You would first use
on the bottom of the box rectangle to
find the diagonal CB =
Then use CB as of the triangle ACB, with AB being the
AB is approximately =
To find the diagonal of the rectangle, you would use the Pythagorean theorem. Let's denote the sides of the rectangle as follows:
AC = 8 (length of the rectangle)
BC = 6 (width of the rectangle)
Using the Pythagorean theorem:
CB^2 = AC^2 + BC^2
CB^2 = 8^2 + 6^2
CB^2 = 64 + 36
CB^2 = 100
CB = 10
So, the diagonal CB of the rectangle is 10 units.
Now, in triangle ACB, using the Pythagorean theorem again, we can find the length of AB (which is the hypotenuse of the triangle):
AB^2 = AC^2 + BC^2
AB^2 = 8^2 + 6^2
AB^2 = 64 + 36
AB^2 = 100
AB = 10
So, the length of AB in the triangle ACB is also 10 units.