A rectangle has a width of 8 meters. The length is twice as long as the width. What is the length of the diagonal? Show your work
W = 8
L = 16
x^2 = w^2 + L^2
x^2 = 8^2 + 16^2 = (2^3)^2 + (2^4)^2
x^2 = 2^6 + 2^8
x^2 = 2^6 + 4*2^6
x^2 = 5 * 2^6
x^2 = sqrt 5 * 2^3
x = 8 sqrt 5
The diagonal is c.
a^2 + b^2 = c^2
8^2 + 16^2 = c^2
64 + 256 = c^2
320 = c^2
17.889 = c
To find the length of the diagonal of a rectangle, we can use the Pythagorean theorem, which states that in a right-angle triangle, the square 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 width of the rectangle is given as 8 meters. Since the length is twice as long as the width, the length would be 2 times 8, which equals 16 meters.
Now we can construct a right-angle triangle using the width, length, and diagonal as the sides, with the diagonal as the hypotenuse. Let's label the width as 'w', the length as 'l', and the diagonal as 'd'.
According to the Pythagorean theorem, we can write the equation as:
w^2 + l^2 = d^2
Substituting the known values, we get:
8^2 + 16^2 = d^2
Simplifying:
64 + 256 = d^2
320 = d^2
To find 'd', we need to take the square root of both sides of the equation:
√320 = √d^2
√320 = d
Calculating the square root of 320, we can approximate the value to be 17.89. Therefore, the length of the diagonal is approximately 17.89 meters.