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In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = ang

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In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = ang [#permalink] New post 08 Jun 2011, 16:18
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In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = angle (QPR). What is the ratio of perimeter of the triangle PRQ to that of the triangle RSQ.

Can somebody please tell me how to find the second identical angles (1st one given in question) in these similar triangles?.
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Re: Triangle [#permalink] New post 08 Jun 2011, 17:29
in triangle PQR, QR=20, SQ=16 and angle(QRS) = angle (QPR).

how is this possible? if above 2 angles are equal, QR = SQ, something mistyped i think.
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Re: Triangle [#permalink] New post 08 Jun 2011, 18:29
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maheshsrini wrote:
In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = angle (QPR). What is the ratio of perimeter of the triangle PRQ to that of the triangle RSQ.

Can somebody please tell me how to find the second identical angles (1st one given in question) in these similar triangles?.

here we know that QR=20, SQ=16, RS=8
From the property of triangle we can say that,
RS² = PS*SQ
8²= PS*16
Thus, PS = 4

Now according to Pythagoras theorem we can calculate PR.
PR² = PS² +RS²
PR² = 4² + 8² = 16 + 64 = 80
So, PR = 8.94 ≈ 9

Now ratio of Perimeter of two triangle is equal to

Perimeter (▲PRQ)/ Perimeter (▲RSQ) = [20+(16+4)+9]/[20+16+8]= 49/44 Ans.
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Re: Triangle [#permalink] New post 08 Jun 2011, 20:49
MrMicrostrip,

In the problem they have not mentioned that RS is perpendicular to PQ. We can't assume to solve the problem right?
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Re: Triangle [#permalink] New post 08 Jun 2011, 22:28
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this is a similar triangle problem
let RP=x,PQ=y
given :- angle QRS= angle QPR,
angle PQR and angle SQR = common ,
so triangle PQR is similar to triangle SQR by AA.
now we will apply the condition for similar triangle
QR/QS=RP/SR=PQ/RQ
20/16=x/8=y/20
therefore x=10,y=25.
perimeter of triangle PRQ=55:perimeter of triangle SRQ=44
ratio=5:4 (ans)
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[#permalink] New post 09 Jun 2011, 00:21
p091987 wrote:
this is a similar triangle problem
let RP=x,PQ=y
given :- angle QRS= angle QPR,
angle PQR and angle SQR = common ,
so triangle PQR is similar to triangle SQR by AA.
now we will apply the condition for similar triangle
QR/QS=RP/SR=PQ/RQ
20/16=x/8=y/20
therefore x=10,y=25.
perimeter of triangle PRQ=55:perimeter of triangle SRQ=44
ratio=5:4 (ans)


This is the most apt way to solve this . My answer is exact same .
has to be 5:4
the trick is understand similar angles
Angle QRS ~ Angle QPR
therefore
QR/QP = QS/QR=RS/PR
Put in the values given , and your good to go
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Re: Triangle [#permalink] New post 09 Jun 2011, 01:55
kichha51 wrote:
MrMicrostrip,

In the problem they have not mentioned that RS is perpendicular to PQ. We can't assume to solve the problem right?

yaa vijandra I saw what I did :shock: ...It's a blunder.. :stupid
I think P091987 did it in correct way.. :thumbup:
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Re: Triangle [#permalink] New post 09 Jun 2011, 01:58
Ferocious wrote:
Nice explanation MrMicrostrip....+1 kudos 4 u...

Sorry dude my explanation is not correct :( .
Follow the explaination :arrow: of "P091987"..
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Re: Triangle [#permalink] New post 09 Jun 2011, 02:27
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Hey you can figure out that they are similar triangles when u separate the diagram PFA the diagram that i ahve made.

in the diagram we can see that angle RPQ = angle SRQ and angle RQP = angle SQP

Since two angles are the same these triangles are similar traingles and u can use ratios of corresponding sides to solve the problem for example RP/20 = 20/16 RP = 10

Similarly PQ = 25

So when we calculate the ratio it is 55:44 so 5:4 is the answer.

