PDF chapter test TRY NOW
1. The diameter of an orange is \(7 \ cm\). Find the surface area of \(50\) oranges.
Solution:
Diameter of an orange, \(d\) \(=\) \(7 \ cm\)
Radius of an orange, \(r\) \(=\)
Surface area of an orange \(=\) Surface area of a sphere
\(=\) \(4 \pi r^2\) sq. units
\(=\)
\(=\)
\(=\) \(154\)
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.
Register for free to see more content