MHB Radian Measure: Show Cone Surface Area is $\pi rl$

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A cone's curved surface area can be derived by transforming it into a sector of a circle with radius equal to the slant height, \( l \), and angle \( \theta \). The relationship between the arc length of the sector and the circumference of the cone's base leads to the equation \( l\theta = 2\pi r \). By calculating the area of the sector, which is given by \( \frac{1}{2}l^2\theta \), and substituting for \( \theta \), it can be shown that the curved surface area of the cone is \( \pi rl \). This derivation highlights the connection between the geometry of the cone and the properties of circular sectors. Understanding this relationship is crucial for solving problems involving cone surface areas.
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A cone with base radius $r$, vertical height $h$ and slant height $l$ has its curved surface slit and flattened out into a sector with radius $l$ and angle $\theta$. By comparing the arc length of this sector with the circumference of the base of the cone, show that $l\theta = 2\pi r$, and deduce by calculating the area of the sector, that the curved surface area of the cone is $\pi rl$.
 
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Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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