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1. Prove the Pythagorean Theorem by a method not used in class.. § 12.1 There are over 260 of them. You should not have had too much trouble finding another one.

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2. On the three sides of a right triangle construct semicircles with centers at the midpoints of the sides. Calculate the area of each of the three semicircles. Do you see a relationship? Do you think it works for other geometric figures? a b c

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3. Find the ratio of the volume to the surface area of a cube. It is good to have an easy one once in a while!

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4. A sphere is circumscribed by a cylinder. Find the ratio of the two surface areas. Find the ratio of the two volumes. Use unit radius. Sphere – Area – 4 πr 2 Volume - Cylinder Area - 6πr 2 Volume - 2 πr 3 The ratios are the same for Sphere/Cylinder = 2/3

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5. Find the volume of a unit regular octagon. Dissect it into two pyramids. The trick is to find the altitude of the pyramid. 1 √2 h 2 = 1 2 – (√2/2) 2 = √2/2 h V = (2) (1/3) (1) (√2/2) = √2/3

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6. What is the volume of the Great Pyramid of Giza that had a side measure of 756 ft and an altitude of 481 feet? If it took 30 years of 6 day weeks working 10 hours a day, how many cubic feet were put in place each hour? V = bh/3 = 100,017,216 cubic feet. 756 481 Time = 93600 hours V/time = 29,370 cubic feet per hour. That is a volume about the size of over two classrooms per hour!!

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7. An auto tunnel through a mountain is being planned. It will be semicircular cylinder with a radius of 30 feet and a length of 5000 feet. How many cubic feet of dirt will have to be removed? If a dump truck has a bed of dimensions 7 feet by 10 feed by 6 feet, how many loads will be required to carry away the dirt? Volume - (1/2)(5000)(π 30 2 ) = 7068583 cubic feet 7068583/420 = 16,830 dump truck loads

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8. Investigate the Archimedean solids. What characteristics do they have in common? Faces are regular polygons.

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6. What is the shape of the cylinder with minimum surface area for a given volume? V = πhr 2 is fixed. Solve for h h = V/ πr 2 r h SA = 2πr 2 + 2πrh and substitute for h. SA = 2πr 2 + (2πr)(V/ πr 2 ) SA = 2πr 2 + 2V/r and take the derivative 0 = 4r - 2V/r 2 and take the derivative

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