In order to improve your ability in solving questions on surface of cones, you are advised to practice more problems.
From the figure below find the curved surface area of the cone.
Examples of curved surface are without the slant height, StudySmarter Originals
Take \(\pi=3.14\)
Solution:
In this problem, you have been given the radius and height but not the slant height.
Recall that the height of a cone is perpendicular to the radius so that with the slant height, a right-angle triangle is formed.
Deriving the slant height of a cone when not given, StudySmarter Originals
By using Pythagoras theorem,
\[l=\sqrt{8^2+3.5^2}\]
\[l=8.73\, m\]
Now you can find the curved surface area
Use \(A_{cs}=\pi rl\). I hope you didn't forget
\[A_{cs}=3.14\times 3.5\, m \times 8.73\, m\]
Thus, the curved surface area of the cone, \(A_{cs}\) is:
\[A_{cs}=95.94\, m^2\]
In Ikeduru palm fruits are arranged in a conical manner, they are required to be covered with palm fronds of average area \(6\, m^2\) and mass \(10\, kg\). If the palm is inclined at an angle \(30°\) to the horizontal, and the base distance of a conical stock of palm fruits is \(100\, m\). Find the mass of palm frond needed to cover the stock of palm fruits. Take \(\pi=3.14\).
Solution:
Make a sketch of the story.
Is that a story or a question? Not sure, just solve it
Finding the area of a cone with a given angle, StudySmarter Originals
So you can use SOHCAHTOA to get your slant height since
\[\cos\theta=\frac{adjacent}{hypotenuse}\]
The \(50\, m\) was gotten from halving the base distance since we need the radius.
\[\cos(30°)=\frac{50\, m}{l}\]
Cross multiply
Note that \[\cos(30°)=0.866\]
\[0.866l=50\, m\]
Divide both sides by \(0.866\) to get the slant height, \(l\)
\[l=57.74\, m\]
Now you can find the total surface area of the conical stock knowing that
\[a=\pi r^2+\pi rl\]
Hence
\[a=(3.14\times (50\, m)^2)+(3.14\times 50\, m \times 57.74\, m)\]
\[a=7850\, m^2+9065.18\, m^2\]
Hence, the area of the conical stock is \(16915.18\, m^2\).
However, your task is to know the weight of palm fronds used to cover the conical stock. To do this, you need to know how many palm fronds would cover the stock since the area of a palm frond is \(6\, m^2\). Thus the number of palm fronds required, \(N_{pf}\) is
\[N_{pf}=\frac{16915.18\, m^2}{6\, m^2}\]
\[N_{pf}=2819.2\, fronds\]
With each palm frond weighing \(10\, kg\), the total mass of frond needed to cover the conical palm fruit stock, \(M_{pf}\) is:
\[M_{pf}=2819.2 \times 10\, kg\]
\[M_{pf}=28192\, kg\]
Therefore the mass of palm frond required to cover an average conical stock of palm fruit in Ikeduru is \(28192\, kg\).