Question

A rectangular box has a length of 8 feet and a width of 2 feet. The length of the three-dimensional diagonal is 10 feet. What is the height of the box?
please answer

Answer 1

Answer: $h=5.66ft$

Step-by-step explanation:

Observe the picture attached.

Find the value of "x" and "y" using the Pythagoren Theorem:

$a^2=b^2+c^2$

If you solve for "a":

$a= \sqrt{b^2+c^2}$

Where "a" is the hypotenuse and "b" and "c" are the legs.

In this case, for "x" you know that:

$a=x\\b=8ft\\c=2ft$

Then, the value of "x" is:

$x=\sqrt{(8ft)^2+(2ft)^2}\\\\x=8.24ft$

For "y" you can see that:

$a=10ft\\b=8.24ft\\c=h$

Subsituting values and solving for h, you get:

$(10ft)^2=(8.24ft)^2+h^2\\\\h=\sqrt{(10ft)^2-(8.24ft)^2} \\\\h=5.66ft$