Area of Sector and Area of Triangle

A

apple2357

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These two areas are the same? It doesn't appear obvious to me why they should be? And it is a little surprising.

11902
Algebraically, it can be seen quite easily? But can anyone help me see geometrically, why they might be? Is it even useful to think of it this way? Thinking graphically/geometrically doesn't occur naturally to me !
 
It is fairly easy to calculate the areas. The circle has radius 5 so circumference \(\displaystyle 10\pi\). The sector has arc length 7 so is fraction \(\displaystyle \frac{7}{10\pi}\) of the circle. The area of the circle is \(\displaystyle 25\pi\) so the area of the sector is \(\displaystyle \frac{7}{10\pi}\left(25\pi\right)= 17.5\). The triangle has side lengths 5, 5, and 7 so has perimeter 17 and semi-perimeter 8.5. so by "Heron's formula" its area is \(\displaystyle \sqrt{8.5(8.5- 5)(8.5- 5)(8.5- 7)}= \sqrt{8.5(3.5)(3.5)(1.5)}= \sqrt{156.1875}= 12.4975\). Clearly the sector has greater area than the triangle.
 
No I disagree.

Area of sector is 0.5x r^2 x theta = 0.5(r)(r)(theta) = 0.5 x5x7 ( since r theta = 7, the arc length)

Area of triangle is 0.5x basexheight = 0.5 x7x5

So the two areas are the same. My question is how can i convince myself geometrically that they are?
 
HallsofIvy said:
It is fairly easy to calculate the areas. The circle has radius 5 so circumference \(\displaystyle 10\pi\). The sector has arc length 7 so is fraction \(\displaystyle \frac{7}{10\pi}\) of the circle. The area of the circle is \(\displaystyle 25\pi\) so the area of the sector is \(\displaystyle \frac{7}{10\pi}\left(25\pi\right)= 17.5\). The triangle has side lengths 5, 5, and 7 so has perimeter 17 and semi-perimeter 8.5. so by "Heron's formula" its area is \(\displaystyle \sqrt{8.5(8.5- 5)(8.5- 5)(8.5- 7)}= \sqrt{8.5(3.5)(3.5)(1.5)}= \sqrt{156.1875}= 12.4975\). Clearly the sector has greater area than the triangle.
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I think you have misinterpreted the dimensions of the triangle. The base is 7 and perpendicular height is 5.
 
Oh bother! Then the area of the triangle is just (1/2)(7)(5)= 35/2= 17.5 the same as the sector. In general, if you have a sector of a circle with radius r, arclength c, then its area is cr/2 the same as the area of a triangle with base c and height r.
 
apple2357 said:
These two areas are the same? It doesn't appear obvious to me why they should be? And it is a little surprising.

View attachment 11902
Algebraically, it can be seen quite easily? But can anyone help me see geometrically, why they might be? Is it even useful to think of it this way? Thinking graphically/geometrically doesn't occur naturally to me !
Click to expand...
To see why they are the same, divide the arc into n equal parts (call them ∆s), turning the sector into n narrow sectors. Each of the latter will be nearly a triangle with base ∆s and height r, so its area is approximately bh/2 = r∆s/2. Add up all n of the areas, and it's rs/2, which is the area of your one triangle. (You can likewise divide the triangle into n equal parts ∆s, and see that the n triangles you get all have the same area, because they have the same base and height.)

In fact, one formula for the area of a circle is that it is the same as the area of a triangle whose height is the radius, and whose base is the circumference: A = 1/2 rC = 1/2 r(2 pi r) = pi r^2. The area of a sector is the same, using just part of the circumference.
 
Dr.Peterson said:
To see why they are the same, divide the arc into n equal parts (call them ∆s), turning the sector into n narrow sectors. Each of the latter will be nearly a triangle with base ∆s and height r, so its area is approximately bh/2 = r∆s/2. Add up all n of the areas, and it's rs/2, which is the area of your one triangle. (You can likewise divide the triangle into n equal parts ∆s, and see that the n triangles you get all have the same area, because they have the same base and height.)

In fact, one formula for the area of a circle is that it is the same as the area of a triangle whose height is the radius, and whose base is the circumference: A = 1/2 rC = 1/2 r(2 pi r) = pi r^2. The area of a sector is the same, using just part of the circumference.
Click to expand...

Nice. So almost like thinking about calculus?
 
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