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

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Math Expert
Joined: 02 Sep 2009
Posts: 64073
The length of sides of triangle PQR are p, q and r units. The length  [#permalink]

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20 Feb 2020, 04:14
10
00:00

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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Joined: 25 Jul 2018
Posts: 712
Re: The length of sides of triangle PQR are p, q and r units. The length  [#permalink]

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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

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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