In this chapter we will discuss questions related to surface area of sphere with solution.
The formula for surface area of sphere is given as;
Surface Area = \mathtt{Surface\ area\ =\ 4.\pi .r^{2}}
Where, r is the radius of sphere.
Surface area of sphere – solved examples
Question 01
Find the surface area of sphere of radius 14 cm.
Solution
r = 14 cm
Surface Area = \mathtt{Surface\ area\ =\ 4.\pi .r^{2}}
Putting the values in formula;
\mathtt{Surface\ area\ =4\times \frac{22}{7} \times ( 14)^{2}}\\\ \\ \mathtt{Surface\ area\ =4\times \frac{22}{7} \times ( 14)^{2} =\ 2464\ cm^{2}}
Hence, 2464 sq. cm is the surface area of given sphere.
Question 02
Find the radius of sphere, if the surface area is 5544 sq. cm.
Solution
Surface area = 5544 sq. cm
\mathtt{4.\pi .r^{2} \ =\ 5544}\\\ \\ \mathtt{4.\frac{22}{7} .r^{2} =\ 5544}\\\ \\ \mathtt{r^{2} \ =\frac{5544\times 7}{88} =\ 441}\\\ \\ \mathtt{r\ =\ 21\ cm}
Hence, radius of sphere is 21 cm
Question 03
A solid sphere made of iron of radius 10 cm is melted into 8 small spherical balls. Find the surface area of small ball.
Solution
Radius of big sphere (R)= 10 cm
Radius of small sphere (r) = ?
Since the material of 8 cm ball is melted into smaller balls, the volume will be the same.
Volume of 10 cm balls = combined volume all 8 balls
\mathtt{\frac{4}{3} .\pi .R^{3} =8\times \frac{4}{3} .\pi .r^{3}}\\\ \\ \mathtt{R^{3} =8.r^{3}}\\\ \\ \mathtt{10^{3} =8.\ r^{3}}\\\ \\ \mathtt{r^{3} =\frac{10^{3}}{2^{3}}}\\\ \\ \mathtt{r=\frac{10}{2} =\ 5\ cm}
Hence, the radius of smaller balls is 5 cm.
Surface area of ball = \mathtt{4.\pi .r^{2}}
Putting the values;
\mathtt{Surface\ area\ =\ 4\times \frac{22}{7} \times 5^{2}}\\\ \\ \mathtt{Surface\ area\ =\ 314.3\ cm^{2}}
Hence, 314.3 sq cm is the surface area of smaller ball.
Example 04
The ratio of volume of sphere is given as 1 : 27. Find the ratio of surface area of the given spheres.
Solution
Let radius of first sphere is r1 and radius of 2nd sphere is r2.
Ratio of the volume is given as 1 : 27.
\mathtt{\frac{V_{1}}{V_{2}} =\frac{1}{27}}\\\ \\ \mathtt{\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}} =\frac{1}{27}}\\\ \\ \mathtt{\frac{r_{1}^{3}}{r_{2}^{3}} =\frac{1}{27}}\\\ \\ \mathtt{\frac{r_{1}}{r_{2}} =\frac{1}{3}}
So, the ratio of the radius is 1 : 3.
Now, let’s calculate the ratio of surface area.
\mathtt{\frac{SA_{1}}{SA_{2}} =\frac{4\pi r_{1}^{2}}{4\pi r_{2}^{2}}}\\\ \\ \mathtt{\frac{SA_{1}}{SA_{2}} =\frac{r_{1}^{2}}{r_{2}^{2}} =\left(\frac{1}{3}\right)^{2}}\\\ \\ \mathtt{\frac{SA_{1}}{SA_{2}} =\frac{1}{9}}
Hence, 1 : 9 is the required ratio.
Example 05
It’s given that the sphere and cube are of same height. Calculate the ratio total surface area of given sphere and cube.
Solution
Let “a” be the height of the cube.
Since both sphere and cube are of same height, the diameter of sphere will be ” a” units.
Diameter of sphere = a
Radius of sphere = a/2
Calculating the ratio of total surface areas.
\mathtt{\frac{SA( cube)}{SA( sphere)} =\frac{6a^{2}}{4\pi \left(\frac{a}{2}\right)^{2}}}\\\ \\ \mathtt{\frac{SA( cube)}{SA( sphere)} =\frac{6}{\pi } =\frac{6\times 7}{22}}\\\ \\ \mathtt{\frac{SA( cube)}{SA( sphere)} =\frac{21}{11}}
Hence, 21 : 11 is the required ratio.
Example 06
Consider the sphere of radius 5 cm and a cone of radius 4 cm. It’s given that surface area of sphere if 5 times the curved surface area of cone. Find the height of cone.
Solution
Radius of sphere (R) = 5 cm
Radius of cone (r) = 4 cm
According to question;
Surface area sphere = 5 ( curved surface area of cone )
\mathtt{4.\pi .R^{2} =5\ ( \ \pi .r.l\ )}\\\ \\ \mathtt{4.( 5)^{2} =( 5) \ ( 4) .\ l}\\\ \\ \mathtt{l\ =\ 5\ cm\ }
Hence, slant height of the cone is 5 cm.
Now let’s calculate the height of cone.
Applying Pythagoras theorem in triangle OAC.
\mathtt{AC^{2} =AO^{2} +OC^{2}}\\\ \\ \mathtt{5^{2} =\ OA^{2} +4^{2}}\\\ \\ \mathtt{OA^{2} =25-16}\\\ \\ \mathtt{h^{2} \ =\ 9\ \ }\\\ \\ \mathtt{h\ =\ 3\ cm}
Hence, the height of given cone is 3 cm.
Example 07
The surface area of sphere is given as 154 sq cm. Calculate its volume.
Solution
Let r = radius of sphere.
The formula for surface area of sphere is given as;
\mathtt{4\pi r^{2} =\ 154}\\\ \\ \mathtt{4\times \frac{22}{7} \times r^{2} =\ 154}\\\ \\ \mathtt{r^{2} =\ \frac{154\times 7}{4\times 22}}\\\ \\ \mathtt{r^{2} =12.25}\\\ \\ \mathtt{r\ =\ 3.5\ cm}
Hence, radius of given sphere is 3.5 cm.
Now let’s calculate the volume.
Volume = \mathtt{\frac{4}{3} \pi r^{3}}
\mathtt{Volume\ =\ \frac{4}{3} \times \frac{22}{7} \times ( 3.5)^{3}}\\\ \\ \mathtt{Volume\ =\ \frac{88}{21} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}}\\\ \\ \mathtt{Volume\ =\ 179.66\ cm^{3}}
Hence, 179.66 cu cm is the required volume