12.5- Volume of Pyramids and Cones

      What I knew before watching the video was the formulas for finding the volume of a pyramid and for finding the volume of a cone. The formula for finding the volume of a pyramid is one third times the area of the base times the height, or V=1/3Bh. The formula for finding the volume of a cone is one third times pi times radius squared times height, or V=1/3πr^2h. You use these two formulas to find the volume of pyramids and cones. This is everything I knew about before watching the video on finding the volumes of pyramids and cones.  

      What I learned about after watching the video was that I learned that you can use volume to find the missing radius in the equation for finding the volume of a cone, which is one third time pi times radius squared times height.  For example, if the volume is 2262, then you just plug in the information that you already know. So, 2226=1/3πr^2(15), then you times three on both sides and divide by 15π. You get your radius to equal about 12 after you square root both sides. Lastly, I learned how to find the volume of a nautical prism. For example if one of the sides of the hexagon on the prism is eight and the height of the prism, not including the hexagon, is six, and the height of the hexagon equals point five, not including the prism, then you plug in the information that you know into the equation, which is 1/3Bh + Bh. But, you need to find the area of the base and you do that by using the equation 6(1/4 s^2 time the square root of three). After you figure that you get the volume is equal about 4156.92.

Word Count: 295

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12.6- Surface Area and Volume of Spheres

     http://www.youtube.com/watch?v=moBGSB3ohc0&feature=channel_video_title

       I didn’t know anything before watching the video besides knowing what a sphere is, but that doesn’t really help with anything. So, that is pretty much everything I knew before watching the video.

      What I learned after watching the video was how to find the surface area of a sphere. The equation is surface area equals four times pi times radius squared, or S.A=4πr^2.  I also learned what great circles are. A great circle is a circle where the circumference is 13.8π. You can use the great circles to find the surface area of spheres. I also learned how to find the surface area of a hemisphere. You use the equation S=1/2(4πr^2) +πr^2.  Another thing that I learned is how to find the volume of a sphere. To find the volume of a sphere you use the equation volume equals four thirds time pi time radius to the square root of three, or V=4/3πr^3. You can use the equations volume of a cylinder (πr^2h) and the volume of a sphere (4/3πr^3) to find the volume of a cylinder with two spheres on both ends. To find the volume you have the volume of a cylinder minus the volume of sphere. So, since the height equals 12 radius equals 5, you get the volume to equal about 418.879.

Word Count: 217

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12.4- Volume of Prisms and Cylinders

     

       What I knew about before watching the video was that I already knew the equations for finding the volume of prisms and cylinders and how to use them to find volume. The equation for finding the volume of prisms is volume equals the area of the base times the height, or V=Bh. The equation for finding the volume of cylinders is volume equals pi times radius squared times the height, or V=πr^2h. This is all I knew about before watching the video.

      What I learned about after watcing the video was that I learned Cavalieri’s Principle.Cavalieri’s Principle is if two solids have the same height h and the same cross-sectional area B at every level, then they have the same volume. I also learned how to use volume to find missing lengths. For example, is the volume of a prism is three hundered ninty six and you need to find x, you just plug in your information and solve. So, you have three hundered ninty six equals eleven x time x, or thirty six equals x squared. Then, you square root both sides and x equals six. You can also use volume to find out the missing lenghts of prims and cylinders when you compare a bigger one to a smaller one. This is everything that I learned after watching the video.

Word Count: 222

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12.3- Surface Area of Pyramids and Cones

      What I knew before watching the lecture was that I knew some the equations like S=B+1/2Pl, A=1/2aP, a^2+b^2=c^2, and opposite of B plus and minus the square root of B squared minus four A,C all over two A. This is everything that I knew about before watching the lecture.

      What I learned after watching the lecture was that I learned how to find the surface area of a pyramid. An example is the you use the equation S=B+1/2Pl and you have the pyramid having a slant height of eight and a perimeter of thirty six. To find the area of the base you use the equation A=¼(s^2)square root of three and plug in your information and get A=¼(36)square root of three, or 54 square root of three. So, you plug in all of your information and get S=54square root of three+½(36)(8), or s=237.5. Another thing that I learned was how to find the surface area of a right cone. An example is that the right cone has a slant height of six and a radius of four. You use the equation S=‘pie’r^2+‘pie’rL and you plug in your information and get S= ‘pie’(4^2)+‘pie’(4)(6),or S=16‘pie’+24‘pie’. So, after you simplify you get S to equal approximately 125.6636. Another thing that I learned about was how to find the surface area of a square pyramid. An example is that you first have to find the slant height and you find that by using the Pythagorean Theorem and get the slant height to equal the square root of 52. Second, you find the perimeter and the area of the base. You do that by first finding the perimeter and you do that by multiplying four times eight or 32 meters. Then, you find the area of the base which is 8 squared, or 64 meters squared. Lastly, you find the surface area where you use the equation s=1/2Pt+B and you plug in the information and you get S=½(32)(square root of 52)+64, or approximately 179.4 meters square. The biggest thing that I learned was the best problem ever. You first need to find the area of the base and you use the equation A=1/2aP and since you need to find a and x you use cosine and sine. So, a=5cos36 and x=5sin36, but since one side equals 10sin36, you have to times 5 by 10, so you really have your perimeter equaling 50sin36. So, the area equals ½(5cos36)(50sin36). The second thing you have to do is to find the slant height and you do that by using the Pythagorean Theorem and so, L^2=(6)^2+(5cos36)^2, or L=the square root of 36+(5cos36)^2. Lastly, step three, you put all of the information together and you have this information so far: b=½(5cos36)(50sin36), L=square root 36+(5cos36)^2, P=50sin36, and S=B+1/2Pl. After you put all the information together you get S=½(5cos(36))(50sin(36))+½(50sin(36)) square root of (36+(5cos(36))^2), or S equals approximately 165.774.  This is everything that I learned about after watching the lecture.

