The length of sides of triangle PQR are p, q and r units. The length

Question

The length of sides of triangle PQR are p, q and r units. The length of sides of triangle STU are s, t and u units. The following relations hold true:

p(q+r-p)=s^2 
 q(p+r-q)=t^2 
 r(p+q-r)=u^2

The type of triangle STU is

(A) Acute angled
(B) Right angled
(C) Obtuse angled
(D) Either A or B
(E) Either A or C

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lacktutor · Answer

We got 3 types of triangles: —> acute— s^{2} + t^{2} < u^{2} —> obtuse — s^{2} + t^{2} > u^{2} —right — s^{2} + t^{2} = u^{2} Let’s say that triangle STU is right angled triangle: s^{2} + t^{2} = u^{2} —> In order STU to be a right angled triangle, we have to prove that triangle PQR is a triangle (p+q > r, p+r > q or r+q > p) —> we got (p —q)^{2} = r^{2} —> p—q = r ( it shouldn’t be a triangle) —> STU is not a right-angled triangle. Let’s say that it’s an obtuse angled triangle. — s^{2} + t^{2} < u^{2} —> we got (p—q)^{2} < r^{2} (p—q—r)(p—q+r) < 0 (p+r—q ) — always greater than zero —> p—q—r < 0 —> p< q+r ( it is one of the features of a triangle) —> STU is an obtuse-angled triangle. Let’s say that it is an acute-angled triangle s^{2} + t^{2} > u^{2} —> we got (p—q)^{2} > r^{2} (p—q—r)( p—q+r) > 0 —> p+r—q is always greater than zero. But —> p—q—r > 0 p > q+r ( it is WRONG) Triangle STU can be Obtuse—angled triangle The answer is C Posted from my mobile device

krahil · Answer

Explanation above is good but with a minor correction in conditions for both acute and obtuse angled triangles. The inequalities needs to be reversed as below:
—> acute -  s^{2} + t^{2}> u^{2} 
—> obtuse -  s^2 + t^2 < u^2

Everything remains same in the explanation, just that obtuse must be replaced with acute and vice-versa.
This will change your posted answer from "C" to "A" (which also matches official answer)

gmealsocks · Answer

It's really easy if you know the law of cosine, which I don't know if that's expected for the gmat. c^2=a^2+b^2-2abcosC.

Not the most rigorous, but... expand S^2 (would work for U^2 and T^2 too), p^2=r^2-2rq+q^2 notice this resembles the above law of cosine expression. Notice also since r and q are positive, since they're magnitudes, then -2rq is negative. Therefore, one can infer that cos<90, as it would be 0 if 90deg and >0 if greater than 90deg.

yashu5279 · Answer

The best way to solve this question is knowing that :

s^2 + t^2 < u^2 --> Obtuse

s^2 + t^2 = u^2 --> Right Angled

s^2 + t^2 > u^2 --> Acute

So here we just try to find out s^2+t^2 - u^2 = r^2 - (p-q)^2

which can be written as (r-p+q)*(r+q-p) --> This is always positive because the p,q,r are sides of triangle and sum of two sides is always greater than 3rd side.

so in effect we just achieved s^2+t^2 > u^2. So the triangle is acute! Option A is correct answer.

cyrwr1 · Answer

Can anyone explain in a different way how A is attained?

bumpbot · Answer

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The length of sides of triangle PQR are p, q and r units. The length

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The length of sides of triangle PQR are p, q and r units. The length

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards