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Solved problems on surface area of spheres
In this lesson you will find typical solved problems on surface area of spheres.
The theoretical base for these problems is the lesson Surface area of spheres under the topic Area and surface area of the section Geometry in this site.
Problem 1Find the surface area of a sphere if its radius is of 5 cm.
Solution
The surface area of the sphere is
 =  =  = 314.159  (approximately).
Answer. The surface area of the sphere is 314.159 (approximately).
Problem 2Find 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 (Figure 1) if both the cylinder diameter and the hemisphere diameter are of 10 cm, and the cylinder height is of 20 cm.
Solution
The full surface area of the composite body under consideration is the sum of
the lateral surface area of the cylinder  , the area of the base of the
cylinder  and the area of the hemisphere  .
So, the total surface area of the composite body is equal to
 = + + = +  =  *( ) =
= 3.14159*(2*5*20 + 3*5^2) = 3.14159*275 = 863.937 (approximately).
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Figure 1. To the Problem 2
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Answer. The surface area of the composite body under consideration is 863.937 (approximately).
Problem 3Find the surface area of a composite body comprised of a cone and a hemisphere attached center-to-center to the cone base (Figure 2) if both the cone base diameter and the hemisphere diameter are of 10 cm and the cone height is of 10 cm.
Solution
The full surface area of the composite body under consideration is the sum of
the lateral surface area of the cone  and the area of the
hemisphere .
So, the total surface area of the composite body is equal to
= + =  + =  *( ) =
=  *( ) = * = * =
= 332.7 (approximately).
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Figure 2. To the Problem 3
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Answer. The surface area of the composite body under consideration is 332.7 (approximately).
Problem 4Find the surface area of a composite body comprised of a cube and a hemisphere attached center-to-center to one of the cube faces (Figure 3) if both the cube edge measure and the hemisphere diameter are of 10 cm.
Solution
The full surface area of the composite body under consideration is the sum of
the surface area of the six cube faces  minus the area of the base of the
hemisphere plus the area of the hemisphere , where  =  = 5 cm.
So, the total surface area of the composite body is equal to
= -  +  = + =  +  =
= 678.54 (approximately).
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Figure 3. To the Problem 4
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Answer. The surface area of the composite body under consideration is 678.54 (approximately).
Problem 5Find the surface area of the sphere inscribed in a cone if the base diameter of the cone is of 80 cm and the height of the cone is of 75 cm (Figure 4a).
Solution
Figure 4a shows 3D view of the cone with the inscribed sphere. Figure 4b
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 Figure 4a is the same
as the radius of the inscribed circle in the triangle in the Figure 4b. So,
instead of determining the radius of the sphere we will find the radius of the
inscribed circle. For it, use the formula =  , where is the radius
of the inscribed circle in a triangle, is the area of the triangle and  is
the perimeter of the triangle. The proof of this formula is in the lesson
Proof of the formula for the area of a triangle via the radius of the inscribed
circle under the topic Area and surface area of the section Geometry
in this site.
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Figure 4a. To the Problem 5
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Figure 4b. To the solution
of the Problem 5
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For our isosceles triangle, we have  =  = 85 cm for its lateral side length, =  = 250 cm for the perimeter and =   = 3000 for the area. Therefore, the radius of the inscribed circle is =  =   in accordance with the formula above.
Hence, the area of the sphere inscribed in the cone is =  = 7238.223 .
Answer. The surface area of the sphere inscribed in the cone is 7238.223 (approximately).
My lessons on surface area of spheres and other 3D solid bodies in this site are
To navigate over all topics/lessons of the Online Geometry Textbook use this file/link GEOMETRY - YOUR ONLINE TEXTBOOK.
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