A standard six-sided die is thrown. What is the probability that the number
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Answer:
Volume of the frustum = ⅓πh(4R² - r²)
Step-by-step explanation:
We are to determine the volume of the frustum.
Find attached the diagram obtained from the given information.
Let height of small cone = h
height of the large cone = H
The height of a small cone is a quarter of the height of the large cone:
h = ¼×H
H = 4h
Volume of the frustum = volume of the large cone - volume of small cone
volume of the large cone = ⅓πR²H
= ⅓πR²(4h) = 4/3 ×π×R²h
volume of small cone = ⅓πr²h
Volume of the frustum = 4/3 ×π×R²h - ⅓πr²h
Volume of the frustum = ⅓(4π×R²h - πr²h)
Volume of the frustum = ⅓πh(4R² - r²)
<u>Solu</u><u>tion</u><u> </u><u>:</u><u>-</u>
Given ratio to us is 60 : 48 and we are required to convert it into simplest form .
For this first find the HCF of 60 and 48 .
= 48 ) 60 ( 1
- 48
__________
12 ) 48 ( 4
- 48
__________
0
Hence the HCF ( 60,48 ) is 12 .
Now divide the given ratio by 12 .
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