60 divided by 70 and multiplied by 100 to get your answer
Assuming that there are N lanes and runners are running in the middle of each lane,
width of each lane = (1/N) [m]turning of first lane = π×(R+(1/N)×(1/2))
turning of second lane = π×(R+((1/N)×(1/2))+(1/N))
you can get the difference in turning lane simply by subtracting turning of first lane from turning of second lane.
If i worked the problem correctly, it should be π×(1/N) meter difference between each lane
Answer:
Part a) The slant height is 
Part b) The lateral area is equal to 
Step-by-step explanation:
we know that
The lateral area of a right pyramid with a regular hexagon base is equal to the area of its six triangular faces
so
![LA=6[\frac{1}{2}(b)(l)]](https://tex.z-dn.net/?f=LA%3D6%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28l%29%5D)
where
b is the length side of the hexagon
l is the slant height of the pyramid
Part a) Find the slant height l
Applying the Pythagoras Theorem

where
h is the height of the pyramid
a is the apothem
we have


substitute



Part b) Find the lateral area
![LA=6[\frac{1}{2}(b)(l)]](https://tex.z-dn.net/?f=LA%3D6%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28l%29%5D)
we have


substitute the values
![LA=6[\frac{1}{2}(6)(3\sqrt{2})]=54\sqrt{2}\ units^{2}](https://tex.z-dn.net/?f=LA%3D6%5B%5Cfrac%7B1%7D%7B2%7D%286%29%283%5Csqrt%7B2%7D%29%5D%3D54%5Csqrt%7B2%7D%5C%20units%5E%7B2%7D)
A-All whole numbers are rational numbers