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Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 00:15
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Is triangle ABC with sides a, b and c acute angled? 1) Triangle with sides \(a^2\), \(b^2\), \(c^2\) has an area of 140 sq cms. 2) Median AD to side BC is equal to altitude AE to side BC.
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Re: Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 01:02
Can you explain why option A is correct ?



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Re: Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 01:15
teaserbae wrote: Can you explain why option A is correct ? I will post the OE after a few discussions. I will PM you the OE, if you are interested.
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Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 02:50
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itisSheldonVery nice question, see my approach. itisSheldon wrote: Is triangle ABC with sides a, b and c acute angled?
1) Triangle with sides \(a^2\), \(b^2\), \(c^2\) has an area of 140 sq cms. 2) Median AD to side BC is equal to altitude AE to side BC.
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Re: Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 03:09
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Hiii For those who are wondering how I relate the angle with the length of sides. Please see the sketch. This is a part of my personal notes. If you like it, let me know by your appreciation of kudos. I will post more of my notes here.
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Re: Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 04:52
itisSheldon wrote: teaserbae wrote: Can you explain why option A is correct ? I will post the OE after a few discussions. I will PM you the OE, if you are interested. Yeah please PM me itisSheldon



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Re: Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 08:04
Hey guys! Is that kind of a joke? S=1/2*h*(a or b or c)^2=140. I completely don’t understand how you found angles relation by given S.
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Re: Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 09:21
Hi The area of triangle given as 140 is not required to solve the question. First statement just show that it is a valid triangle with sides \(a^2, b^2, c^2.\) As for any triangle the sum of 2 sides > third side we get \(a^2 + b^2 > c^2\) Now consider triangle ABC with sides a, b, c as \(a^2 + b^2 > c^2\), hence angle between sides a and b is acute, and \(b^2 + c^2 >a^2\) similarly,angle between sides b and c is acute, and \(c^2 + a^2 > b^2\) similarly,angle between sides c and a is acute, Since all 3 angles of triangle is acute, Hence ABC is an acute angled triangle.Also see the attached sketch to find how we get that the angle is acute from the relation between sides. Iamnowjust wrote: Hey guys! Is that kind of a joke? S=1/2*h*(a or b or c)^2=140. I completely don’t understand how you found angles relation by given S.
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Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 09:27
OE What kind of an answer will the question fetch?The question is an "Is" question. Answer to an "is" questions is either YES or NO. When is the data sufficient?The data is sufficient if we are able to get a DEFINTE YES or DEFINITE NO as the answer. If the statements independently or together do not provide a DEFINITE YES or DEFINITE NO, the data is NOT sufficient. What do we know from the question stem?The question stem states that a, b, and c are the measures of the sides of a triangle. Key properties of a triangleIf a, b, and c are the measures of the sides of a triangle, and if 'a' is the longest side of the triangle, then i) the triangle is acute angled if \(a^2\) < \(b^2\) + \(c^2\) ii) right angled if \(a^2\) = \(b^2\) + \(c^2\) iii) obtuse angled if \(a^2\) > \(b^2\) + \(c^2\) Quote: Statement 1: Triangle with sides \(a^2\), \(b^2\), \(c^2\) has an area of 140 sq cms. The statement provides us with one valuable information: we can form a triangle with sides \(a^2\), \(b^2\), \(c^2\) For any triangle we know that sum of two sides is greater than the third side. So, we can infer that \(a^2\) < \(b^2\) + \(c^2\). The inequality above is the condition to be met if the triangle with sides a, b and c were to be an acute triangle. So Statement 1 itself is sufficient. Quote: Statement 2: Median AD to side BC is equal to altitude AE to side BC. Equilateral and Isosceles triangle propertiesi) For an equilateral triangle, medians to the sides of the triangle are the corresponding altitudes. i.e., the median and altitude of all 3 sides are coincident lines. ii) For an isosceles triangle, the median to the side whose measure is different is the altitude to that side. i.e., only one median is the same as the altitude. From statement 2, we can infer that the triangle is either equilateral or isosceles. An equilateral triangle is definitely an acute angled triangle. However, an isosceles triangle need not be an acute angled triangle. So Statement 2 is not sufficient gmatbusters has explained it well
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Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 13:22
i dont agree with the above explanations for A. here's why:
we're given a triangle with sides \(a^2, b^2, c^2\) so \(a^2+b^2>c^2\) just means that the sum of 2 legs of the triangle is greater than the 3rd leg, which is true for every triangle (acute, obtuse, right). this specific triangle is acute if \(a^4+b^4>c^4\)
Onto my answer...
Statement 1 Sides are \(a^2, b^2, c^2\) and area is 140 We can make our lives simpler and assume the sides are x,y,z. I think the exponents are only there to distract us. We can vary the sides/angles as we like, while keeping the same area. Not sufficient. As an example, if the triangle is equilateral (and thus acute), then the area is \(140 = x^2\sqrt{3}/4\). There's some x that fits the equation, thus it's possible that the triangle is acute. If the triangle is a right isosceles triangle (and thus not acute), then the area is \(140=\frac{1}{2}x^2\). There's some x that fits the equation, thus it's possible that the triangle is not acute.
Statement 2 I like the explanations above. Triangle can be isosceles (and acute, obtuse, or right) or equilateral. Not sufficient.
Combining both statements We can use the same 2 examples in Statement 1. Not sufficient
Answer: E



