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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  [#permalink]

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New post 20 Feb 2020, 04:14
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

25% (01:56) correct 75% (02:07) wrong based on 29 sessions

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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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Re: The length of sides of triangle PQR are p, q and r units. The length  [#permalink]

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New post 20 Feb 2020, 07:15
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

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Re: The length of sides of triangle PQR are p, q and r units. The length   [#permalink] 20 Feb 2020, 07:15

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

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