SOLUTION: We want to place the largest rod we can in a box. To figure this out we will need to do
the Pythagorean Theorem two times. If the box is 4 x 3 x 2, find the longest rod that wo
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-> SOLUTION: We want to place the largest rod we can in a box. To figure this out we will need to do
the Pythagorean Theorem two times. If the box is 4 x 3 x 2, find the longest rod that wo
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Question 1067982: We want to place the largest rod we can in a box. To figure this out we will need to do
the Pythagorean Theorem two times. If the box is 4 x 3 x 2, find the longest rod that would fit in the box.
You can put this solution on YOUR website! let the length of the box be 4 units long.
let the width of the box be 3 units long.
leg the height of the box be 2 units long.
find the diagonal of the length and the width.
that's your first right triangle.
the length is 4 and the width is 3.
the hypotenuse of the right triangle is the diagonal.
sqrt(4^2 + 3^2) = sqrt(16 + 9) = sqrt(25) = 5
that's the length of the diagonal of the bottom base.
you now need to find the length of the diagonal using the diagonal of the base and the height as the legs.
this forms another right triangle with the legs of 5 and 2 units long.
the hypotenuse is equal to sqrt(5^2 + 2^2) = sqrt(25 + 4) = sqrt(29).
that's your longest diagonal, whose length is sqrt(29) which is approximately equal to 5.3852 units long rounded to 4 decimal places.
You can put this solution on YOUR website!
We want to place the largest rod we can in a box. To figure this out we will need to do
the Pythagorean Theorem two times. If the box is 4 x 3 x 2, find the longest rod that would fit in the box.
The longest rod represents the HYPOTENUSE of the triangle formed by the DIAGONAL of the base of the box and the height of the box.
Let's name that diagonal AB
We then get:
AB, or length of longest rod =