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Re: Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 14:56
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Hi I would be happy to clear your doubt... You are right that for the triangle given in statement 1, a^2+b^2>c^2, this is true for all triangle whether it is acute angled, right angled or obtuse angled. But we are using this inequality for triangle ABC which has sides a,b,c.Now in a triangle with sides a,b,c, if a^2+b^2>c^2, we know that the angle between a and b is acute. Similarly we can prove that angle between b and C is acute. And finally angle between C and a is acute. Hence the triangle ABC is acute angled triangle. aserghe1 wrote: i dont agree with the above explanations for A. here's why:
we're given a triangle with sides \(a^2, b^2, c^2\) so \(a^2+b^2>c^2\) just means that the sum of 2 legs of the triangle is greater than the 3rd leg, which is true for every triangle (acute, obtuse, right). this specific triangle is acute if \(a^4+b^4>c^4\)
Onto my answer...
Statement 1 Sides are \(a^2, b^2, c^2\) and area is 140 We can make our lives simpler and assume the sides are x,y,z. I think the exponents are only there to distract us. We can vary the sides/angles as we like, while keeping the same area. Not sufficient. As an example, if the triangle is equilateral (and thus acute), then the area is \(140 = x^2\sqrt{3}/4\). There's some x that fits the equation, thus it's possible that the triangle is acute. If the triangle is a right isosceles triangle (and thus not acute), then the area is \(140=\frac{1}{2}x^2\). There's some x that fits the equation, thus it's possible that the triangle is not acute.
Statement 2 I like the explanations above. Triangle can be isosceles (and acute, obtuse, or right) or equilateral. Not sufficient.
Combining both statements We can use the same 2 examples in Statement 1. Not sufficient
Answer: E
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Re: Is triangle ABC with sides a, b and c acute angled? [#permalink]
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15 Apr 2018, 17:46
gmatbusters wrote: Hi I would be happy to clear your doubt... You are right that for the triangle given in statement 1, a^2+b^2>c^2, this is true for all triangle whether it is acute angled, right angled or obtuse angled. But we are using this inequality for triangle ABC which has sides a,b,c.Now in a triangle with sides a,b,c, if a^2+b^2>c^2, we know that the angle between a and b is acute. Similarly we can prove that angle between b and C is acute. And finally angle between C and a is acute. Hence the triangle ABC is acute angled triangle. aserghe1 wrote: i dont agree with the above explanations for A. here's why:
we're given a triangle with sides \(a^2, b^2, c^2\) so \(a^2+b^2>c^2\) just means that the sum of 2 legs of the triangle is greater than the 3rd leg, which is true for every triangle (acute, obtuse, right). this specific triangle is acute if \(a^4+b^4>c^4\)
Onto my answer...
Statement 1 Sides are \(a^2, b^2, c^2\) and area is 140 We can make our lives simpler and assume the sides are x,y,z. I think the exponents are only there to distract us. We can vary the sides/angles as we like, while keeping the same area. Not sufficient. As an example, if the triangle is equilateral (and thus acute), then the area is \(140 = x^2\sqrt{3}/4\). There's some x that fits the equation, thus it's possible that the triangle is acute. If the triangle is a right isosceles triangle (and thus not acute), then the area is \(140=\frac{1}{2}x^2\). There's some x that fits the equation, thus it's possible that the triangle is not acute.
Statement 2 I like the explanations above. Triangle can be isosceles (and acute, obtuse, or right) or equilateral. Not sufficient.
Combining both statements We can use the same 2 examples in Statement 1. Not sufficient
Answer: E Thanks  I see where i went wrong. I got caught up in the info in statement 1 that I thought we had to find out if a triangle with sides a^2, b^2, c^2 is acute. But as the question stem clearly states, we need to find out if a triangle with sides a,b,c is acute.




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