BlindVision wrote:
Can someone please show me how:
pi*[x*(sqrt 2)*(1/2)]^2 BECOMES pi*(x^2 / 2)
How did the (sqrt 2) in the first numerator cancel out to BECOME x^2 in the second numerator??
Sure. The key is to remember to apply the exponent (in this case 2) to each term in the expression (in this case [x*(sqrt 2)*(1/2)]. I happened to blog about just this topic
earlier today.
So, let's actually write out what happens when we apply the exponent:
[x*(sqrt 2)*(1/2)]
^2 = [x*(sqrt 2)*(1/2)]*[x*(sqrt 2)*(1/2)
Now, let's pull the similar pieces together: x*x*(sqrt 2)*(sqrt 2)*(1/2)*(1/2)
x*x = x^2
sqrt(2) * sqrt(2) = 2
(1/2)*(1/2) = 1/4
Put it all together: (x^2)*2*(1/4) = (x^2)/2
Since the original expression was
pi*[x*(sqrt 2)*(1/2)]^2, the final expression is pi*(x^2)/2
Having said that,
there is a MUCH FASTER way to approach this question:
Once you determine that the ratio of diameter of T:diameter of S is root(2):1, you can immediately jump to the conclusion that the ratio of area of T:area of S is [root(2)]^2:[1^2], which is 2:1.
This is true because the ratio of the areas of two circles is always equal to the square of the ratio of any of their linear parts (e.g. radius, diameter, circumference).
This is true more broadly for any pair of similar figures, including any two squares (the linear parts could be side length, diagonal or perimeter), any two equilateral triangles (side length, perimeter, height), any two similar rectangles or triangles, etc.
Furthermore, if this were a solid geometry question, a similar fact would be true: the ratio of the
volumes of two [spheres, cubes, similar rectangular prisms, etc.] is always equal to the
cube (third power) of the ratio of any of their linear parts.
Hope this helps, let me know what you think.
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