Diagonal walk

[Denisel]

If you walk diagonally down a street, does it cut the distance in half?

 

[Steven G]

Have you heard of the pathagorian theorem?
In any right triangles a^2 + b^2 = c^2. a and b are the length of the 'streets' and c is the length of the diagonal.

So you are suggesting that (a + b)/2 = c. Does this work in the formula above?

Also note that 3^2 + 4^2 = 5^2. Is (3+4)/2 = 5??

 

[Denisel]

Have you heard of the pathagorian theorem?
In any right triangles a^2 + b^2 = c^2. a and b are the length of the 'streets' and c is the length of the diagonal.

So you are suggesting that (a + b)/2 = c. Does this work in the formula above?

Also note that 3^2 + 4^2 = 5^2. Is (3+4)/2 = 5??

7/2 is not 5. Forgive me but I'm not familiar with arrows between the 4&2 and 5&2! I'm still uncertain

 

[Cubist]

Forgive me but I'm not familiar with arrows between the 4&2 and 5&2!

4^2 is equivalent to 4² or "four squared". The word "squared" means to multiply the number by itself, and in this case 4*4=16. You can remember this because the up arrow suggests that the next number should be higher up (and smaller).

I'm still uncertain

What are you uncertain of?

The diagonal will be less distance to travel than going along and across. However, it will not be half the distance (it's a bit farther than that). It will be somewhere between. Would you like to work out the maximum distance that can be saved (with our help)?

 

[pka]

7/2 is not 5. Forgive me but I'm not familiar with arrows between the 4&2 and 5&2! I'm still uncertain

Please tell us what level (grade, year) of schooling you are doing?
Suppose that you have a rectangle [imath]ABCD[/imath] in which the length of side [imath]\overline{AD}[/imath] is six.
The length of side [imath]\overline{AB}[/imath] is two. do you understand that the diagonal [imath]\overline{AC}[/imath] has length [imath]\sqrt{6^2+2^2}=\sqrt{40}~?[/imath]
Is it the case that the length of side [imath]\overline{AC}=\dfrac{6+2}{2}~?[/imath]

[imath][/imath][imath][/imath]