If the lengths of the sides of triangle ABC are a^2, b^2 and c^2 and t

Question

Competition Mode Question

If the lengths of the sides of triangle ABC are a^2, b^2 and c^2 and the lengths of the sides of triangle DEF are a, b, c, then triangle DEF could be:

I. Right-angled
II. Acute-angled
III. Obtuse-angled

A. I only
B. II only 
C. III only
D. I and II only
E. I, II, and III

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jpan · Answer

IMO the answer is Obtuse only. For sides of a triangle, 
a^2 +b^2 = c^2 => right triangle
a^2 +b^2 < c^2 => acute
a^2 +b^2 > c^2 => obtuse
Since the squares of sides of DEF are already sides of another triangle, using the property that sum of any 2 sides of ANY triangle is ALWAYS greater than thrid side, we can rule out right triangle and acute triangle. So DEF can be only Obtuse.

madgmat2019 · Answer

..

exc4libur · Answer

acute max angle < 90o: a,b,c = equilateral sides 60o
right max angle = 90o: a,b,c = 6-8-10, squared = 36-64-100 (not possible)
obtuse max angle > 90o: a,b,c was obtuse, then squared it wouldn't be possible

Ans (B)

freedom128 · Answer

Sides of triangle ABC are a^2, b^2 and c^2.
Sides of triangle DEF are a, b and c.

Triangle DEF could be:

I. Right-angled --> NO!  
If triangle DEF (a,b,c) is right-angled, then it must be true that a^2+b^2=c^2. However, this never satisfies the basic property of triangle ABC (i.e. a^2+b^2>c^2).
ELIMINATE OPTIONS (A),(D),(E)

II. Acute-Angled --> YES! 
Consider acute-angled triangle DEF (a=2,b=2,c=2). Then, triangle ABC (a^2=4, b^2=4, c^2=4)
ELIMINATE OPTIONS (C)

Just for cross check (not necessary), 
 III. Obtuse-angled --> NO!  
If triangle DEF (a,b,c) is obtuse-angled, then it must be true that a^2+b^2<c^2. However, this never satisfies the basic property of triangle ABC (i.e. a^2+b^2>c^2).

FINAL ANSWER IS (B) II only

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funkyakki · Answer

Answer should be E.

All of the sides of DEF are proportional to all of the sides of ABC, and there is no information given about the type of triangle ABC is, therefore DEF could be any of the triangle ABC could be which is all of the three.

ostrick5465 · Answer

The lengths of the sides of triangle ABC are a^2, b^2 and c^2 => a^2 + b^2 > c^2
=> triangle DEF couldn't be a Right-angled triangle.
 => Eliminate A D E

If DE = a, EF = b, FD = c => DE > EF + FD
If  DE^2 > EF^2 + FD^2 => \angle DFE > 90^{\circ}

=> Choice C

lacktutor · Answer

Let's say that c is the longest side of the triangle

In order ∆ DEF to be:
---> Right-angled triangle:  a^{2} +b^{2}= c^{2} 
---> Obtuse-angled triangle :  a^{2} +b^{2} < c^{2} 
---> Acute-angled triangle :  a^{2} +b^{2} > c^{2}

But  a^{2} ,  b^{2}  and  c^{2}  are also the lengths of the sides of ∆ ABC
--it could be only  a^{2} +b^{2} > c^{2} .

Acute-angled triangle  
Answer (B)

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If the lengths of the sides of triangle ABC are a^2, b^2 and c^2 and t

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If the lengths of the sides of triangle ABC are a^2, b^2 and c^2 and t

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