Lesson Surface area of spheres
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<H2>Surface area of spheres</H2> A <B>sphere</B> is a surface in a <B>3D</B> space all points of which are <B>equidistant</B> from one point called the <B>center of the sphere</B>. <B>Figure 1a</B> shows the sphere. A <B>radius of a sphere</B> is a straight segment connecting the center of the sphere with a point on the sphere surface (<B>Figure 1b</B>). A <B>diameter of a sphere</B> is the straight segment passing through the center of the sphere and connecting the opposite points of the sphere (<B>Figure 1c</B>). <TABLE> <TR> <TD> {{{drawing( 180, 180, -4.5, 4.5, -4.5, 4.5, circle( 0.0, 0.0, 4.0, 4.0), circle( 0.0, 0.0, 0.1, 0.1), locate(-0.2, -0.1, O), green(ellipse(0.0, 0.0, 8, 4)) )}}} <B>Figure 1a</B>. A sphere </TD> <TD> {{{drawing( 180, 180, -4.5, 4.5, -4.5, 4.5, circle(0.0, 0.0, 4, 4), circle( 0.0, 0.0, 0.1, 0.1), locate(-0.2, -0.1, O), green(ellipse( 0.0, 0.0, 8, 4)), green(line( 0.0, 0.0, 2.4, 1.5)), locate( 1.2, 0.9, r) )}}} <B>Figure 1b</B>. A sphere and its radius </TD> <TD> {{{drawing( 180, 180, -4.5, 4.5, -4.5, 4.5, circle(0.0, 0.0, 4, 4), circle( 0.0, 0.0, 0.1, 0.1), locate(-0.2, -0.1, O), green(ellipse( 0.0, 0.0, 8, 4)), green(line(-2.4, -1.5, 2.4, 1.5)), locate( 1.2, 0.9, d) )}}} <B>Figure 1c</B>. A sphere and its diameter </TD> </TR> </TABLE> <H3>Properties of spheres</H3> 1. Every section of a sphere by a plane is a circle. 2. A tangent segment to a sphere released from a point in <B>3D</B> space outside the sphere is perpendicular to the radius of the sphere drawn from its center to the tangent point. 3. All tangent segment to a sphere released from one fixed point outside the sphere have the same length. <H3>Formula for calculating the surface area of spheres</H3> The surface area of a sphere</B> is {{{S}}} = {{{4pi}}}{{{r^2}}} = {{{pi}}}{{{d^2}}}, where {{{r}}} is the radius of the sphere and {{{d}}} is the diameter of the sphere. <H3>Example 1</H3>Find the surface area of a sphere if its radius is of 10 cm. <B>Solution</B> The surface area of the sphere is {{{4pi*r^2}}} = {{{4*3.14159*10^2}}} = {{{3.14159*400}}} = 1256.636 {{{cm^2}}} (approximately). <B>Answer</B>. The surface area of the sphere is 1256.636 {{{cm^2}}} (approximately). <H3>Example 2</H3>Find the surface area of a composite body comprised of a right circular cylinder and a hemisphere attached center-to-center to one of the cylinder bases (<B>Figure 1</B>) if both the cylinder diameter and the hemisphere diameter are of 20 cm, and the cylinder height is of 40 cm. <TABLE> <TR> <TD> <B>Solution</B> The full surface area of the composite body under consideration is the sum of the lateral surface area of the cylinder {{{2pi*r*h}}}, the area of the base of the cylinder {{{pi*r^2}}} and the area of the hemisphere {{{2pi*r^2}}}. So, the total surface area of the composite body is equal to {{{S}}} = {{{2pi*r*h}}} + {{{pi*r^2}}} + {{{2pi*r^2}}} = {{{2pi*r*h}}} + {{{3pi*r^2}}} = {{{pi}}}*({{{2r*h + 3r^2}}}) = = 3.14159*(2*10*40 + 3*10^2) = 3.14159*1100 = 3455.749 {{{cm^2}}} (approximately). </TD> <TD> {{{drawing( 210, 195, -3.5, 3.5, -1.0, 5.5, ellipse( 0.5, 3.5, 3.0, 1.0), ellipse( 0.5, 0.0, 3.0, 1.0), line( -1, 3.5, -1, 0.0), line( 2, 3.5, 2, 0.0), arc ( 0.5, 3.5, 3.00, 3.00, 180, 360), arc ( 0.5, 3.5, 3.06, 3.06, 180, 360) )}}} <B>Figure 1</B>. To the <B>Example 2</B> </TD> </TR> </TABLE> <B>Answer</B>. The surface area of the composite body under consideration is 3455.749 {{{cm^2}}} (approximately). <H3>Example 3</H3>Find the surface area of a composite body comprised of a cone and a hemisphere attached center-to-center to the cone base (<B>Figure 2</B>) if both the cone base diameter