Answer
Verified
451.2k+ views
Hint – In this question first draw the pictorial representation of the problem it will give us a clear picture of what we have to find out later on apply the property of similar triangles and calculate the base radius of smaller cones, so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Let us consider the cone as shown in figure the base radius (R) of the cone is 10 cm (see figure.)
Now the cone cut into two parts through the midpoint of its height (at point B see figure).
Let the height of the cone be h cm.
$ \Rightarrow AB = BC = \dfrac{h}{2}$ cm.
Let the base radius of the smaller cone is r cm.
$ \Rightarrow BE = r$ cm (see figure).
Now the triangle ABE and the triangle ACD is concurrent by the property of angle-angle-angle (AAA).
$ \Rightarrow \dfrac{{AB}}{{AC}} = \dfrac{{BE}}{{CD}}$
Now substitute the values in above equation we have,
$ \Rightarrow \dfrac{{\dfrac{h}{2}}}{h} = \dfrac{r}{{10}}$
On simplifying we get.
$ \Rightarrow r = 5$ cm.
As we know that the volume of the cone is $ = \dfrac{1}{3}\pi {\left( {{\text{radius}}} \right)^2}\left( {{\text{height}}} \right)$
So the volume (${V_1}$) of the upper part of the bigger cone which is also a cone is
$ \Rightarrow {V_1} = \dfrac{1}{3}\pi {\left( {BE} \right)^2}\left( {AB} \right) = \dfrac{1}{3}\pi {\left( r \right)^2}\left( {\dfrac{h}{2}} \right)$…………………. (1)
Let the volume of the lower part is (${V_2}$) so the volume of the lower part is calculated as subtraction of volume of bigger cone and volume of smaller cone.
Volume (V) of bigger cone is $ = \dfrac{1}{3}\pi {\left( {{\text{radius}}} \right)^2}\left( {{\text{height}}} \right)$
$ \Rightarrow V = \dfrac{1}{3}\pi {\left( R \right)^2}h$
So the lower part volume is
$ \Rightarrow {V_2} = V - {V_1}$
$ \Rightarrow {V_2} = \dfrac{1}{3}\pi {\left( R \right)^2}h - \dfrac{1}{3}\pi {\left( r \right)^2}\left( {\dfrac{h}{2}} \right)$
So, the required ratio (RT) of two parts of the cone is
$ \Rightarrow RT = \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{\dfrac{1}{3}\pi {{\left( r \right)}^2}\left( {\dfrac{h}{2}} \right)}}{{\dfrac{1}{3}\pi {{\left( R \right)}^2}h - \dfrac{1}{3}\pi {{\left( r \right)}^2}\left( {\dfrac{h}{2}} \right)}}$
Now substitute the values in the above equation we have,
$ \Rightarrow RT = \dfrac{{\dfrac{1}{3}\pi {{\left( 5 \right)}^2}\left( {\dfrac{h}{2}} \right)}}{{\dfrac{1}{3}\pi {{\left( {10} \right)}^2}h - \dfrac{1}{3}\pi {{\left( 5 \right)}^2}\left( {\dfrac{h}{2}} \right)}}$
Now cancel out the common term $\dfrac{1}{3}\pi \left( {\dfrac{h}{2}} \right)$ from, numerator and denominator and simplify we have,
$ \Rightarrow RT = \dfrac{{{{\left( 5 \right)}^2}}}{{\left[ {2{{\left( {10} \right)}^2} - {5^2}} \right]}} = \dfrac{{25}}{{\left( {200 - 25} \right)}} = \dfrac{{25}}{{175}} = \dfrac{1}{7}$
So, this is the required ratio of the two parts of the cone.
So, this is the required answer.
Note – In such types of questions always remember the formula of cone which is stated above then first of all using the property of similar triangles calculate the value of base radius of smaller cone as above and the cone is cut from the midpoint so the height of the smaller cone is equal to the height of lower part of the bigger solid cone which is equal to half of the height of bigger solid cone, then take the ratio of volume of smaller cone to the volume of lower part of the big solid cone as above and simplify which is the required ratio in the volumes of two parts of the cone.
