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

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Director
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Joined: 01 Oct 2017
Posts: 931
WE: Supply Chain Management (Energy and Utilities)
The length of sides of triangle PQR are p,q and r units. The length of  [#permalink]

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27 Jun 2018, 21:15
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Difficulty:

95% (hard)

Question Stats:

33% (03:37) correct 67% (02:27) 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
Math Expert
Joined: 02 Aug 2009
Posts: 7107
The length of sides of triangle PQR are p,q and r units. The length of  [#permalink]

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28 Jun 2018, 05:22
1
PKN wrote:
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

Difficult to understand ..
Then $$s^2, t^2 ,u^2$$ should tell you something

Now the observation when you square the three sides..

1) RIGHT angle-
If it is right angled triangle, the sum of square of two sides will be equal to square of third(hypotenuse)
$$(side_1)^2+(side_2)^2=(side_3)^2$$

2) ACUTE angle-
If you keep the two sides same and REDUCE the hypotenuse, what will happen??
Sum of square of the two sides will be more than the square of third side.
$$(side_1)^2+(side_2)^2>(side_3)^2$$

3) OBTUSE angle-
If you keep the two sides same and INCREASE the hypotenuse, what will happen??
Sum of square of the two sides will be less than the square of third side.
$$(side_1)^2+(side_2)^2<(side_3)^2$$

Now back to the question

$$p(q+r-p)=s^2$$..(i)
$$q(p+r-q)=t^2$$...(ii)
$$r(p+q-r)=u^2$$...(iii)

$$q(p+r-q)+p(q+r-p)=s^2+t^2.............qr+rp-(p^2+q^2-2pq)=s^2+t^2...........rp+rq-(p-q)^2=s^2+t^2$$...

from (iii)
$$r(p+q-r)=u^2......rp+rq-r^2=u^2.......rp+rq=u^2+r^2$$..

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

since p,q and r are three sides of a triangle and as per triangle INEQUALITY rule, any side of the triangle is more than the difference of other two sides. Hence r>|p-q| ..
therefore r^2>(p-q)^2.....r^2-(p-q)^2>0 and let that be equal to A

this further tells us that $$u^2+r^2-(p-q)^2=s^2+t^2........u^2+A=s^2+t^2$$...
thus $$s^2+t^2>r^2$$

also all equations are similar so we will get similar results for other sides THAT is SUM of square of two sides is greater than the third side..
hence it is ACUTE angled triangle

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

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28 Jun 2018, 07:49
chetan2u wrote:
PKN wrote:
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

Difficult to understand ..
Then $$s^2, t^2 ,u^2$$ should tell you something

Now the observation when you square the three sides..

1) RIGHT angle-
If it is right angled triangle, the sum of square of two sides will be equal to square of third(hypotenuse)
$$(side_1)^2+(side_2)^2=(side_3)^2$$

2) ACUTE angle-
If you keep the two sides same and REDUCE the hypotenuse, what will happen??
Sum of square of the two sides will be more than the square of third side.
$$(side_1)^2+(side_2)^2>(side_3)^2$$

3) OBTUSE angle-
If you keep the two sides same and INCREASE the hypotenuse, what will happen??
Sum of square of the two sides will be less than the square of third side.
$$(side_1)^2+(side_2)^2<(side_3)^2$$

Now back to the question

$$p(q+r-p)=s^2$$..(i)
$$q(p+r-q)=t^2$$...(ii)
$$r(p+q-r)=u^2$$...(iii)

$$q(p+r-q)+p(q+r-p)=s^2+t^2.............qr+rp-(p^2+q^2-2pq)=s^2+t^2...........rp+rq-(p-q)^2=s^2+t^2$$...

from (iii)
$$r(p+q-r)=u^2......rp+rq-r^2=u^2.......rp+rq=u^2+r^2$$..

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

since p,q and r are three sides of a triangle and as per triangle INEQUALITY rule, any side of the triangle is more than the difference of other two sides. Hence r>|p-q| ..
therefore r^2>(p-q)^2.....r^2-(p-q)^2>0 and let that be equal to A

this further tells us that $$u^2+r^2-(p-q)^2=s^2+t^2........u^2+A=s^2+t^2$$...
thus $$s^2+t^2>r^2$$

also all equations are similar so we will get similar results for other sides THAT is SUM of square of two sides is greater than the third side..
hence it is ACUTE angled triangle

A

Thank you for an excellent explanation!!
Should we solve such a question in GMAT or take a guess to save time? Because this will definitely eat up a lot of time.
_________________

Only those who risk going too far, can possibly find out how far one can go

Re: The length of sides of triangle PQR are p,q and r units. The length of &nbs [#permalink] 28 Jun 2018, 07:49
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