UPSKILL MATH PLUS

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Theory:

1. The diameter of an orange is \(7 \ cm\). Find the surface area of \(50\) oranges.
 
orange.png
 
Solution:
 
Diameter of an orange, \(d\) \(=\) \(7 \ cm\)
 
Radius of an orange, \(r\) \(=\) d2=72=3.5cm
 
Surface area of an orange \(=\) Surface area of a sphere
 
\(=\) \(4 \pi r^2\) sq. units
 
\(=\)  4×227×3.52
 
\(=\) 4×227×12.25
 
\(=\) \(154\)
 
Surface area of an orange \(=\) \(154\) \(cm^2\).
 
Surface area of \(50\) oranges:
 
\(=\) \(154 \times 50\)
 
\(=\) \(7700\)
 
Therefore, the surface area of \(50\) oranges is \(7700 \ cm^2\).
 
 
2. The radii of the frustum of a cone are \(6 \ cm\) and \(2 \ cm\), and the height of the cone is \(5 \ cm\). Find the total surface area of a frustum of a cone.
 
Solution:
 
Let \(R = 6 \ cm\), \(r = 2 \ cm\) and \(h = 5 \ cm\).
 
Slant height, \(l = \sqrt{h^2 + (R - r)^2}\)
 
\(l = \sqrt{5^2 + (6 - 2)^2}\)
 
\(l = \sqrt{25 + 16}\)
 
\(l = \sqrt{41}\)
 
\(l = 6.4 \ cm\) (approximately)
Total surface area of frustum of a cone \(=\) \(\pi l(R + r)\) \(+\) \(\pi R^2 + \pi r^2\) sq. units
T. S. A. \(=\) \(\frac{22}{7} \times 6.4 (6 + 2)\) \(+\) \(\frac{22}{7} (6^2 + 2^2)\)
 
\(=\) \(\frac{22}{7} \times 6.4 \times 8\) \(+\) \(\frac{22}{7} (36 + 4)\)
 
\(=\) \(\frac{22}{7} \times 51.2\) \(+\) \(\frac{22}{7} \times 40\)
 
\(=\) \(\frac{22}{7} \times (51.2 + 40)\)
 
\(=\) \(\frac{22}{7} \times  91.2\)
 
\(=\) \(286.63\) (approximately)
 
Therefore, the total surface area of the frustum of a cone is \(286.63 \ cm^2\).
 
Important!
The value of \(\pi\) should be taken as \(\frac{22}{7}\) unless its value is shared in the problem.