Two similar cones have volumes \[12\pi \] cu. units and \[96\pi \] cu. units. If the curved surface area of the smaller cone is \[15\pi \] sq. units, then what is the curved surface area of the larger cone?
Answer
Verified467.1k+ views
Hint: Here, we need to find the curved surface area of the larger cone. We will use the fact that the ratio of the heights and radii of two similar cones is equal. We will write the radius and height of one cone in terms of the radius and height of the other cone. Then, we will use the given information, formula for volume, formula for slant heights, and finally, the formula for the curved surface area of a cone to find the required area.
Formula used:
We will use the following formulas:
1.The volume of a cone is given by the formula \[\dfrac{1}{3}\pi {r^2}h\], where \[r\] is the radius of the base of the cone, and \[h\] is the height of the cone.
2.The slant height of a cone is given by the formula \[l = \sqrt {{r^2} + {h^2}} \], where \[r\] is the radius of the base of the cone, and \[h\] is the height of the cone.
3.The curved surface area of a cone is given by the formula \[\pi rl\], where \[r\] is the radius of the base of the cone, and \[l\] is the slant height of the cone.
Complete step-by-step answer:
Let the height of the smaller and larger cone be \[{h_1}\] and \[{h_2}\] units respectively.
Let the radius of the smaller and larger cone be \[{r_1}\] and \[{r_2}\] units respectively.
So, now we will draw the diagrams of the cone.
It is given that the two cones are similar. This means that the ratios of the dimensions of the cones are equal.
Thus, the ratio of the heights of the cones is equal to the ratio of the radii of the cone.
Therefore, we get
\[ \Rightarrow \dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{{h_1}}}{{{h_2}}}\]
Let \[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{{h_1}}}{{{h_2}}} = n\].
Therefore, we get the equations
\[{r_1} = n{r_2}\] and \[{h_1} = n{h_2}\]
Now, we will calculate the volumes of the two cones.
Substituting \[r = {r_1}\] and \[h = {h_1}\] in the formula for volume of a cone \[\dfrac{1}{3}\pi {r^2}h\], we get
Volume of the smaller cone \[ = \dfrac{1}{3}\pi {\left( {{r_1}} \right)^2}{h_1}\]
Substituting \[{r_1} = n{r_2}\] and \[{h_1} = n{h_2}\] in the equation, we get
\[ \Rightarrow \] Volume of the smaller cone \[ = \dfrac{1}{3}\pi {\left( {n{r_2}} \right)^2}n{h_2}\]
Simplifying the expression, we get
\[ \Rightarrow \] Volume of the smaller cone \[ = \dfrac{1}{3}\pi {n^2}{\left( {{r_2}} \right)^2}n{h_2} = \dfrac{1}{3}{n^3}\pi {\left( {{r_2}} \right)^2}{h_2}\]
Substituting \[r = {r_2}\] and \[h = {h_2}\] in the formula for volume of a cone \[\dfrac{1}{3}\pi {r^2}h\], we get
Volume of the larger cone \[ = \dfrac{1}{3}\pi {\left( {{r_2}} \right)^2}{h_2}\]
It is given that the volume of the smaller and larger cone is \[12\pi \] cu. units and \[96\pi \] cu. units respectively.
Thus, we get the equations
\[ \Rightarrow \] \[\dfrac{1}{3}{n^3}\pi {\left( {{r_2}} \right)^2}{h_2} = 12\pi \]
and
\[ \Rightarrow \] \[\dfrac{1}{3}\pi {\left( {{r_2}} \right)^2}{h_2} = 96\pi \]
We will divide the two equations for volumes to form an equation in terms of \[n\].
Dividing the equation \[\dfrac{1}{3}{n^3}\pi {\left( {{r_2}} \right)^2}{h_2} = 12\pi \] by the equation \[\dfrac{1}{3}\pi {\left( {{r_2}} \right)^2}{h_2} = 96\pi \], we get
\[ \Rightarrow \] \[\dfrac{{\dfrac{1}{3}{n^3}\pi {{\left( {{r_2}} \right)}^2}{h_2}}}{{\dfrac{1}{3}\pi {{\left( {{r_2}} \right)}^2}{h_2}}} = \dfrac{{12\pi }}{{96\pi }}\]
Simplifying the expression, we get
\[ \Rightarrow {n^3} = \dfrac{1}{8}\]
Taking the cube root of both sides of the equation, we get
\[ \Rightarrow n = \dfrac{1}{2}\]
Substituting \[n = \dfrac{1}{2}\] in the equation \[{r_1} = n{r_2}\], we get
\[ \Rightarrow {r_1} = \dfrac{1}{2}{r_2}\]
Substituting \[n = \dfrac{1}{2}\] in the equation \[{h_1} = n{h_2}\], we get
\[ \Rightarrow {h_1} = \dfrac{1}{2}{h_2}\]
Now, we will calculate the slant heights of the two cones.
