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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).

Though I just realized something. I considered the area given to be 40 by mistake. It is actually 20.

Although the logic will be same, the calculations will be harder since we will have to use the real value of \(\sqrt{3}\). (no margin for approximations!)

Sorry for the mistake guys. However, as long as you understand the logic, it shouldn't matter. At least you'll know how to go about approaching such questions in the future!

Cheers.

Ps. Any trick for the calculations Bunuel?

I would go backward.

Let's assume the perimeter is 20. The largest area with given perimeter will have the equilateral triangle, so side=20/3. Let's calculate the area and if the area will be less than 20 it'll mean that perimeter must be more than 20.

will go with A 1. BC = 10+AC. If BC is the longest side then going by the rule that sum of the 2 smaller sides of triangle will be more than the larger side we will get perimeter more than 20. suff 2. we can get sides as 8,5,4 and 10,4,7 satisying that area is 20. for 8,5,4 perimeter is 17 and for 10,4,7 perimeter is 21. hence insuff

basing S1 on the rule that "the length of a triangle is always greater than the absolute difference of the lengths of other two sides" we know that other side must be > 10 and from S1 we know that atleast one other side is 10, hence sufficient.

S2) not sufficient. ab=40, a,b can be 5,8 or 2, 20 _________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

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).

Yes, the OA is D. It was the hard one.

This problem could be solved knowing the properties sriharimurthy mentioned. +1.

For (1): The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.

For (2): A. For a given perimeter equilateral triangle has the largest area. B. For a given area equilateral triangle has the smallest perimeter. _________________

Geometry Is the perimeter of triangle ABC greater than 20? (1) BC-AC=10. (2) The area of the triangle is 20.

Please help to solve this best

Stmt1 says: BC-AC=10, which implies BC>10, and by property of triangles AB>10. This means AB+BC > 20, So Perimeter>20 ie. stmt1 is suff. Stmt2 says: Area = 20, Area > Perimeter, implies perimter <20, i.e stmt2 is suff.

OA is C. _________________

The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!

for statement 2 u r using properties of equilateral triangle but no where in the q it is mentioned it is equilateral triangle

We are not told that ABC is an equilateral triangle.

Let's assume the perimeter is 20. The largest area with given perimeter will have the equilateral triangle, so side=20/3. Then we calculated the area of this hypothetical equilateral triangle and get that its area<20 but statement (2) says area=20 so as p=20 is not enough to produce area=20 even for the best case (for equilateral triangle) then perimeter must be more than 20.

The explanation for why Statement 1 is sufficient is already given above

For Statement 2 it is stated that Area = 20

=> 1/2*B*H = 20 or B*H = 40.

If we look at the possible twin factors that could yield 40 we have {1,40}, {2,20}...{5,8}..

If we consider the base to be any one of the numbers occuring in the twin factors, and the height to be the other number, both the sides other than the base will always have a length that is greater than the height.

This implies that the perimeter will always be greater than 20 if the area is 20.

I go for D each statement alone is suff. (1) is evident (2) is more tricky but going for one extrem (base, height) (40,1) to the other (1, 40) and the mid-couple (20, 2), the perimeter will seemingly always be greater than 20.

Though I just realized something. I considered the area given to be 40 by mistake. It is actually 20.

Although the logic will be same, the calculations will be harder since we will have to use the real value of \(\sqrt{3}\). (no margin for approximations!)

Sorry for the mistake guys. However, as long as you understand the logic, it shouldn't matter. At least you'll know how to go about approaching such questions in the future!

Cheers.

Ps. Any trick for the calculations Bunuel? _________________

Let's assume the perimeter is 20. The largest are with given perimeter would have the equilateral triangle, so side=20/3. Let's calculate the area and if the area will be less than 20 it'll mean that perimeter must be more than 20.

Think this way is easier.

Yes. You are right. It will be easier. Thanks! _________________

Though I just realized something. I considered the area given to be 40 by mistake. It is actually 20.

Although the logic will be same, the calculations will be harder since we will have to use the real value of \(\sqrt{3}\). (no margin for approximations!)

Sorry for the mistake guys. However, as long as you understand the logic, it shouldn't matter. At least you'll know how to go about approaching such questions in the future!

Cheers.

Ps. Any trick for the calculations Bunuel?

I would go backward.

Let's assume the perimeter is 20. The largest area with given perimeter will have the equilateral triangle, so side=20/3. Let's calculate the area and if the area will be less than 20 it'll mean that perimeter must be more than 20.

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