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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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The largest triangle will be right angle triangle with sides 6 and 8, so the maximum area will be 1/2*6*8 = 24. And the minimum area will be when one of the angle approaches 180 degree. When when of the angle approaches 180 degree, the area will approach 0. Hence all three choice are valid.
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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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Hi-

so is the answer not B and C?

I thought that the third side must be at least 2+ (based on third side rule), if we multiply 2 *6/2, the smallest area I computed was 6.

Are we just supposed to assume that this third side is not the same as the height of the triangle? and we can make the area very small if the angle between the two larger sides is approaching zero?

Thanks,


Rahul
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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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Please, someone help me...

If two sides of a triangle have length 6 and 8 third side (x) has to between 2 < x < 14, which means answer A cannot be correct.
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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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yell2012prime wrote:
Please, someone help me...

If two sides of a triangle have length 6 and 8 third side (x) has to between 2 < x < 14, which means answer A cannot be correct.



When the lengths of two sides of a triangle are given, there is no description on type of triangle you are dealing with. It may be a right triangle, isosceles or a very small triangle but not a straight line

and as such, you have to find the maximum and minimum value for the area

the maximum area will occur if the two sides form the legs of a right triangle.

Therefore Area \(= (\frac{1}{2})*(6)*(8) = 24 =\) maximum possible area

Now the range of possible areas is \(0 < Area\leq{24}\)

So all three values are possible.
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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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Holy Cow! This problem is so devious! I also had B and C as my answers, but then realized that the height of the triangle can be chosen arbitrarily to make the area of the triangle equal to 2.

Originally posted by gremather on 24 Feb 2018, 05:32.
Last edited by gremather on 25 Feb 2018, 10:34, edited 4 times in total.
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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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Please avoid expressions such as Holy Christ. Please. For a stupid question, I do think is not necessary

Crap is already more than enough. Or similar to this.

Regards
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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
Just relax there, dude. Not meant to offend or anything. Anyway, I've changed it.

Carcass wrote:
Please avoid expressions such as Holy Christ. Please. For a stupid question, I do think is not necessary

Crap is already more than enough. Or similar to this.

Regards
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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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if there are only 2 unidentified sides,
we could put 2 or 3 values into A=1/2 b*h to test and
quickly assume the answers are all of the above and move on! XD
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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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[quote="arc601"][/quote]

Great explanation
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Re: QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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UPDATED for further discussions
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QOTD #6 Two sides of a triangle have length 6 and 8 [#permalink]
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ANSWER: C )II and III only.
The possible areas of the triangle with side lengths 6 and 8 are II and III, which means the correct answer is C. II and III only.

To determine the possible areas of the triangle, we can use the formula for the area of a triangle given its side lengths. Let's denote the two given side lengths as a = 6 and b = 8. The area of the triangle can be calculated using Heron's formula:

Area = √(s(s-a)(s-b)(s-c))

where s is the semiperimeter of the triangle and c is the remaining side length.

The semi perimeter s is calculated as s = (a + b + c) / 2.

For a triangle to exist, the sum of any two sides must be greater than the third side. In this case, the remaining side c must satisfy the following inequality:

c < a + b = 6 + 8 = 14.

Given that a = 6 and b = 8, we can calculate the semi perimeter as s = (6 + 8 + c) / 2 = (14 + c) / 2 = 7 + c/2.

Using this information, we can calculate the possible areas for different values of c:

For c = 2:

Area = √(7(7-6)(7-8)(7-2)) = √(7(1)(-1)(5)) = √(-35), which is not a valid area for a triangle since the square root of a negative number is not defined.

For c = 12:

Area = √(7(7-6)(7-8)(7-12)) = √(7(1)(-1)(-5)) = √(35) = 5.92, which is a possible area for the triangle.

For c = 24:

Area = √(7(7-6)(7-8)(7-24)) = √(7(1)(-1)(-17)) = √(119) = 10.92, which is also a possible area for the triangle.

Therefore, the possible areas of the triangle are II (12) and III (24), and the correct answer is C. II and III only.
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