adding 2 inches to the width:
Length = x + 2
Length = -1 + 10√2 + 2
Length = 1 + 10√2
To find the area of the rectangle, we multiply the width by the length:
Area = width * length
Area = (-1 + 10√2)(1 + 10√2)
Area = -1 + 10√2 + 10√2 - 100
Area = -101 + 20√2
Therefore, the area of the rectangle is -101 + 20√2 square inches.
Step 1: Let's assume the width of the rectangle is "x" inches.
Step 2: According to the given information, the length of the rectangle is 2 inches more than the width, so the length would be (x + 2) inches.
Step 3: We can use the Pythagorean theorem to find the relationship between the length, width, and diagonal of the rectangle. According to the theorem:
Diagonal^2 = Length^2 + Width^2
(20 - √ inches)^2 = (x + 2)^2 + x^2
Step 4: Simplify the equation by expanding and combining like terms:
400 - 40√ inches + 1 = x^2 + 4x + 4 + x^2
2x^2 + 4x - 403 + 40√ inches = 0
Step 5: To find the value of "x", we can use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / 2a
In this equation, a = 2, b = 4, and c = -403 + 40√ inches.
Step 6: Using the quadratic formula, we can find the two possible values of "x".
x = [-4 ± √(4^2 - 4(2)(-403 + 40√)] / 2(2)
x = [-4 ± √(16 + 3212 - 320√)] / 4
x = [-4 ± √(3228 - 320√)] / 4
Step 7: We can simplify the expression further by factoring out 4 from the numerator:
x = [ -4 ± 2√(807 - 80√) ] / 4
x = [ -1 ± √(807 - 80√) ] / 2
Since the width of a rectangle can't be negative, we ignore the negative solution.
Step 8: Simplify the expression for "x":
x = [ -1 + √(807 - 80√) ] / 2
Step 9: To find the area of the rectangle, we multiply the length and width:
Area = Length * Width
Area = (x+2) * x
Area = ( [ -1 + √(807 - 80√) ] / 2 + 2) * ( [ -1 + √(807 - 80√) ] / 2 )
Thus, the area of the rectangle is ( [ -1 + √(807 - 80√) ] / 2 + 2) * ( [ -1 + √(807 - 80√) ] / 2 ).