and the hemisphere diameter are of 20 cm and the cone height is of 20 cm. <TABLE> <TR> <TD> <B>Solution</B> The full surface area of the composite body under consideration is the sum of the lateral surface area of the cone {{{pi*r*slant_height}}} and the area of the hemisphere {{{2pi*r^2}}}. So, the total surface area of the composite body is equal to {{{S}}} = {{{pi*r*slant_height}}} + {{{2pi*r^2}}} = {{{pi*r*sqrt(r^2 + h^2)}}} + {{{2pi*r^2}}} = {{{pi*r}}}*({{{sqrt(r^2+h^2) + 2r}}}) = = {{{3.14159*10}}}*({{{sqrt(10^2 + 20^2) + 2*10}}}) = {{{3.14159*10}}}*{{{(sqrt(500) + 20)}}} = {{{3.14159*10}}}*{{{(10sqrt(5) + 20)}}} = = 1330.8 {{{cm^2}}} (approximately). </TD> <TD> {{{drawing( 195, 195, -3.0, 3.5, -1.0, 5.5, ellipse( 0.5, 3.5, 3.0, 1.0), line( -1.0, 3.5, 0.5, 0.0), line( 2.0, 3.5, 0.5, 0.0), arc ( 0.5, 3.5, 3.00, 3.00, 180, 360), arc ( 0.5, 3.5, 3.06, 3.06, 180, 360) )}}} <B>Figure 2</B>. To the <B>Example 3</B> </TD> </TR> </TABLE> <B>Answer</B>. The surface area of the composite body under consideration is 1330.8 {{{cm^2}}} (approximately). <H3>Example 4</H3>Find the surface area of a composite body comprised of a cube and a hemisphere attached center-to-center to one of the cube faces (<B>Figure 3</B>) if both the cube edge measure and the hemisphere diameter are of 20 cm. <TABLE> <TR> <TD> <B>Solution</B> The full surface area of the composite body under consideration is the sum of the surface area of the six cube faces {{{6a^2}}} minus the area of the base of the hemisphere {{{pi*r^2}}} plus the area of the hemisphere {{{2pi*r^2}}}, where {{{r}}} = {{{20/2}}} = 10 cm. So, the total surface area of the composite body is equal to {{{S}}} = {{{6a^2}}} - {{{pi*(a/2)^2}}} + {{{2pi*(a/2)^2}}} = {{{6a^2}}} + {{{pi*(a/2)^2}}} = {{{6*20^2}}} + {{{3.14159*10^2}}} = = 2714.159 {{{cm^2}}} (approximately). </TD> <TD> {{{drawing( 180, 180, -2.5, 3.5, -0.5, 5.5, line ( 0.0, 0.0, 3.0, 0.0), line ( 0.0, 0.0, 0.0, 3.0), line ( 0.0, 0.0, -2.0, 0.8), line ( 0.0, 3.0, 3.0, 3.0), line ( 3.0, 3.0, 3.0, 0.0), line ( 0.0, 3.0, -2.0, 3.8), line ( -2.0, 3.8, 1.0, 3.8), line ( -2.0, 3.8, -2.0, 0.8), green(line ( 1.0, 3.8, 1.0, 0.8)), line ( 1.0, 3.8, 3.0, 3.0), green(line ( -2.0, 0.8, 1.0, 0.8)), green(line ( 1.0, 0.8, 3.0, 0.0)), locate( 1.2, 0.1, a), locate(-1.5, 0.6, a), locate(-0.4, 2.1, a), ellipse( 0.52, 3.43, 2.54, 0.77), arc ( 0.52, 3.43, 2.54, 2.54, 175, 354) )}}} <B>Figure 3</B>. To the <B>Example 4</B> </TD> </TR> </TABLE> <B>Answer</B>. The surface area of the composite body under consideration is 2714.159 {{{cm^2}}} (approximately). <H3>Example 5</H3>Find the surface area of the sphere inscribed in a cone if the base diameter of the cone is of 24 cm and the height of the cone is of 16 cm (<B>Figure 4a</B>). <TABLE> <TR> <TD> <B>Solution</B> <B>Figure 4a</B> shows <B>3D</B> view of the cone with the inscribed sphere. <B>Figure 4b</B> shows the axial section of this cone and the inscribed sphere as the isosceles triangle with the inscribed circle. The radius of the inscribed sphere in the cone in the <B>Figure 4a</B> is the same as the radius of the inscribed circle in the triangle in the <B>Figure 4b</B>. So, instead of determining the radius of the sphere we will find the radius of the inscribed circle. For it, use the formula {{{r}}} = {{{2S/P}}}, where {{{r}}} is the radius of the inscribed circle in a triangle, {{{S}}} is the area of the triangle and {{{P}}} is the perimeter of the triangle. The proof of this formula is in the lesson <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Proof-of-the-formula-for-the-area-of-a-triangle-via-the-radius-of-the-inscribed-circle.lesson>Proof of the formula for the area of a triangle via the radius of the