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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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3-4-5 triangle

let qr = x
then rs = 5-x

let pr = h

2 equations:

3^2 = h^2 + x^2
and
4^2 = h^2 + (5-x)^2

expand and substract them to get x = 9/5

substitute 9/5 into the first equation (or second, your choice)
to get
9 = h^2 + 81/25

solve the equation to get h = 12/5
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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This is quite a well-known diagram in mathematics: it can be used to prove the Pythagorean Theorem (using a and b for the two legs, and not 3 and 4). There are (at least) three entirely different ways to solve this problem, two of which were described above. Call the length we're looking for 'd':

-The area of the triangle must be the same no matter which base you choose. Thus 3*4/2 = 5*d/2 --> d = 12/5.

-Let QR = c; then QS = 5-c. We have two right angled triangles, and can use Pythagoras to set up two equations, two unknowns (c and d; this is the most time-consuming approach).

-The approach not mentioned above: notice that the two smaller triangles in the diagram are each similar to the 3-4-5 triangle. Because PQR is similar to QSP, we have d/3 = 4/5 --> d = 12/5.

It's the similarity of the triangles that can let you prove Pythagoras, incidentally.
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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PQ =3 and PS = 4,
So QS = 5

area of the triangle = 1/2 * PQ * PS = 1/2*QS*PR

OR, 1/2 * 3 * 4 = 1/2 * QS * PR
OR PR = 12/5
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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Be patient. I have not fully understood

Why 5/2 * PQ...should be QS ??

Secondly why my reasoning do not lead me to the answer and is flawed, completely wrong ???

We have a right triangle PQS where angle P is 90. Also two sides PS=3 and PQ= 4 so from this we have a 30-60-90 triangle.

So, if we look at angle R is 90, from this we can see that angle P is shared between PQS and PRS so P for PRS should be 60 (likewise angle P for triangle PRQ should be 30, so angle P is 60+30=90).

So, we have: for triangle PRS angle R is 90 (PS is the hypotenuse), angle P is 60 (RS long leg) and angle S is 30 (short leg PR).

If PS is 3 (hypotenuse opposite 90 angle) PR should be 1.5 (short leg opposite angle S that is 30). This based on ratio 30:60:90.

I know that it does not hold anywater, but is useful to understand :)

Thanks
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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carcass wrote:
Be patient. I have not fully understood

Why 5/2 * PQ...should be QS ??

Secondly why my reasoning do not lead me to the answer and is flawed, completely wrong ???

We have a right triangle PQS where angle P is 90. Also two sides PS=3 and PQ= 4 so from this we have a 30-60-90 triangle.

So, if we look at angle R is 90, from this we can see that angle P is shared between PQS and PRS so P for PRS should be 60 (likewise angle P for triangle PRQ should be 30, so angle P is 60+30=90).

So, we have: for triangle PRS angle R is 90 (PS is the hypotenuse), angle P is 60 (RS long leg) and angle S is 30 (short leg PR).

If PS is 3 (hypotenuse opposite 90 angle) PR should be 1.5 (short leg opposite angle S that is 30). This based on ratio 30:60:90.

I know that it does not hold anywater, but is useful to understand :)

Thanks


First of all 5/2 * PQ does not equal to QS. We equate the areas, which can be found in two ways:
1. 1/2*Leg1*Leg2 --> \(area=\frac{1}{2}*PQ*PS=6\);
2. 1/2*Perpendicular to hypotenuse*Hypotenuse --> \(area=\frac{1}{2}*PR*QS=\frac{5}{2}*PR\) (since hypotenuse QS=5);

Now, equate the areas: \(6=\frac{5}{2}*PR\) --> \(PR=\frac{12}{5}\).

Next, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\). In PQS sides PQ and PS are NOT in the ratio \(1 : \sqrt{3}\), so PQS is not a 30°, 60°, and 90° right triangle. I think you are mixing 3-4-5 Pythagorean Triples triangle with 30°-60°-90° triangle.

Hope it's clear.
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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As mentioned by Bunuel, there are other methods as well to solve this problem

One of the method is Similarity
Triangle QPS is similar to PRS (because one common side & angle is 3 & 90 degree)
So,
QS/QP = PS/PR
5/4 = 3/x
x = 12/5

Hope it helps.
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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prasannar wrote:


In ΔPQS above, if PQ =3 and PS = 4, then PR =?

