Answer:
Part a) The slant height is ![3\sqrt{2}\ units](https://tex.z-dn.net/?f=3%5Csqrt%7B2%7D%5C%20units)
Part b) The lateral area is equal to ![54\sqrt{2}\ units^{2}](https://tex.z-dn.net/?f=54%5Csqrt%7B2%7D%5C%20units%5E%7B2%7D)
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
![l^{2}=h^{2} +a^{2}](https://tex.z-dn.net/?f=l%5E%7B2%7D%3Dh%5E%7B2%7D%20%2Ba%5E%7B2%7D)
where
h is the height of the pyramid
a is the apothem
we have
![h=3\ units](https://tex.z-dn.net/?f=h%3D3%5C%20units)
![a=3\ units](https://tex.z-dn.net/?f=a%3D3%5C%20units)
substitute
![l^{2}=3^{2} +3^{2}](https://tex.z-dn.net/?f=l%5E%7B2%7D%3D3%5E%7B2%7D%20%2B3%5E%7B2%7D)
![l^{2}=18](https://tex.z-dn.net/?f=l%5E%7B2%7D%3D18)
![l=3\sqrt{2}\ units](https://tex.z-dn.net/?f=l%3D3%5Csqrt%7B2%7D%5C%20units)
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
![b=6\ units](https://tex.z-dn.net/?f=b%3D6%5C%20units)
![l=3\sqrt{2}\ units](https://tex.z-dn.net/?f=l%3D3%5Csqrt%7B2%7D%5C%20units)
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)
Answer:
$0.22/inch
Step-by-step explanation:
6ft=72 inches
15.84/72
=0.22
Answer:
6 and 9
Step-by-step explanation:
Just took it
Answer:
1) c=100
2) x^4-40x^3+400x^2
Step-by-step explanation:
When I write ^# that means it's squared and has a number above it :)
You're welcome :)
A. 8
B. 10
C. 3,000
D. 1,440
E. 16