Hello everyone!

I've been trying to solve this one problem the whole day, and even after all of the hints in the book I still can't solve it.

It says a ribbon is run around the box so that it makes a complete loop as shown in the diagram. The ribbon rests tightly against the box all way round. If the box is 60cm*30cm*15cm, how long will the ribbon be?

I have attached the diagram, and a picture of the ribbon rolled out into a straight line below.

I feel like I'm almost there, but I can never find the exact lenght for the second leg of the triangle.

2. Originally Posted by Coach
Hello everyone!

It says a ribbon is run around the box so that it makes a complete loop as shown in the diagram. The ribbon rests tightly against the box all way round. If the box is 60cm*30cm*15cm, how long will the ribbon be?
...
I've attached a sketch of the box. It is cut open and layed out into the plane.

Equal letters indicate the same vertex that means $B_1, B_2, B_3$ are labels of the same vertex B.

The black lines are the actual ribbon around the box. The black lines have the same length as the red one.

The length of the red line is calculated by:

$r = \sqrt{(30+15)^2+60^2} = 75$

Therefore the total length is 150

3. Originally Posted by earboth
I've attached a sketch of the box. It is cut open and layed out into the plane.

Equal letters indicate the same vertex that means $B_1, B_2, B_3$ are labels of the same vertex B.

The black lines are the actual ribbon around the box. The black lines have the same length as the red one.

The length of the red line is calculated by:

$r = \sqrt{(30+15)^2+60^2} = 75$

Therefore the total length is 150

Thank you, but according to my lesson notes the right answer is 191 cm.

Won't there be two pieces left in the corners of each of the large rectangles?

Because I thought that one could first unravel the ribbon into a single line, and then use Pythagoras to calculate the hypotenuse.

4. Originally Posted by Coach
Thank you, but according to my lesson notes the right answer is 191 cm.
A) Won't there be two pieces left in the corners of each of the large rectangles?
B) Because I thought that one could first unravel the ribbon into a single line, and then use Pythagoras to calculate the hypotenuse.
to A): The complete ribbon consists of 8 parts. I've numerated these parts in your drawing of the box and in my sketch of the flattened surface of the box.

to B) It is possible of course to stretch out the ribbon in one piece. For me it was easier to draw if I take 2 halfs of the ribbon.

I've made a mistake when I calculated the length of the legs of the right triangle. The length of the red line is:

$l = \sqrt{(30+15)^2+(60+15)^2} =\sqrt{7650}\approx 87.464..$

So to me the total length of the ribbon is 174.929 cm