Question

The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.

Answer 1

Answer:

1:7

Step-by-step explanation:

If the cone is divided in two Equal parts from height ,

The Triangle AMD and ANC are similar ( by AAA similarity)

So AM/AN = MD/NC

Or 5/10 = r/R

or R =2r

The Volume of the upper small cone of height 5cm

=1/3(πr²h)

=πr²5/3

And Volume of the frustum BCDE

=Volume of whole cone of height 10 cm - volume of small cone height 5cm

=1/3[πR²10 -πr²5]

=1/3[5π(2R² -r²)]

=5π[2(2r)² -r²]/3

=5π[8r²-r²]/3

=5π(7r²)/3

=35πr²/3

So the ratio of Volumes

=(5πr²/3)/(35πr²/3)

=5/35

=1/7

=1:7

Answer 2

Heyy mate ❤✌✌❤

Here's your Answer ......

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AQ = AP/2

with the help of similarity theory,

QD / PC = AQ / AP

so, QD / PC = 1/2

so, rdius of QD = PC/2 = R/2

now,the volume of frustum = 1/3πR²H - 1/3π(R/2)²(H/2)
                                                = 1/3πR²H*7/8
compare the two parts in cone

1st is the volume of small cone and 2nd the volume of frustum

i.e. (1/3π(R/2)²(H/2))/(1/3πR²H*7/8) = (1/8)/(7/8)=1/7.
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