erikvm wrote:
Is this a valid way of approaching the question, or just a lucky coincidence:
Area of equilateral triangle is \((S^2\sqrt{3})/4\)
So \((S^2\sqrt{3})/4\) = \(64\sqrt{3}\)
Simplify this to\(S = 16\sqrt{3}/\sqrt{3}\) -> = 16. One side is 16, divide by 2 to get 8.
I simply ignored the "-32" - is this a valid way of doing it?
Hi
erikvm,
The best approach would be to write down the equations for the area of shaded region and then equate both sides. Your method worked because the final expression was written in the form where you could equate two expressions from LHS and RHS. This may not be the case always and hence is not the best practice to follow.
Also to answer
bgpower's question, the final equation can be written as
\(\sqrt{3}r^2 - \frac{πr^2}{2} = 64\sqrt{3} - 32π\)
There are two easy ways to solve the equation for r.
Equating the expressionYou can see the similarity in expression on both sides of the equation and can equate \(\frac{πr^2}{2} = 32π\) or \(\sqrt{3}r^2 = 64\sqrt{3}\) . This would give you the value of \(r = 8\).
Solving the equationAlternatively the equation can also be solved very quickly. You just need to be aware of the possibilities of the terms canceling out.
The equation can be written as \(r^2 =\) (\(64\sqrt{3} - 32π)/(\sqrt{3}- \frac{π}{2})\)
If we take \(64\) common from the numerator we will have the expression as \(r^2 =\) \(64 *( \sqrt{3} - \frac{π}{2})/(\sqrt{3}- \frac{π}{2})\) thus resulting in \(r^2 = 64\) and \(r = 8\)
So with the above approach, it is possible to solve the question in less than 2 minutes.
Hope this helps
Regards
Harsh