Geometry: Effect of dimension changes on volume
This is a free lesson from our course in Geometry
In this lesson youll explore the concepts of Volume and how the Surface Areas and Volume of solids are influenced when the Dimensions Change. As a first step, youll need to review earlier learning about the solids, surface areas, volumes and use of the formulas for them. Moving forward youll be able to find out the effects of changes and solve several problems involving such effects. The presentation covering such content will be done by the instructor in own handwriting, watching video and with the help of several examples with solution. This will help you apply important geometry properties and relationships to solve problems from day to day life situations and also doing geometry home work. (More text below video...)
<h2> Geometry- Effect of dimension changes on volume- Watch video</h2> <p> volume, problems, solution, figure, volume of solids, video, radius, sphere, doubled, solution, geometry, times, Effect of dimension changes on volume, cylinder, example, math help, practice questions, quizzes.</p> <p> when radius of a sphere is doubled, by what factor the volume will be increased, E.g. in a rectangular prism, if the length and width are doubled, then volume is increased 4 times</p>
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(Continued from above) You have noticed in earlier learning on surface area and volumes that multiplying all the dimensions of a solid by a common factor results in changes to the surface area and volume of the solid as well. E.g. if a solid is dilated by a scale factor of 3/4.What effect will this have on the surface area of a solid? E.g. If a prism is reduced by a scale factor of 2/3, then the surface area of the dilated prism is multiplied by the square of the scale factor i.e. (2/3)2.
Similarly you may explore: if a solid is dilated by a scale factor of 2/3, what effect will this have on the volume of a solid? It would conclude to that the volume of the dilated solid is multiplied by the cube of the scale factor, i.e. (2/3)3.

Look at some of the examples to develop skills to apply the concepts and formulae for solving problems related to Surface Area and Volume, when Dimensions Change.
Example 1: Given- if two square pyramids (Figure, below) are similar, find out the ratio of the volume of pyramid 2 to Pyramid 1.
Step 1: find the scale factor / common ratio of the dimensions corresponding in pyramid 1 and pyramid 2.
Step 2: height of pyramid 2 / height of pyramid 1 = 15/10 = 1.5, and
            base of pyramid 2/ base of pyramid 1 = 10/6.7 = 1.5
Step 3: Notice that scale factor is 1.5 and; therefore, ratio of volumes will be (1.5)3 = 3.375
i.e. volume of pyramid 2, will be 3.375 times the volume of pyramid 1.
The ratio of the volume of pyramid 2 to the volume of pyramid 1 is 3.375, as the final answer.
Example 2: Given- the solid 4 x 4 x 4 cube weighs 7 pounds and is worth $300. How many dollars is a 5 x 5 x 5 cube of the same material worth?
Step 1: The scale factor is 5/4.
Step 2: Volume of the dilated solid = (5/4)3 and weight will also go up correspondingly over 7 pounds.
Step 3: The multiplication factor is thus = (5/4)3 = 1.95
Step 4: The weight will be = 1.95 * 7 = 13.67 pounds
Step 5: The material worth is (13.67/7) * 300 = $596, as the final answer.
Example 3: Given- A rectangle rectangular platform measuring 27 cm long and 24 cm wide is to be covered with square tiles that are 2 cm on each side. How many tiles are needed? Another plan is if this platform to be covered with tiles that are 3 centimeters on each side, then how many tiles would be needed?
Step 1: Area of the rectangular platform = 27 * 24 = 648 cm2
Step 2: Area of one tile = 2 * 2 = 4 cm2
Step 3: No. of tiles required = 648/4 = 162 tiles.
Step 4: In second case, scale factor is 3/2=1.5. The area covered will be more by (1.5)2=2.25
Step 5: Thus number of tiles can be worked out: area / area of one tile = 648 / (4 * 2.25)
Step 6: Simplifying it gives: = 72 tiles, as the final answer.
Example 4: Given- If one side of a square is doubled in length and the adjacent side is decreased by two centimeters, the area of the resulting rectangle is 96 square centimeters larger than that of the original square. Find the dimensions of the rectangle.
Step 1: Say side length of example square is a
Step 2: When one side is doubled; it equals 2a and other side becomes (a - 2)
Step 3: Area of square is a2, and area of rectangle = 2a (a - 2) = 2a2 4a
Step 4: The new area is 96 more than old area i.e. 2a2 4a = a2 + 96.
Step 5: Simplify it, a2 4a 96 = 0. Further simplification gives a= 12 OR a= -8
Step 6: Negative side length doesnt find logic, and hence a=12.
Step 7: As given, the side is doubled i.e. 12 * 2 = 24 and adjacent side is decreased by two units i.e. (12 - 2) = 10
Thus the rectangle measures 24 cm by 10 cm, as the final answer.
Example 5: Given- find the largest possible rectangular area that can be enclosed, if you are given 72 m long fencing.
Step 1: Say l is the length and w is the width of rectangular area. Also given is 72 m length of fencing.
Step 2: You know perimeter of the area equals to = (2l + 2w) = 72. Thus (l + w) = 36, and l = (36 - w)
Step 3: Area = l  *  w. In turn A = (36 w) * w = 36w -w2 (by substituting value of l)
Step 4: You know it is a quadratic equation of the form y = ax2 + bx + c , and you need to find the maximum. As it is a negative quadratic equation, it will represent an upside down parabola i.e. vertex is maximum.
Step 5: Assuming coordinates of vertex of the parabola (h, k), h = -b/2a = - (36 /-(2 * 1)) = 18.
Step 6: To determine the value of k, substitute the value of w = 18.
So k is given by K = - (18)2 + 36(18)= - 324 + 648 = 324
Notice that the points in the equation are (width and area) i.e. if plug in w and figure out the area, h value of coordinate maximizes width and k reflects maximum area i.e. 324.
The largest possible rectangular area that can be enclosed is 324 m, as the final answer.
The video above will explain more in detail about Effect of Dimension Changes on Surface Area and Volume, and how to apply the concepts in solving real-world problems. This is explained with the help of several examples and done watching video. This helps you to deal with solving problems and help doing the Geometry home work.  
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