Hi we have an example in our book The problem is to find the area of the sector and the area of the segment if the radius of the circle is 12 cm and the central angle is 60°. And there's a thing we don't understand: Area of 🔺AOC= ½ • (12)(6√3) We do not understand where the 6√3 came from, it wasn't stated in the book can you please help?
What you are finding is the area of the triangle formed by the sector and the triangle in this case is equilateral(all sides the same-12). THe 6sqrt(3) is the height of this triangle if you split the triangle in half with an altitude. To find the height of this triangle you need to use any of Pythagorean theorem (h^2+6^2=12^2), trigonometric function (sin(60)=h/12), or the shortcut method for finding side lengths of a 30-60-90 right triangle. All of these methods would give you 6sqrt(3) for the height. Hope this helps. 🙏 I can make a video but would take longer. Let me know.
got a mock tommorow and was just watching.Your way of teaching is perfect and concise so thanks a lot
Man thank you for saving me from painfully long homework sessions
You're welcome
Hi we have an example in our book
The problem is to find the area of the sector and the area of the segment if the radius of the circle is 12 cm and the central angle is 60°.
And there's a thing we don't understand:
Area of 🔺AOC= ½ • (12)(6√3)
We do not understand where the 6√3 came from, it wasn't stated in the book can you please help?
What you are finding is the area of the triangle formed by the sector and the triangle in this case is equilateral(all sides the same-12). THe 6sqrt(3) is the height of this triangle if you split the triangle in half with an altitude. To find the height of this triangle you need to use any of Pythagorean theorem (h^2+6^2=12^2), trigonometric function (sin(60)=h/12), or the shortcut method for finding side lengths of a 30-60-90 right triangle. All of these methods would give you 6sqrt(3) for the height. Hope this helps. 🙏 I can make a video but would take longer. Let me know.