Mixing
A simple problem from 3d math primer book
Clash Royale CLAN TAG#URR8PPP
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The question
A man is boarding a plane. The airline has a rule that no carry-on item may
be more than two feet long, two feet wide, or two feet tall. He has a very
valuable sword that is three feet long, yet he is able to carry the sword
on board with him.9 How is he able to do this? What is the longest possible
item that he could carry on?
And the answer is
The man buys a box or has a piece of luggage that is 2 feet long, 2 feet
wide, and 2 feet tall. If the object is very thin, such as a sword, then he
can put the object diagonally in the box or luggage. The longest such
object he could carry on is
$sqrt2^2 + 2^2 + 2^2$ ≈ 3.46 feet
What i don't understand is how the maximum object length is 3.46 feet. we can imagine a box then length of diagonal by Pythagorus theorem is $sqrt2^2 + 2^2$ ≈ 2.8284271 feet. why did the author took extra $2^2$.
geometry
asked Aug 3 at 6:18
Atul
1106
add a comment |Â
up vote
0
down vote
favorite
The question
A man is boarding a plane. The airline has a rule that no carry-on item may
be more than two feet long, two feet wide, or two feet tall. He has a very
valuable sword that is three feet long, yet he is able to carry the sword
on board with him.9 How is he able to do this? What is the longest possible
item that he could carry on?
And the answer is
The man buys a box or has a piece of luggage that is 2 feet long, 2 feet
wide, and 2 feet tall. If the object is very thin, such as a sword, then he
can put the object diagonally in the box or luggage. The longest such
object he could carry on is
$sqrt2^2 + 2^2 + 2^2$ ≈ 3.46 feet
What i don't understand is how the maximum object length is 3.46 feet. we can imagine a box then length of diagonal by Pythagorus theorem is $sqrt2^2 + 2^2$ ≈ 2.8284271 feet. why did the author took extra $2^2$.
geometry
asked Aug 3 at 6:18
Atul
1106
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question
A man is boarding a plane. The airline has a rule that no carry-on item may
be more than two feet long, two feet wide, or two feet tall. He has a very
valuable sword that is three feet long, yet he is able to carry the sword
on board with him.9 How is he able to do this? What is the longest possible
item that he could carry on?
And the answer is
The man buys a box or has a piece of luggage that is 2 feet long, 2 feet
wide, and 2 feet tall. If the object is very thin, such as a sword, then he
can put the object diagonally in the box or luggage. The longest such
object he could carry on is
$sqrt2^2 + 2^2 + 2^2$ ≈ 3.46 feet
What i don't understand is how the maximum object length is 3.46 feet. we can imagine a box then length of diagonal by Pythagorus theorem is $sqrt2^2 + 2^2$ ≈ 2.8284271 feet. why did the author took extra $2^2$.
geometry
asked Aug 3 at 6:18
Atul
1106
The question
A man is boarding a plane. The airline has a rule that no carry-on item may
be more than two feet long, two feet wide, or two feet tall. He has a very
valuable sword that is three feet long, yet he is able to carry the sword
on board with him.9 How is he able to do this? What is the longest possible
item that he could carry on?
And the answer is
The man buys a box or has a piece of luggage that is 2 feet long, 2 feet
wide, and 2 feet tall. If the object is very thin, such as a sword, then he
can put the object diagonally in the box or luggage. The longest such
object he could carry on is
$sqrt2^2 + 2^2 + 2^2$ ≈ 3.46 feet
What i don't understand is how the maximum object length is 3.46 feet. we can imagine a box then length of diagonal by Pythagorus theorem is $sqrt2^2 + 2^2$ ≈ 2.8284271 feet. why did the author took extra $2^2$.
geometry
asked Aug 3 at 6:18
Atul
1106
asked Aug 3 at 6:18
Atul
1106
asked Aug 3 at 6:18
Atul
1106
asked Aug 3 at 6:18
Atul
1106
1106
add a comment |Â
add a comment |Â
1 Answer
1
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1
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accepted
You are only thinking 2D, not 3D.