Let the slant height of the smaller and larger cone be \[{l_1}\] and \[{l_2}\] units respectively.
The slant height of a cone is given by the formula \[l = \sqrt {{r^2} + {h^2}} \], where \[r\] is the radius of the base of the cone, and \[h\] is the height of the cone.
Substituting \[l = {l_1}\], \[r = {r_1}\] and \[h = {h_1}\] in the formula for slant height, we get
\[ \Rightarrow {l_1} = \sqrt {{r_1}^2 + {h_1}^2} \]
Substituting \[l = {l_2}\], \[r = {r_2}\] and \[h = {h_2}\] in the formula for slant height, we get
\[ \Rightarrow {l_2} = \sqrt {{r_2}^2 + {h_2}^2} \]
Now, we will calculate the curved surface areas of the two cones.
The curved surface area of a cone is given by the formula \[\pi rl\], where \[r\] is the radius of the base of the cone, and \[l\] is the slant height of the cone.
Substituting \[l = {l_1}\] and \[r = {r_1}\] in the formula for curved surface area of a cone, we get
Curved surface area of smaller cone \[ = \pi {r_1}{l_1}\]
Substituting \[{l_1} = \sqrt {{r_1}^2 + {h_1}^2} \] in the equation, we get
Curved surface area of smaller cone \[ = \pi {r_1}\sqrt {{r_1}^2 + {h_1}^2} \]
It is given that the curved surface area of the smaller cone is \[15\pi \] sq. units.
Thus, we get the equation
\[ \Rightarrow \pi {r_1}\sqrt {{r_1}^2 + {h_1}^2} = 15\pi \]
Substituting \[{r_1} = \dfrac{1}{2}{r_2}\] and \[{r_1} = \dfrac{1}{2}{r_2}\] in the equation, we get
\[ \Rightarrow \pi \left( {\dfrac{1}{2}{r_2}} \right)\sqrt {{{\left( {\dfrac{1}{2}{r_2}} \right)}^2} + {{\left( {\dfrac{1}{2}{h_2}} \right)}^2}} = 15\pi \]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow \pi \left( {\dfrac{1}{2}{r_2}} \right)\sqrt {\dfrac{1}{4}{r_2}^2 + \dfrac{1}{4}{h_2}^2} = 15\pi \\ \Rightarrow \pi \left( {\dfrac{1}{2}{r_2}} \right)\sqrt {\dfrac{1}{4}\left( {{r_2}^2 + {h_2}^2} \right)} = 15\pi \\ \Rightarrow \pi \left( {\dfrac{1}{2}{r_2}} \right)\dfrac{1}{2}\sqrt {{r_2}^2 + {h_2}^2} = 15\pi \end{array}\]
Thus, we get
\[ \Rightarrow \dfrac{1}{4}\pi {r_2}\sqrt {{r_2}^2 + {h_2}^2} = 15\pi \]
Multiplying both sides of the equation by 4, we get
\[ \Rightarrow \pi {r_2}\sqrt {{r_2}^2 + {h_2}^2} = 60\pi \]
Substituting \[l = {l_2}\] and \[r = {r_2}\] in the formula for curved surface area of a cone, we get
Curved surface area of larger cone \[ = \pi {r_2}{l_2}\]
Substituting \[{l_2} = \sqrt {{r_2}^2 + {h_2}^2} \] in the equation, we get
Curved surface area of larger cone \[ = \pi {r_2}\sqrt {{r_2}^2 + {h_2}^2} \]
Substituting \[\pi {r_2}\sqrt {{r_2}^2 + {h_2}^2} = 60\pi \] in the equation, we get
Curved surface area of larger cone \[ = 60\pi \]
Therefore, we get the curved surface area of the larger cone as \[60\pi \] sq. units.
Note: We have not substituted \[\pi = \dfrac{{22}}{7}\] because the volumes of the two cones, and the curved surface area of the smaller cone is given in the question in terms of \[\pi \]. Therefore, substituting the value \[\pi = \dfrac{{22}}{7}\] and multiplying the expression is not needed.
We calculated the curved surface area of the cone in the question. The curved surface area of a cone is the area of the rounded surface of a cone. It is equal to the difference in the total surface area of the cone, and the area of the circular base of the cone.