inscribed circle</A> under the topic <B>Area and surface area</B> of the section <B>Geometry</B> in this site. </TD> <TD> {{{drawing( 210, 249, -3.5, 3.5, -3.8, 4.5, ellipse (0.0, -2.0, 6.0, 3.0), green(ellipse (0.0, 0.8, 3.2, 1.6)), line ( -2.75, -1.5, -1.375, 1.25), line ( 2.82, -1.5, 1.410, 1.25), line ( -1.375, 1.25, 0.0, 4.0), line ( 1.410, 1.25, 0.0, 4.0), green(line (0, -2, 0, 4)), circle( 0.0, -0.12, 1.88, 1.88) )}}} <B>Figure 4a</B>. To the <B>Example 5</B> </TD> <TD> {{{drawing( 210, 249, -3.5, 3.5, -3.8, 4.5, line (-3.0, -2.0, 3.0, -2.0), line ( -3.0, -2.0, 0.0, 4.0), line ( 3.0, -2.0, 0.0, 4.0), green(line (0, -2, 0, 4)), circle( 0.0, -0.12, 1.88, 1.88), line ( 0.0, -0.2, -1.6, 0.8), line ( 0.0, -0.2, 1.6, 0.8), line ( 0.0, -0.2, 0.0, -2.0) )}}} <B>Figure 4b</B>. To the solution of the <B>Example 5</B> </TD> </TR> </TABLE>For our isosceles triangle, we have {{{l}}} = {{{sqrt((24/2)^2 + 16^2)}}} = 20 cm for its lateral side length, {{{P}}} = {{{20 + 20 + 24}}} = 64 cm for the perimeter and {{{S}}} = {{{1/2}}}{{{16*12}}} = 96 {{{cm^2}}} for the area. Therefore, the radius of the inscribed circle is {{{r}}} = {{{2*96/64}}} = {{{3}}} {{{cm}}} in accordance with the formula above. Hence, the area of the sphere inscribed in the cone is {{{4pi*r^2}}} = {{{4*3.14159*3^2}}} = 113.097 {{{cm^2}}}. <B>Answer</B>. The surface area of the sphere inscribed in the cone is 113.097 {{{cm^2}}} (approximately). My lessons on surface area of spheres and other 3D solid bodies in this site are <TABLE> <TR> <TD> <B>Lessons on surface area of prisms</B> <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-prisms.lesson>Surface area of prisms</A> <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-surface-area-of-prisms.lesson>Solved problems on surface area of prisms</A> <A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-prisms.lesson>Overview of lessons on surface area of prisms</A> </TD> <TD> <B>Lessons on surface area of pyramids</B> <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-pyramids.lesson>Surface area of pyramids</A> <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-surface-area-of-pyramids.lesson>Solved problems on surface area of pyramids</A> <A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-pyramids.lesson>Overview of lessons on surface area of pyramids</A> </TD> </TR> </Table><TABLE> <TR> <TD> <B>Lessons on surface area of cylinders</B> <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-cylinders.lesson>Surface area of cylinders</A> <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-surface-area-of-cylinders.lesson>Solved problems on surface area of cylinders</A> <A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-cylinders.lesson>Overview of lessons on surface area of cylinders</A> </TD> <TD> <B>Lessons on surface area of cones</B> <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Surface-area-of-cones.lesson>Surface area of cones</A> <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-surface-area-of-cones.lesson>Solved problems on surface area of cones</A> <A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-cones.lesson>Overview of lessons on surface area of cones</A> </TD> <TD> <B>Lessons on surface area of spheres</B> Surface area of spheres <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-surface-area-of-spheres.lesson>Solved problems on surface area of spheres</A> <A HREF=http://www.algebra.com/algebra/homework/Surface-area/OVERVIEW-of-LESSONS-on-surface-area-of-spheres.lesson>Overview of lessons on surface area of spheres</A> </TD> </TR> </Table> To navigate over all topics/lessons of the Online Geometry Textbook use this file/link <A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.