Word Count: 487 

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12.2- Surface Area of Cylinders and Prisms

 

      Before watching the lecture the only thing that I knew about was what Surface Area, Prisms, and Cylinders were. I somewhat know how to find it, but not really. This is all I knew before watching the lecture.

      After watching the lecture I learned how to find the surface area of prisms and cylinders and how to find the lateral area of prisms and cylinders. To find the lateral area of prisms the equation is l=Ph, or the length equals the perimeter times the height. To find the surface area of a right prisms you use the equations S=2B+Ph, or the surface area equals two times the base plus perimeter times the area. To find the surface area of a cylinder you use the equations S=2B+Ch and S=πr^2+πrH, or the surface area equals two times the base plus the circumference time the height and surface area equals pie times radius squared plus pie times radius times height. You can use these equations to find lateral area of prisms and cylinders and surface area of prisms and cylinder, but you have to make sure you use the right equation because they are closely alike and if you don’t use the right one, your answer will be wrong. This is everything that I learned after watching the lecture.

Word Count: 216

Questions:

I commented on Michelle K's blog and Adam B's blog.

11.4- Area of Regular Polygons

      What I knew about area of regular polygons before watching the video was that I knew how to find the area of triangles, circles, etc., but not the way that the video explains it. This is all I knew about before watching the video.

      What I learned about after watching the video was about how to find the area of an equilateral triangle and how to find the area of a regular polygon. To find the area of an equilateral triangle you use the equation Area equals one half times side squared times the square root of three (A=½{s^2} <!--[if !vml]--><!--[endif]-->). To find the area of a regular polygon you use the equation Area equals one half time apothem times perimeter (A=½aP). The apothem of a regular polygon is the measure of the line that goes from the center of the polygon to the middle of a side. To find the perimeter of a regular polygon you divide 360 and the number of sides the polygon has, and then you use either cosine, sine, or tangent to find the half of the side length. Lastly, you times that number by two to get the whole side length and times that by the number of sides you have. You use these two equations to find the area of equilateral triangles and the area of a regular polygon.  This is everything that I learned about in the video.

Word Count: 234

Questions:

I commented on Kollin Heimke's Blog and Rebecca Phipps's blog.

11.3- Areas of Circles and Sectors

      What I knew before watching the video was that I knew how to find the area of a circle. The equation is area equals pie time radius squared. So, for example if the radius of a circle is seven the equation would be area equals pie times seven square or forty nine. So, the area would equal 153.94. I also knew somewhat how to find the area of only part of a circle because it is like what we learned last chapter, but I didn’t know the exact formula.

      What I learned after watching the video was that I learned the formula of how to find the area of only part of a circle. It is the area equals the measure of the arc that is part of your sector over three hundred sixty multiplied by pie times radius squared.  For example, if the measure of the arc is thirty five and the radius equals eight, your equation is area equals thirty five over three hundred sixty times pie times eight squared or sixty four. So, the area of the small sector of the circle equals 19.54. If you wanted to find the large sector you would minus 360 minus 65 or 325. You do the same thing, but instead of using 35, you use 325 and your answer is 181.54. This is everything that I learned about after watching the video.

Word Count: 323

Questions: 

I commented on Cheyanne Schleis's blog and Rebecca Phipps's blog.

6.1- Angle Measures in Polygons

      Before watching the lecture I already knew about polygons. Some examples are triangles, quadrilaterals, pentagons, and hexagons. They have different numbers of sides, different number of triangles that can fit inside, and their sum of their interior angles. Triangles have three sides, one triangle can fit inside, and the sum of their interior angles is one hundred eighty. Quadrilaterals have four sides, two triangles can fit inside, and the sum is three sixty. Pentagons have five sides, they can fit three triangles inside, and the sum is five hundred forty. Lastly, hexagons have six sides, four triangles can fit inside, and the sum of their interior angles is seven hundred twenty. This is all that I knew about before watching the lecture.
      After watching the lecture I learned about interior angles and exterior angles. The theorem for interior angles is the sum of interior angles of an n-gon is (n-2)180. The corollary is the measure of each interior angle of a regular n-gon is (n-2)180 divided by n. The exterior angle theorem is the sum of the exterior angles of a convex polygon, one angle each vertex, is 360. The corollary is the measure of each exterior angle of a regular n-gon is 360/n.  You can use both of these to find may different things dealing with angle measurement in polygons. This is everything that I learned about after watching the lecture.

Word Count: 232

Questions:

I commented on Josh Cohen's and Michelle Krueger's blog.