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An insect is located at one corner (point A) on the surface of a cube

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An insect is located at one corner (point A) on the surface of a cube  [#permalink]

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New post 28 Oct 2016, 12:57
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An insect is located at one corner (point A) on the surface of a cube that measures 3 x 4 x 5 inches, as shown in the diagram.
Image
Note: Figure is not drawn to scale.

If the insect crawls along the surface of the cube to the opposite corner (point B), what is the shortest possible length, in inches, of the insect’s path from point A to point B?

(A)5\(\sqrt{2}\)
(B)\(\sqrt{74}\)
(C)4\(\sqrt{5}\)
(D)3\(\sqrt{10}\)
(E)10
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Re: An insect is located at one corner (point A) on the surface of a cube  [#permalink]

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New post 28 Oct 2016, 16:33
The shortest distance between plane A and B is 5 inches.

The shortest distance from point A to the point where shortest distance from point B meets plane A = sqrt(3^2+4^2)

= 5 inches

Total = 10 inches.

IMO E


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An insect is located at one corner (point A) on the surface of a cube  [#permalink]

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New post 28 Oct 2016, 18:07
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AbdurRakib wrote:
An insect is located at one corner (point A) on the surface of a cube that measures 3 x 4 x 5 inches, as shown in the diagram.
Image
Note: Figure is not drawn to scale.

If the insect crawls along the surface of the cube to the opposite corner (point B), what is the shortest possible length, in inches, of the insect’s path from point A to point B?

(A)5\(\sqrt{2}\)
(B)\(\sqrt{74}\)
(C)4\(\sqrt{5}\)
(D)3\(\sqrt{10}\)
(E)10


Hi

Two points..
1) A cube is supposed to have same dimensions so it will be 3*3*3 or 4*4*4... here it is cuboid..
2) Now the answer..
a) if it can fly and is inside the box, it will be DIAGONAL.
b) but here it is crawling, so open the two rectangle faces adjacent.. these faces are 3*5 and 4*5..
So when you open it, it becomes rectangle with sides 3+4 and 5..
So the hypotenuse of this triangle will be our ANSWER and it is √(7^2+5^2)=√(49+25)=√74
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Re: An insect is located at one corner (point A) on the surface of a cube  [#permalink]

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New post 29 Oct 2016, 00:01
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AbdurRakib wrote:
An insect is located at one corner (point A) on the surface of a cube that measures 3 x 4 x 5 inches, as shown in the diagram.
Image
Note: Figure is not drawn to scale.

If the insect crawls along the surface of the cube to the opposite corner (point B), what is the shortest possible length, in inches, of the insect’s path from point A to point B?

(A)5\(\sqrt{2}\)
(B)\(\sqrt{74}\)
(C)4\(\sqrt{5}\)
(D)3\(\sqrt{10}\)
(E)10


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Hope it helps.
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An insect is located at one corner (point A) on the surface of a cube  [#permalink]

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New post 22 Jun 2017, 00:30
chetan2u wrote:
AbdurRakib wrote:
An insect is located at one corner (point A) on the surface of a cube that measures 3 x 4 x 5 inches, as shown in the diagram.
Image
Note: Figure is not drawn to scale.

If the insect crawls along the surface of the cube to the opposite corner (point B), what is the shortest possible length, in inches, of the insect’s path from point A to point B?

(A)5\(\sqrt{2}\)
(B)\(\sqrt{74}\)
(C)4\(\sqrt{5}\)
(D)3\(\sqrt{10}\)
(E)10


Hi

Two points..
1) A cube is supposed to have same dimensions so it will be 3*3*3 or 4*4*4... here it is cuboid..
2) Now the answer..
a) if it can fly and is inside the box, it will be DIAGONAL.
b) but here it is crawling, so open the two rectangle faces adjacent.. these faces are 3*5 and 4*5..
So when you open it, it becomes rectangle with sides 3+4 and 5..
So the hypotenuse of this triangle will be our ANSWER and it is √(7^2+5^2)=√(49+25)=√74


Hi chetan2u,

As it is a cuboid, I tried to solve in below way, but it is incorrect

first sqrt(3^2+4^2)=sqrt(9+16)=5 --- find the diagonal that will work the edge for second triangle
then diagonal it needs to travel = sqrt(25+25)=5*sqrt(2)

However, if we solve in the following way, it is correct

sqrt((3+4)^2 +5^2)

Why is first approach incorrect?
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Re: An insect is located at one corner (point A) on the surface of a cube  [#permalink]

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New post 22 Jun 2017, 02:33
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As attached in the picture..

Questions like these are best done by opening up the cube/cuboid (which makes a rectangle), and then applying pythagoras theorem.

Shortest path for the insect will pass from A, to somewhere along the left edge of the given cuboid, and then to B.

Or we could also go from point A to somewhere along right edge of given cuboid, and then to B. Answer will be same.

Hence B answer
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An insect is located at one corner (point A) on the surface of a cube  [#permalink]

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New post 22 Jun 2017, 05:01
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AR15J wrote:
chetan2u wrote:
AbdurRakib wrote:
An insect is located at one corner (point A) on the surface of a cube that measures 3 x 4 x 5 inches, as shown in the diagram.
Image
Note: Figure is not drawn to scale.

If the insect crawls along the surface of the cube to the opposite corner (point B), what is the shortest possible length, in inches, of the insect’s path from point A to point B?

(A)5\(\sqrt{2}\)
(B)\(\sqrt{74}\)
(C)4\(\sqrt{5}\)
(D)3\(\sqrt{10}\)
(E)10


Hi

Two points..
1) A cube is supposed to have same dimensions so it will be 3*3*3 or 4*4*4... here it is cuboid..
2) answer as mentioned above will be 10...
It has to travel on diagonal of rectangle of size 3*4 and then along length/edge of size 5..



Hi chetan2u,

As it is a cuboid, I tried to solve in below way, but it is incorrect

first sqrt(3^2+4^2)=sqrt(9+16)=5 --- find the diagonal that will work the edge for second triangle
then diagonal it needs to travel = sqrt(25+25)=5*sqrt(2)

However, if we solve in the following way, it is correct

sqrt((3+4)^2 +5^2)

Why is first approach incorrect?



Hi..
In first case after you take √(3^2+4^2)=5, you get a diagonal of the BASE of the given cuboid.
By taking the √(5^2+5^2), you are finding the DIAGONAL of the cuboid and this would be correct if the insect is flying BUT the insect is moving along the surface.
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An insect is located at one corner (point A) on the surface of a cube   [#permalink] 22 Jun 2017, 05:01

An insect is located at one corner (point A) on the surface of a cube

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