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Is the perimeter of triangle ABC greater than 20? (1)

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Re: Perimeter of triangle ABC. [#permalink] New post 17 Dec 2010, 09:01
Bunuel wrote:
sriharimurthy wrote:
Thanks Bunuel.

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.

Area=s^2*\frac{\sqrt{3}}{4}=(\frac{20}{3})^2*\frac{\sqrt{3}}{4}=\frac{100*\sqrt{3}}{9}=~\frac{173}{9}<20

Think this way is easier. \sqrt{3}\approx{1.73}.


Bunuel,

is it necessary to solve it? what if you just recognized that you could solve it and the answer is either > 20 or < 20? I'm just thinking in terms of timing strategy...
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Re: Perimeter of triangle ABC. [#permalink] New post 17 Dec 2010, 09:51
psirus wrote:
Bunuel wrote:
sriharimurthy wrote:
Thanks Bunuel.

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.

Area=s^2*\frac{\sqrt{3}}{4}=(\frac{20}{3})^2*\frac{\sqrt{3}}{4}=\frac{100*\sqrt{3}}{9}=~\frac{173}{9}<20

Think this way is easier. \sqrt{3}\approx{1.73}.


Bunuel,

is it necessary to solve it? what if you just recognized that you could solve it and the answer is either > 20 or < 20? I'm just thinking in terms of timing strategy...


You have to solve it. If we get that the minimum perimeter possible for a triangle with an area of 20 is less than 20 then we won't be able to answer the question. Similarly if we get that the maximum area possible for a triangle with a perimeter of 20 is more than 20 (for example 25) then knowing that area is 20 won't mean that perimeter must be more than 20.

Hope it's clear.
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Re: Perimeter of triangle ABC. [#permalink] New post 14 Jun 2011, 22:07
A bc-ac < ab < bc + ac

ab > 10 and bc > 10 thus adding ab+bc = 20 already.Hence sufficient.

B base * altitude = 40
base = bc
altitude = ad

now assuming values bc = 0.1,1 and 4 , ad = 400,40 and 10. all giving values of perimeter > 20.

sufficient.

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Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink] New post 13 Jan 2012, 03:08
Hi Bunuel , Is there any way to find out the other two wxtreme in those two statement . Like :
A. For a given perimeter equilateral triangle has the largest area: What is the lowest area ?

B. For a given area equilateral triangle has the smallest perimeter : What is the largest area ?
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Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink] New post 13 Jan 2012, 12:06
Good one:

TRIANGLE RULES:
1. Third side < Sum of other two sides
2. Third side > Difference of other two sides

1. Starting with BC=11, AC=1, we get AB=9. Rule 1 is violated because BC > AB + AC. Keep on going BC = 12, 13... it gets worst. Therefore, the perimeter of the triangle has to be greater than 20. Suff.
2. This one I had to semi-guess. I picked 3-4-5 triangle and found that area < perimeter. So, perimeter has to be > 20. Suff.

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Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink] New post 07 May 2012, 21:56
While solving statement-2 in this problem, I realized that –
For a right triangle with a given area, perimeter will be minimum when that right triangle is an "isosceles" right triangle.

I feel this property makes sense. Still, can someone confirm it please?

Using this property the problem can be solved as follows:
Assume that the given right triangle with area 40, is an isosceles right triangle and then find the perimeter. If the perimeter of isosceles right triangle is greater than 20, that will mean any perimeter for that right triangle is definitely greater than 20.

Lets call base of right triangle = b
Lets call height of right triangle = h
Area given = 20. ==> (1/2)b.h = 20 ==> b.h = 40

If we assume this right triangle to be isosceles, then
b=h= sq.root(40)
=> hypotenuse = sq.root(80)
=> perimeter = 2. sq.root(40) + sq.root(80)
=> This is greater than 20.
=> So minimum perimeter is more than 20. So any perimeter for this right triangle with the area as 20, will be more than 20.
=> Statement#2 is sufficient.
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Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink] New post 29 Dec 2012, 08:21
Great qs. Bunuel...!And very nice back calculations shown :)

Could you please share some more like it ?
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Re: Is the perimeter of triangle ABC greater than 20? (1) [#permalink] New post 03 Jan 2013, 22:46
very good question. thanks a lot for the explanation
Re: Is the perimeter of triangle ABC greater than 20? (1)   [#permalink] 03 Jan 2013, 22:46
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