Answer:
Step-by-step explanation:
we know that
The surface area of the square pyramid is equal to the area of the square base plus the area of its four lateral triangular faces
so
![SA=b^{2}+4[\frac{1}{2}bh]](https://tex.z-dn.net/?f=SA%3Db%5E%7B2%7D%2B4%5B%5Cfrac%7B1%7D%7B2%7Dbh%5D)
where
b is the length side of the square base
h is the height of the triangular face
Remember that
Each lateral triangular face is a triangle 45°-90°-45°
so
we have
so
substitute in the formula
![SA=6^{2}+4[\frac{1}{2}(6)(3)]=72\ ft^{2}](https://tex.z-dn.net/?f=SA%3D6%5E%7B2%7D%2B4%5B%5Cfrac%7B1%7D%7B2%7D%286%29%283%29%5D%3D72%5C%20ft%5E%7B2%7D)
Answer:
10 and 4
Step-by-step explanation:
2/5 of 10 is 4 so that is it
For 12, you would do 90=4x+42, and 48=4x, and x=12
for 15, you would do 90=8x+66, and 24=8x, and x=3
for 18, you would do 90=3x+57, and 33=3x, and x=11
Answer:
about 8.2 cm
Step-by-step explanation:
The key is to realize that the volume never changes.
The formula for the volume of a sphere is (4/3)*pi*radius^3
22/7 is being substituted in place of pi in this problem
The volume of one of the small spheres is
(4/3)*(22/7)*2^3 is about 33.524 cm^3
64 of those spheres would have a volume of 33.524*64, or about 2145 cm^3
Now, the problem is to find a sphere with a volume of 2145 cm^3
Volume = (4/3)*(22/7)*radius^3
plug in and solve
2145 = 4.1905*r^3
r^3=511.87
r is about equal to 8.2 centimeters