Formula used:
We will use the following formulas:
1.The volume of a cone is given by the formula \[\dfrac{1}{3}\pi {r^2}h\], where \[r\] is the radius of the base of the cone, and \[h\] is the height of the cone.
2.The slant height of a cone is given by the formula \[l = \sqrt {{r^2} + {h^2}} \], where \[r\] is the radius of the base of the cone, and \[h\] is the height of the cone.
3.The curved surface area of a cone is given by the formula \[\pi rl\], where \[r\] is the radius of the base of the cone, and \[l\] is the slant height of the cone.
Complete step-by-step answer:
Let the height of the smaller and larger cone be \[{h_1}\] and \[{h_2}\] units respectively.
Let the radius of the smaller and larger cone be \[{r_1}\] and \[{r_2}\] units respectively.
So, now we will draw the diagrams of the cone.
It is given that the two cones are similar. This means that the ratios of the dimensions of the cones are equal.
Thus, the ratio of the heights of the cones is equal to the ratio of the radii of the cone.
Therefore, we get
\[ \Rightarrow \dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{{h_1}}}{{{h_2}}}\]
Let \[\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{{h_1}}}{{{h_2}}} = n\].
Therefore, we get the equations
\[{r_1} = n{r_2}\] and \[{h_1} = n{h_2}\]
Now, we will calculate the volumes of the two cones.
Substituting \[r = {r_1}\] and \[h = {h_1}\] in the formula for volume of a cone \[\dfrac{1}{3}\pi {r^2}h\], we get
Volume of the smaller cone \[ = \dfrac{1}{3}\pi {\left( {{r_1}} \right)^2}{h_1}\]
Substituting \[{r_1} = n{r_2}\] and \[{h_1} = n{h_2}\] in the equation, we get
\[ \Rightarrow \] Volume of the smaller cone \[ = \dfrac{1}{3}\pi {\left( {n{r_2}} \right)^2}n{h_2}\]
Simplifying the expression, we get
\[ \Rightarrow \] Volume of the smaller cone \[ = \dfrac{1}{3}\pi {n^2}{\left( {{r_2}} \right)^2}n{h_2} = \dfrac{1}{3}{n^3}\pi {\left( {{r_2}} \right)^2}{h_2}\]
Substituting \[r = {r_2}\] and \[h = {h_2}\] in the formula for volume of a cone \[\dfrac{1}{3}\pi {r^2}h\], we get
Volume of the larger cone \[ = \dfrac{1}{3}\pi {\left( {{r_2}} \right)^2}{h_2}\]
It is given that the volume of the smaller and larger cone is \[12\pi \] cu. units and \[96\pi \] cu. units respectively.
Thus, we get the equations
\[ \Rightarrow \] \[\dfrac{1}{3}{n^3}\pi {\left( {{r_2}} \right)^2}{h_2} = 12\pi \]
and
\[ \Rightarrow \] \[\dfrac{1}{3}\pi {\left( {{r_2}} \right)^2}{h_2} = 96\pi \]
We will divide the two equations for volumes to form an equation in terms of \[n\].
Dividing the equation \[\dfrac{1}{3}{n^3}\pi {\left( {{r_2}} \right)^2}{h_2} = 12\pi \] by the equation \[\dfrac{1}{3}\pi {\left( {{r_2}} \right)^2}{h_2} = 96\pi \], we get
\[ \Rightarrow \] \[\dfrac{{\dfrac{1}{3}{n^3}\pi {{\left( {{r_2}} \right)}^2}{h_2}}}{{\dfrac{1}{3}\pi {{\left( {{r_2}} \right)}^2}{h_2}}} = \dfrac{{12\pi }}{{96\pi }}\]
Simplifying the expression, we get
\[ \Rightarrow {n^3} = \dfrac{1}{8}\]
Taking the cube root of both sides of the equation, we get
\[ \Rightarrow n = \dfrac{1}{2}\]
Substituting \[n = \dfrac{1}{2}\] in the equation \[{r_1} = n{r_2}\], we get
\[ \Rightarrow {r_1} = \dfrac{1}{2}{r_2}\]
Substituting \[n = \dfrac{1}{2}\] in the equation \[{h_1} = n{h_2}\], we get
\[ \Rightarrow {h_1} = \dfrac{1}{2}{h_2}\]
Now, we will calculate the slant heights of the two cones.
Let the slant height of the smaller and larger cone be \[{l_1}\] and \[{l_2}\] units respectively.
The slant height of a cone is given by the formula \[l = \sqrt {{r^2} + {h^2}} \], where \[r\] is the radius of the base of the cone, and \[h\] is the height of the cone.
Substituting \[l = {l_1}\], \[r = {r_1}\] and \[h = {h_1}\] in the formula for slant height, we get
\[ \Rightarrow {l_1} = \sqrt {{r_1}^2 + {h_1}^2} \]
Substituting \[l = {l_2}\], \[r = {r_2}\] and \[h = {h_2}\] in the formula for slant height, we get
\[ \Rightarrow {l_2} = \sqrt {{r_2}^2 + {h_2}^2} \]
Now, we will calculate the curved surface areas of the two cones.
The curved surface area of a cone is given by the formula \[\pi rl\], where \[r\] is the radius of the base of the cone, and \[l\] is the slant height of the cone.
Substituting \[l = {l_1}\] and \[r = {r_1}\] in the formula for curved surface area of a cone, we get
Curved surface area of smaller cone \[ = \pi {r_1}{l_1}\]
Substituting \[{l_1} = \sqrt {{r_1}^2 + {h_1}^2} \] in the equation, we get
Curved surface area of smaller cone \[ = \pi {r_1}\sqrt {{r_1}^2 + {h_1}^2} \]
It is given that the curved surface area of the smaller cone is \[15\pi \] sq. units.
Thus, we get the equation
\[ \Rightarrow \pi {r_1}\sqrt {{r_1}^2 + {h_1}^2} = 15\pi \]
Substituting \[{r_1} = \dfrac{1}{2}{r_2}\] and \[{r_1} = \dfrac{1}{2}{r_2}\] in the equation, we get
\[ \Rightarrow \pi \left( {\dfrac{1}{2}{r_2}} \right)\sqrt {{{\left( {\dfrac{1}{2}{r_2}} \right)}^2} + {{\left( {\dfrac{1}{2}{h_2}} \right)}^2}} = 15\pi \]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow \pi \left( {\dfrac{1}{2}{r_2}} \right)\sqrt {\dfrac{1}{4}{r_2}^2 + \dfrac{1}{4}{h_2}^2} = 15\pi \\ \Rightarrow \pi \left( {\dfrac{1}{2}{r_2}} \right)\sqrt {\dfrac{1}{4}\left( {{r_2}^2 + {h_2}^2} \right)} = 15\pi \\ \Rightarrow \pi \left( {\dfrac{1}{2}{r_2}} \right)\dfrac{1}{2}\sqrt {{r_2}^2 + {h_2}^2} = 15\pi \end{array}\]
Thus, we get
\[ \Rightarrow \dfrac{1}{4}\pi {r_2}\sqrt {{r_2}^2 + {h_2}^2} = 15\pi \]
Multiplying both sides of the equation by 4, we get
\[ \Rightarrow \pi {r_2}\sqrt {{r_2}^2 + {h_2}^2} = 60\pi \]
Substituting \[l = {l_2}\] and \[r = {r_2}\] in the formula for curved surface area of a cone, we get
Curved surface area of larger cone \[ = \pi {r_2}{l_2}\]
Substituting \[{l_2} = \sqrt {{r_2}^2 + {h_2}^2} \] in the equation, we get
Curved surface area of larger cone \[ = \pi {r_2}\sqrt {{r_2}^2 + {h_2}^2} \]
Substituting \[\pi {r_2}\sqrt {{r_2}^2 + {h_2}^2} = 60\pi \] in the equation, we get
Curved surface area of larger cone \[ = 60\pi \]
Therefore, we get the curved surface area of the larger cone as \[60\pi \] sq. units.
Note: We have not substituted \[\pi = \dfrac{{22}}{7}\] because the volumes of the two cones, and the curved surface area of the smaller cone is given in the question in terms of \[\pi \]. Therefore, substituting the value \[\pi = \dfrac{{22}}{7}\] and multiplying the expression is not needed.
We calculated the curved surface area of the cone in the question. The curved surface area of a cone is the area of the rounded surface of a cone. It is equal to the difference in the total surface area of the cone, and the area of the circular base of the cone.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE
Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE
With reference to graphite and diamond which of the class 11 chemistry CBSE
A certain household has consumed 250 units of energy class 11 physics CBSE
The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE
What is the formula mass of the iodine molecule class 11 chemistry CBSE
Trending doubts
Worlds largest producer of jute is aBangladesh bIndia class 9 social science CBSE
Distinguish between Conventional and nonconventional class 9 social science CBSE
What was the Treaty of Constantinople of 1832 class 9 social science CBSE
What is a legitimate government class 9 social science CBSE
Describe the 4 stages of the Unification of German class 9 social science CBSE
What was the main aim of the Treaty of Vienna of 1 class 9 social science CBSE