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.
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