A. 9/4
B. 12/5
C. 16/5
D. 15/4
E. 20/3

Project PS Butler : Question #88


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Attachment:
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the triangle can be figured out to be a 3:4:5 triangle..

when you see that there is a line from one of vertices to a side of the triangle (90deg) it cuts the triangle in half..
therefore,
triangle PSR is now 45 : 45 : and 90 (where line bisects the side QS)
and
QPR too, is a 45 : 45 : 90
as we know, the 45 : 45 sides will be of same length...
as QR is 5
we can say QR = RS = 2.5
2.5 is side RS
so PR will be 2.5

only answer closest to that is option B
feel free to correct my opinion
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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Shrey9 wrote:
when you see that there is a line from one of vertices to a side of the triangle (90deg) it cuts the triangle in half..


You are not cutting the triangle in half here - that would only be true if the large triangle was isosceles, and it's not. That's why your answer doesn't match any of the answer choices.
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
We we cannot solve it like this?

PRS will become a 45 - 45 - 90 triangele, so \(x\sqrt{2}= 4\) and solving it will lead us to:

PR = RS = \(2\sqrt{2}\)

What is wrong with it?

Kind regards!
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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jfranciscocuencag wrote:
PRS will become a 45 - 45 - 90 triangele, so \(x\sqrt{2}= 4\) and solving it will lead us to:


PRS is not a 45-45-90 triangle. The line PR would only cut the 90 degree angle at P perfectly in half if the big triangle was isosceles, and it is not isosceles.

You can see PRS can't be a 45-45-90 triangle by looking at the angles. In the big 3-4-5 triangle, the smallest angle is opposite the smallest side, so the smallest angle is at S. You'd need trigonometry to find that angle but it's definitely less than 45 degrees (it's actually roughly 37 degrees) if it's the smallest angle in a right triangle. But that angle at S is also one of the angles in triangle PRS, so PRS is not a 45-45-90 triangle.
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
Hi @Bunnel,

Just wanted to understand That is it not true that a line from the right-angle triangle to the hypotenuse is half of the Hypotenuse. Hence if we know that 5 is the Hypotenuse why is length not equal to 5/2?

Please let me know the flaw in my understanding, Thanks.
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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kavach wrote:
Hi @Bunnel,

Just wanted to understand That is it not true that a line from the right-angle triangle to the hypotenuse is half of the Hypotenuse. Hence if we know that 5 is the Hypotenuse why is length not equal to 5/2?

Please let me know the flaw in my understanding, Thanks.


The length of MEDIAN from a right angle to the hypotenuse is half the length of the hypotenuse. In our question, PR is not the median, it's perpendicular and perpendicular to the hypotenuse equals the median to the hypotenuse if and only a right triangle is isosceles.

Hope it helps.
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
One tip here if one is in trouble and cannot do it properly.

To narrow down the options, consider that QRP also needs to be a right triangle (PR is perpendicular to QS and forms with QS two adjacent 90 degrees angles).

So QP is the hypotenuse of QPR. Therefore, PR must be less than 3 (by definition), and that excludes the last 3 solutions, which are more than 3. Now you have a 50-50% chance that is better than a 20% chance of getting it right.

Feel free to comment if you find something wrong in my approach :)
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
Bunuel wrote:
The length of MEDIAN from a right angle to the hypotenuse is half the length of the hypotenuse. In our question, PR is not the median, it's perpendicular and perpendicular to the hypotenuse equals the median to the hypotenuse if and only a right triangle is isosceles.

Hope it helps.


Hey Bunuel - is there any post/thread where all such properties have been noted for reference?
Would be of great help!
Thanks!
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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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RajatJ79 wrote:
Bunuel wrote:
The length of MEDIAN from a right angle to the hypotenuse is half the length of the hypotenuse. In our question, PR is not the median, it's perpendicular and perpendicular to the hypotenuse equals the median to the hypotenuse if and only a right triangle is isosceles.

Hope it helps.


Hey Bunuel - is there any post/thread where all such properties have been noted for reference?
Would be of great help!
Thanks!




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Re: In PQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
This question can be solved in an easier way as follows—
ΔPQS and ΔPRS are similar angle triangle. So, we can express their analogous arms proportionately:
5/4=PR/3
PR=12/5
Ans. B

Posted from my mobile device
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