Complete step-by-step answer:
Let us consider the cone as shown in figure the base radius (R) of the cone is 10 cm (see figure.)
Now the cone cut into two parts through the midpoint of its height (at point B see figure).
Let the height of the cone be h cm.
$ \Rightarrow AB = BC = \dfrac{h}{2}$ cm.
Let the base radius of the smaller cone is r cm.
$ \Rightarrow BE = r$ cm (see figure).
Now the triangle ABE and the triangle ACD is concurrent by the property of angle-angle-angle (AAA).
$ \Rightarrow \dfrac{{AB}}{{AC}} = \dfrac{{BE}}{{CD}}$
Now substitute the values in above equation we have,
$ \Rightarrow \dfrac{{\dfrac{h}{2}}}{h} = \dfrac{r}{{10}}$
On simplifying we get.
$ \Rightarrow r = 5$ cm.
As we know that the volume of the cone is $ = \dfrac{1}{3}\pi {\left( {{\text{radius}}} \right)^2}\left( {{\text{height}}} \right)$
So the volume (${V_1}$) of the upper part of the bigger cone which is also a cone is
$ \Rightarrow {V_1} = \dfrac{1}{3}\pi {\left( {BE} \right)^2}\left( {AB} \right) = \dfrac{1}{3}\pi {\left( r \right)^2}\left( {\dfrac{h}{2}} \right)$…………………. (1)
Let the volume of the lower part is (${V_2}$) so the volume of the lower part is calculated as subtraction of volume of bigger cone and volume of smaller cone.
Volume (V) of bigger cone is $ = \dfrac{1}{3}\pi {\left( {{\text{radius}}} \right)^2}\left( {{\text{height}}} \right)$
$ \Rightarrow V = \dfrac{1}{3}\pi {\left( R \right)^2}h$
So the lower part volume is
$ \Rightarrow {V_2} = V - {V_1}$
$ \Rightarrow {V_2} = \dfrac{1}{3}\pi {\left( R \right)^2}h - \dfrac{1}{3}\pi {\left( r \right)^2}\left( {\dfrac{h}{2}} \right)$
So, the required ratio (RT) of two parts of the cone is
$ \Rightarrow RT = \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{\dfrac{1}{3}\pi {{\left( r \right)}^2}\left( {\dfrac{h}{2}} \right)}}{{\dfrac{1}{3}\pi {{\left( R \right)}^2}h - \dfrac{1}{3}\pi {{\left( r \right)}^2}\left( {\dfrac{h}{2}} \right)}}$
Now substitute the values in the above equation we have,
$ \Rightarrow RT = \dfrac{{\dfrac{1}{3}\pi {{\left( 5 \right)}^2}\left( {\dfrac{h}{2}} \right)}}{{\dfrac{1}{3}\pi {{\left( {10} \right)}^2}h - \dfrac{1}{3}\pi {{\left( 5 \right)}^2}\left( {\dfrac{h}{2}} \right)}}$
Now cancel out the common term $\dfrac{1}{3}\pi \left( {\dfrac{h}{2}} \right)$ from, numerator and denominator and simplify we have,
$ \Rightarrow RT = \dfrac{{{{\left( 5 \right)}^2}}}{{\left[ {2{{\left( {10} \right)}^2} - {5^2}} \right]}} = \dfrac{{25}}{{\left( {200 - 25} \right)}} = \dfrac{{25}}{{175}} = \dfrac{1}{7}$
So, this is the required ratio of the two parts of the cone.
So, this is the required answer.
Note – In such types of questions always remember the formula of cone which is stated above then first of all using the property of similar triangles calculate the value of base radius of smaller cone as above and the cone is cut from the midpoint so the height of the smaller cone is equal to the height of lower part of the bigger solid cone which is equal to half of the height of bigger solid cone, then take the ratio of volume of smaller cone to the volume of lower part of the big solid cone as above and simplify which is the required ratio in the volumes of two parts of the cone.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Why Are Noble Gases NonReactive class 11 chemistry CBSE
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE