
The height of a cone is 21 cm. Find the area of the base if the slant height is 28 cm.
Answer
132.9k+ views
Hint: Here, we will use the formula of slant height, \[l = \sqrt {{r^2} + {h^2}} \], where \[r\] is the radius of a cone and \[h\] is the height of the cone to find the radius of the base of a cone. Then we will substitute the value of radius in the formula of area of base of cone, \[\pi {r^2}\] to find the required value.
Complete step by step answer:
Given that the height \[h\] of a cone is 21 cm and the slant height \[l\] of a cone is 28 cm.
Let the radius of a cone is \[r\].
We know that the slant height is \[l = \sqrt {{r^2} + {h^2}} \], where \[r\] is the radius of a cone and \[h\] is the height of a cone.
Substituting the values of \[r\]and \[l\] in the above formula of \[l\], we get
\[
\Rightarrow 28 = \sqrt {{r^2} + {{21}^2}} \\
\Rightarrow {28^2} = {r^2} + {21^2} \\
\Rightarrow 784 = {r^2} + 441 \\
\]
Subtracting the above equation by 441 on each of the sides, we get
\[
\Rightarrow 784 - 441 = {r^2} + 441 - 441 \\
\Rightarrow 343 = {r^2} \\
\]
Taking the square root on both sides in the above equation, we get
\[
\Rightarrow r = \pm \sqrt {343} \\
\Rightarrow r = \pm 7\sqrt 7 {\text{ cm}} \\
\]
Since the value of the radius can never be negative, so negative value of \[r\] is discarded.
Thus, the radius of the base of a cone is \[7\sqrt 7 \] cm.
We know that the area of the base of a cone is \[\pi {r^2}\], where \[r\]is the radius of the cone.
Substituting the above value of \[r\] in \[\pi {r^2}\] to find the area of the base of cone, we get
\[
{\text{Area of base of a cone}} = \pi {\left( {7\sqrt 7 } \right)^2} \\
= \pi \left( {49 \times 7} \right) \\
\]
Using the value of \[\pi \] in the above equation, we get
\[
\dfrac{{22}}{7}\left( {49 \times 7} \right) = 22 \times 49 \\
= 1078{\text{ c}}{{\text{m}}^2} \\
\]
Thus, the area of the base of a cone is 1078 cm\[^2\].
Note: In this question, students should know the formulae of the slant height of cone, \[l = \sqrt {{r^2} + {h^2}} \], where \[r\] is the radius of a cone and \[h\] is height of cone and the area of the cone, area of the base of a cone is \[\pi {r^2}\], where \[r\]is the radius of the cone properly. Also, we are supposed to write the values properly to avoid any miscalculation.
Complete step by step answer:
Given that the height \[h\] of a cone is 21 cm and the slant height \[l\] of a cone is 28 cm.
Let the radius of a cone is \[r\].
We know that the slant height is \[l = \sqrt {{r^2} + {h^2}} \], where \[r\] is the radius of a cone and \[h\] is the height of a cone.
Substituting the values of \[r\]and \[l\] in the above formula of \[l\], we get
\[
\Rightarrow 28 = \sqrt {{r^2} + {{21}^2}} \\
\Rightarrow {28^2} = {r^2} + {21^2} \\
\Rightarrow 784 = {r^2} + 441 \\
\]
Subtracting the above equation by 441 on each of the sides, we get
\[
\Rightarrow 784 - 441 = {r^2} + 441 - 441 \\
\Rightarrow 343 = {r^2} \\
\]
Taking the square root on both sides in the above equation, we get
\[
\Rightarrow r = \pm \sqrt {343} \\
\Rightarrow r = \pm 7\sqrt 7 {\text{ cm}} \\
\]
Since the value of the radius can never be negative, so negative value of \[r\] is discarded.
Thus, the radius of the base of a cone is \[7\sqrt 7 \] cm.
We know that the area of the base of a cone is \[\pi {r^2}\], where \[r\]is the radius of the cone.
Substituting the above value of \[r\] in \[\pi {r^2}\] to find the area of the base of cone, we get
\[
{\text{Area of base of a cone}} = \pi {\left( {7\sqrt 7 } \right)^2} \\
= \pi \left( {49 \times 7} \right) \\
\]
Using the value of \[\pi \] in the above equation, we get
\[
\dfrac{{22}}{7}\left( {49 \times 7} \right) = 22 \times 49 \\
= 1078{\text{ c}}{{\text{m}}^2} \\
\]
Thus, the area of the base of a cone is 1078 cm\[^2\].
Note: In this question, students should know the formulae of the slant height of cone, \[l = \sqrt {{r^2} + {h^2}} \], where \[r\] is the radius of a cone and \[h\] is height of cone and the area of the cone, area of the base of a cone is \[\pi {r^2}\], where \[r\]is the radius of the cone properly. Also, we are supposed to write the values properly to avoid any miscalculation.
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