Cheers!
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Re: Triangle [#permalink] New post 10 Jun 2011, 20:17
maheshsrini wrote:
In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = angle (QPR). What is the ratio of perimeter of the triangle PRQ to that of the triangle RSQ.

Can somebody please tell me how to find the second identical angles (1st one given in question) in these similar triangles?.


Are these two angles QRS and QPR - congruent or equal?
if they are equal as mentioned in the Q then i am not sure how is Q itself correct. I guess everyone is assuming them congruent and solving. Please confirm.
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Re: Triangle [#permalink] New post 10 Jun 2011, 20:21
agdimple333 wrote:
maheshsrini wrote:
In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = angle (QPR). What is the ratio of perimeter of the triangle PRQ to that of the triangle RSQ.

Can somebody please tell me how to find the second identical angles (1st one given in question) in these similar triangles?.


Are these two angles QRS and QPR - congruent or equal?
if they are equal as mentioned in the Q then i am not sure how is Q itself correct. I guess everyone is assuming them congruent and solving. Please confirm.


How can they not be equal??? and most definitely i have solved this based on the concept of equal angles and not congruency.

Please check the first diagram and relate the diagram to what i have drawn in my previous post I'm sure it helps.
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Re: Triangle [#permalink] New post 14 Jul 2011, 00:44
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It's not even necessary to calculate all the sides, nor the actual value of the perimeter.

Once you recognize the two similar triangles, you see QR (value=20) of the "big triangle" corresponds to SQ (value=16) of the "small triangle." Similarity rules dictate that the ratio of perimeters will be the same as the ratio of corresponding sides. 20/16 equals 5/4.
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Re: Triangle [#permalink] New post 14 Jul 2011, 01:35
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Hi, can anyone help in solvin this:
How many triangles can be constructed so that lengths of the sides are three consecutive odd integers and the perimeter is less than 1000?
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Re: Triangle [#permalink] New post 14 Jul 2011, 02:29
maheshsrini wrote:
In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = angle (QPR). What is the ratio of perimeter of the triangle PRQ to that of the triangle RSQ.

Can somebody please tell me how to find the second identical angles (1st one given in question) in these similar triangles?.


1st is given in question - 2nd is the common angle (q) and third is the remaing one.
to find sides - just take the opposite sides.
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Re: Triangle [#permalink] New post 14 Jul 2011, 04:09
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kamnlesh wrote:
Hi, can anyone help in solvin this:
How many triangles can be constructed so that lengths of the sides are three consecutive odd integers and the perimeter is less than 1000?


I'll try to reason this out:
The perimeter of any such triangle would be 3 times the length of the medium side, as it is also the average side length. 3 times the middle side length must be less than 1000, meaning the 331, 333, 335 triangle is the biggest possible. The smallest triangle is composed of 3, 5, 7 (anything smaller would be impossible).

How many odd numbers are from 5-333, inclusive?

By my count, the difference (328) divided by 2 (164) adding in the extra one (the smallest triangle) equals:
165 triangles

