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.
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?