Suppose you want to draw a rectangle where the width is 7 inches less than the length and the diagonal is 7 inches longer than the length. What are the dimensions of the rectangle?
if the width is w, then the length is w+7, so the diagonal is w+14:
w^2 + (w+7)^2 = (w+14)^2
right triangle with hypotenuse = diagonal
x^2 + (x-7)^2 = (x+7)^2
x^2 + x^2 - 14 x + 49 = x^2 +14 x +
49
x^2 -28 x = 0 = x(x-28)
x = 0 or x = 28
so
28 by 21
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try that length =
diagonal^2 = 28^2 + 21^2
= 784 + 411 = 1225
so diagonal = sqrt(1225) = 35
yes, that works
Let's break down the problem step by step:
Step 1: Let's assume the length of the rectangle as "x" inches.
Step 2: According to the problem, the width is 7 inches less than the length. So, the width of the rectangle would be (x - 7) inches.
Step 3: Now, let's find the length of the diagonal. The diagonal of a rectangle can be found using the Pythagorean theorem formula, which states that the square of the length plus the square of the width is equal to the square of the diagonal.
Using the Pythagorean theorem, we can write the equation as:
x^2 + (x - 7)^2 = (x + 7)^2
Step 4: Now, let's solve this equation to find the value of "x":
Expanding the equation:
x^2 + (x^2 - 14x + 49) = (x^2 + 14x + 49)
Combining like terms:
2x^2 - 14x + 49 = x^2 + 14x + 49
Subtracting x^2 + 14x + 49 from both sides:
2x^2 - x^2 - 14x - 14x = 0
Simplifying:
x^2 - 28x = 0
Factoring out an "x":
x(x - 28) = 0
Set each factor equal to zero:
x = 0 or x - 28 = 0
Ignoring the x = 0 solution as it is not practical in this context, we have:
x - 28 = 0
x = 28
Hence, the length of the rectangle is 28 inches.
Step 5: Now, let's calculate the width of the rectangle using the length we found:
Width = Length - 7 = 28 - 7 = 21 inches.
So, the dimensions of the rectangle are:
Length = 28 inches
Width = 21 inches.
To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the length and width of the rectangle form the two sides of a right triangle, while the diagonal of the rectangle is the hypotenuse. We can set up the equation as follows:
(length)^2 + (width)^2 = (diagonal)^2
Let's denote the length of the rectangle as "L" and the width as "W". According to the given information, the width is 7 inches less than the length, so we can write:
W = L - 7
Also, the diagonal is 7 inches longer than the length, so we have:
diagonal = L + 7
Substituting these values into the Pythagorean theorem equation, we have:
L^2 + (L - 7)^2 = (L + 7)^2
Expanding and simplifying this equation will help us find the values of L and W.
L^2 + (L^2 - 14L + 49) = L^2 + 14L + 49
Combining like terms:
2L^2 - 14L + 49 = L^2 + 14L + 49
Subtracting L^2 and 49 from both sides of the equation:
L^2 - 14L = 14L
Adding 14L to both sides:
L^2 = 28L
Dividing both sides by L:
L = 28
Now that we know the value of L, we can substitute it into the equation we previously obtained for W:
W = L - 7
W = 28 - 7
W = 21
Therefore, the dimensions of the rectangle are length = 28 inches and width = 21 inches.