Did anyone else get that answer?
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Re: Triangle [#permalink] New post 14 Nov 2013, 01:37
NY's solution procedure is correct but he did not explain as to how did he get the sides of the triangle with the largest perimeter. Well here goes. Firstly you have to remember that the sum of odds always is odd. As there are three consecutive odds their sum should also be odd. The greatest odd number less than 1000 is 999. Now, the question is what are the three consecutive odds whose sum is 999? Let the three consecutive odds be 2n+1, 2n+3, 2n+5. All these three are of the form (even+odd) which equals odd. Now, odd+odd+odd=even+odd=odd=999. Now make the equation as 2n+1+2n+3+2n+5=999. We get 6n=990--->3*2n=990---->2n=330. Therefore the three consecutive odds are 331, 333 and 335.
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Re: Triangle [#permalink] New post 14 Nov 2013, 02:23
3,5,7; 5, 7, 9; 7,9,11; 9,11,13; ................327,329,331; 329,331,333; 331,333,335. The list here represents the entire set of possible triangles. Observe that 3 is being displayed only one time. Also, the last number 335 is also displayed one time. 5 on the other hand appears two times. 7, 9, 11.......331 appear three times. Also in the end 333- two times and 335-one time. Now, as three CONSECUTIVE odds making a triplet you have to take into consideration the repetitions when arriving at the number of triangles. Total odds between 3 and 335 inclusive is [(335-3)/2]+1=167. Now we have to isolate the number of odds which repeat themselves 3 times, 2 times and which do not repeat(3 and 335) at all. Therefore 3 timers=167-2(1 timers)-2(2 timers)= 163. Now, 163 is to be multiplied with 3 to get all repetitions------------->163*3=489. To this we add 2(1 timers) and 4(as two 2 timers are being repeated 2 times). Hence 489+2+4=495. You must be wondering what does 495 represent? Well, it represents total number of elements in the triplet combinations. Each triangle consists of 3 sides. So, we divide 495 by 3 to get the possible triangles which is 165.
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Re: In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = ang [#permalink] New post 02 Jan 2014, 09:56
maheshsrini wrote:
Attachment:
Tri1.jpg
In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = angle (QPR). What is the ratio of perimeter of the triangle PRQ to that of the triangle RSQ.

Can somebody please tell me how to find the second identical angles (1st one given in question) in these similar triangles?.


What is the trick to understand what sides are in proportion and set up the ratios? I'm always having to draw the two triangles to visualize what side matches with the other.

Does anybody know how to best approach these problems when dealing with similar triangles and know what sides are the correct to compare?

Thanks a lot!
Will provide kudos for good answers!!

Cheers
J :)
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Re: In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = ang [#permalink] New post 03 Jan 2014, 03:18
Expert's post
jlgdr wrote:
maheshsrini wrote:
Attachment:
Tri1.jpg
In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = angle (QPR). What is the ratio of perimeter of the triangle PRQ to that of the triangle RSQ.

Can somebody please tell me how to find the second identical angles (1st one given in question) in these similar triangles?.


What is the trick to understand what sides are in proportion and set up the ratios? I'm always having to draw the two triangles to visualize what side matches with the other.

Does anybody know how to best approach these problems when dealing with similar triangles and know what sides are the correct to compare?

Thanks a lot!
Will provide kudos for good answers!!

Cheers
J :)


In similar triangles corresponding sides are opposite similar angles.

For more check here: http://www.mathopenref.com/similartriangles.html

Hope this helps.
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Re: In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = ang [#permalink] New post 29 Jan 2014, 05:03
Bunuel wrote:
jlgdr wrote:
maheshsrini wrote:
Attachment:
Tri1.jpg
In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = angle (QPR). What is the ratio of perimeter of the triangle PRQ to that of the triangle RSQ.

Can somebody please tell me how to find the second identical angles (1st one given in question) in these similar triangles?.


What is the trick to understand what sides are in proportion and set up the ratios? I'm always having to draw the two triangles to visualize what side matches with the other.

Does anybody know how to best approach these problems when dealing with similar triangles and know what sides are the correct to compare?

Thanks a lot!
Will provide kudos for good answers!!

Cheers
J :)


In similar triangles corresponding sides are opposite similar angles.

For more check here: http://www.mathopenref.com/similartriangles.html

Hope this helps.


Got it. I think all one has to do is recognize the three angles and their corresponding sides.

That is, for this problem we have that RQS is the common angle for both triangles. We also have an unstated angle that is equal in both triangles (SRQ = QPR) Therefore, one can establish the following relationships

RQ/PQ = SQ/RQ = RS/PR ---> 20/y = 16/20 = 8/x

Given that x = PR and y=PQ

Or as NY pointed out if we know one relationship SQ/RQ = 4/5 then we know that the perimeters are also in the ratio 4:5, or 5:4 since we are trying to find the bigger/smaller.

Just my 2c

Cheers!!
J :)
Re: In the triangle PQR, QR=20, SQ=16, RS=8 and angle(QRS) = ang   [#permalink] 29 Jan 2014, 05:03
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