Thread: Insect

A L by B by H cm box is lying on the floor on one of its B by H faces. An insect, located at one of the bottom corners of the box, must crawl along the outside of the box to reach the opposite bottom corner. What is the length of the shortest such path?

Find the diagonal from one corner to another corner (because it the shortest path inside the box) and use pythagoras theorem to get the answer a^2 + b^2 = c^2... if u give some values for L, B and H then i will give u the answer...

Thanks
Manoj

a 12 by 25 by 36 box
and its lying on 25 by 36 side face

mind u manoj!!
its an insect...and insect cannot crawl in air!!
it has to move on the surface of box...and it cannot move under the box!!

Yeah i know Smart Coder the shortest distance over the box for these values is

Square Root((25+12)^2 + (36+12)^2) ...

Answer is Square Root(3673)...

Thanks
Manoj

Answer is correct but gimme a detailed explantion..

Flatten the box and join all the possible opposite corner of the box diagonally then apply the pythagoras theorem to find the distance from outside the box....

These kind of problem we use to do in school days... But pretty difficult to explain... need more 3D thinking...

answer is wrong i suppose!!

correct answer is:
12 + square root( 36^2 + 12^2) = 55. sumthing

Sorry Smart Coder i think u r wrong....

Work out ur values... I worked out it and ur values are crossing only one plane...

Thanks
Manoj

I want ur detailed solution Manoj!!!
Coz i m not able to think d way ur thinkin..also tell me wtz flaw in my solution?

Kindly gimme a detailed solution for ur answer manoj!!

I will give u the detailed solution tomorrow... Once i go home i will draw one small image and explain u...

Thanks
Manoj

that wud b fine!!!