Question Stem : Is AB + BC + AC > 20?
St. (1) : BC = AC + 10
Triangle Property : The sum of any two sides of a triangle is always greater than the third.
Since we are given that one of the sides is greater than 10, the sum of the other two sides must also be greater than 10.
Hence the perimeter will always be greater than 20.
Statement is sufficient.
St. (2) : A = 40
Triangle Property : For triangles with same area, the perimeter is smallest for an equilateral triangle.
Area of equilateral triangle with side x = \(\frac{\sqrt{3}}{4}x^2\)
Therefore, \(\frac{\sqrt{3}}{4}x^2\) = 40
\(x^2 = \frac{160}{\sqrt{3}}\)
Now, in order to speed up calculations, I will assume \(\sqrt{3}\) to be equal to 2.
If the condition is satisfied with \(\sqrt{3}\) equal to 2 then it will definitely be satisfied with the actual value of \(\sqrt{3}\) which is less than 2.
Therefore, \(x^2 = \frac{160}{2}\) = 80
This tells us that x is almost 9. More importantly, it tells us that x is greater than 8. Thus perimeter will be 3*x = 24.
Since this is the minimum perimeter possible (actually it is still less than what the actual minimum would be due to our approximations), we can conclude that the question stem will always be true.
Hence Sufficient.
Answer : D
Another interesting triangle property : For triangles with same perimeter, the area is maximum for an equilateral triangle. (If you think about it, this property goes hand in hand with the one we used in St. 2).
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