That $2.82$ number is the length of the diagonal along the box's bottom. Run along that diagonal and then up two feet along one of the boxes vertical corner edges to make another right triangle whose hypotenuse length is $$sqrt(2.82ldots)^2+2^2=3.46ldots$$ Since the $2.82ldots$ came from $sqrt2^2+2^2$, it's the same as $$sqrt2^2+2^2+2^2=3.46ldots$$
answered Aug 3 at 6:22
alex.jordan
36.9k459117
1
so one side of sword is touching the bottom corner of the box and other side is touching the opposite top corner of the box. right.
– Atul
Aug 3 at 6:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are only thinking 2D, not 3D.
That $2.82$ number is the length of the diagonal along the box's bottom. Run along that diagonal and then up two feet along one of the boxes vertical corner edges to make another right triangle whose hypotenuse length is $$sqrt(2.82ldots)^2+2^2=3.46ldots$$ Since the $2.82ldots$ came from $sqrt2^2+2^2$, it's the same as $$sqrt2^2+2^2+2^2=3.46ldots$$
answered Aug 3 at 6:22
alex.jordan
36.9k459117
1
so one side of sword is touching the bottom corner of the box and other side is touching the opposite top corner of the box. right.
– Atul
Aug 3 at 6:27
add a comment |Â
up vote
1
down vote
accepted
You are only thinking 2D, not 3D.
That $2.82$ number is the length of the diagonal along the box's bottom. Run along that diagonal and then up two feet along one of the boxes vertical corner edges to make another right triangle whose hypotenuse length is $$sqrt(2.82ldots)^2+2^2=3.46ldots$$ Since the $2.82ldots$ came from $sqrt2^2+2^2$, it's the same as $$sqrt2^2+2^2+2^2=3.46ldots$$
answered Aug 3 at 6:22
alex.jordan
36.9k459117
1
so one side of sword is touching the bottom corner of the box and other side is touching the opposite top corner of the box. right.
– Atul
Aug 3 at 6:27
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are only thinking 2D, not 3D.
That $2.82$ number is the length of the diagonal along the box's bottom. Run along that diagonal and then up two feet along one of the boxes vertical corner edges to make another right triangle whose hypotenuse length is $$sqrt(2.82ldots)^2+2^2=3.46ldots$$ Since the $2.82ldots$ came from $sqrt2^2+2^2$, it's the same as $$sqrt2^2+2^2+2^2=3.46ldots$$
answered Aug 3 at 6:22
alex.jordan
36.9k459117
You are only thinking 2D, not 3D.
That $2.82$ number is the length of the diagonal along the box's bottom. Run along that diagonal and then up two feet along one of the boxes vertical corner edges to make another right triangle whose hypotenuse length is $$sqrt(2.82ldots)^2+2^2=3.46ldots$$ Since the $2.82ldots$ came from $sqrt2^2+2^2$, it's the same as $$sqrt2^2+2^2+2^2=3.46ldots$$
answered Aug 3 at 6:22
alex.jordan
36.9k459117
answered Aug 3 at 6:22
alex.jordan
36.9k459117
answered Aug 3 at 6:22
alex.jordan
36.9k459117
answered Aug 3 at 6:22
alex.jordan
36.9k459117
36.9k459117
1
so one side of sword is touching the bottom corner of the box and other side is touching the opposite top corner of the box. right.
– Atul
Aug 3 at 6:27
add a comment |Â
1
so one side of sword is touching the bottom corner of the box and other side is touching the opposite top corner of the box. right.
– Atul
Aug 3 at 6:27
1
1
so one side of sword is touching the bottom corner of the box and other side is touching the opposite top corner of the box. right.
– Atul
Aug 3 at 6:27
so one side of sword is touching the bottom corner of the box and other side is touching the opposite top corner of the box. right.
– Atul
Aug 3 at 6:27
add a comment |Â
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