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In the diagram above, <PQR is a right angle, and QS is

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In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post Updated on: 06 May 2012, 02:25
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Attachment:
Geometry.jpg
Geometry.jpg [ 3.7 KiB | Viewed 24641 times ]
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined

Originally posted by Joy111 on 05 May 2012, 19:37.
Last edited by Bunuel on 06 May 2012, 02:25, edited 1 time in total.
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 06 May 2012, 02:43
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6
BhaskarPaul wrote:
Attachment:
Geometry.jpg
Geometry.jpg [ 3.7 KiB | Viewed 24581 times ]
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined


Perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular QS divides right triangle PRQ into two similar triangles PQS and QRS (which are also similar to the big triangle PRQ). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).

So, PS/QS=QS/SR --> QS^2=PS*SR=25*4=100 --> QS=10 --> area of PQR equals to 1/2*PR*QS=1/2*(PS+SR)*QS=1/2*29*10=145.

Answer: B.

For more on check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.
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Re: Geometry  [#permalink]

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New post 06 May 2012, 02:16
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1
is it true?

25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **

from * and ** we have:

pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 06 May 2012, 15:58
Bunuel wrote:
BhaskarPaul wrote:
Attachment:
Geometry.jpg
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined


Perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular QS divides right triangle PRQ into two similar triangles PQS and QRS (which are also similar to the big triangle PRQ). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).

So, PS/QS=QS/SR --> QS^2=PS*SR=25*4=100 --> QS=10 --> area of PQR equals to 1/2*PR*QS=1/2*(PS+SR)*QS=1/2*29*10=145.

Answer: B.

For more on check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.



ya that helps , great
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Re: Geometry  [#permalink]

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New post 14 May 2012, 04:23
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mehdi2012 wrote:
is it true?

25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **

from * and ** we have:

pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145



This is also a good way to solve this problem if we find it difficult to apply similarity of triangles, though we have to find the squares of some numbers .
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Re: Geometry  [#permalink]

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New post 31 Jul 2013, 07:00
mehdi2012 wrote:
is it true?

25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **

from * and ** we have:

pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145






How do we figure out which side is proportional to which using angles? :S
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 01 Aug 2013, 20:37
Bunuel wrote:
BhaskarPaul wrote:
Attachment:
Geometry.jpg
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined


Perpendicular to the hypotenuse always divides the triangle into two triangles with the same properties as the original triangle.

Thus, the perpendicular QS divides right triangle PRQ into two similar triangles PQS and QRS (which are also similar to the big triangle PRQ). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles).

So, PS/QS=QS/SR --> QS^2=PS*SR=25*4=100 --> QS=10 --> area of PQR equals to 1/2*PR*QS=1/2*(PS+SR)*QS=1/2*29*10=145.

Answer: B.

For more on check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.


Corresponding angles are QPS and QRS right.
In that case, it must be PS/QS=SR/QS..

Please let me know where I am wrong...
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 24 May 2015, 20:27
Why do we write the proportion inversely as PS/QS=QS/QR? I wrote it as QS/SR=QS/PS and then got stuck from there but wondering why we choose to write the proportion that way.
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 24 May 2015, 21:54
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healthjunkie wrote:
Why do we write the proportion inversely as PS/QS=QS/QR? I wrote it as QS/SR=QS/PS and then got stuck from there but wondering why we choose to write the proportion that way.


What made you think that QS/SR=QS/PS is the correct relation?
I assume you found that the two small triangles are similar to each other. The point is how are they similar to each other? They are similar because they are both similar to the big triangle PQR.
Angle PQR = Angle QSP = angle QSR = 90 degrees
Ange P is common to PQR and PSQ so by AA, triangle PQR is similar to triangle PSQ - note the naming of the triangles. The angles which are equal are placed in corresponding positions. Angle P is common so it is the first vertex of each triangle. Then angle Q = angle S so we have Q and S as second vertices and the leftover as third vertices to get PQR and PSQ.

Similarly, angle R is common to triangle PQR and triangle QSR so by AA, triangle PQR is similar to triangle QSR - the naming of the triangles should be in order to ensure that you get the corresponding sides correctly.

So triangle PQR is similar to triangles PSQ and QSR. Now you know the corresponding sides:

PS/QS (Sides made by first two underlined letters)= SQ/SR (sides made by next two letters) = PQ/QR (sides made by first and third letters)
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 25 May 2015, 09:35
Ah that makes sense, I was assuming the smaller triangles were similar to eachother as opposed them both being similar because they're similar to the larger triangle. Makes sense now!
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 02 Jun 2015, 20:32
healthjunkie wrote:
Ah that makes sense, I was assuming the smaller triangles were similar to eachother as opposed them both being similar because they're similar to the larger triangle. Makes sense now!


Hi healthjunkie ,

All 3 triangles are similar. PQR ~ PSQ ~ QSR . U can try to prove it using AAA. Lemme know if u have any doubts on that.

Thanks!
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 02 Jun 2015, 22:02
when we draw perpendicular then the sides of the two triangle are in same ratio
small side of large triangle/ large side of large triangle = small side of small triangle/large side of small triangle
we know PQS is large triangle and QSR is smaller triangle
PQS and QSR share the same height QS and angle QSR = QSP which confirms that the sides must be in same ratio
thus we know PS = 25 and RS = 4
we can write
25/QS = QS/4
100 = QS^2
10 = QS
now we know height = 10 and PR = 25+4 = 29
area of the triangle = 1/2(10)(29) = 145
Its B
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 11 Apr 2016, 19:09
Joy111 wrote:
Attachment:
Geometry.jpg
In the diagram above, <PQR is a right angle, and QS is perpendicular to PR. If PS has a length of 25 and SR has a length of 4, what is the area of triangle PQR?

A. 125
B. 145
C. 240
D. 290
E. It cannot be determined


right from the start I knew there's something with the similar triangles...
P is the same, and 90 degree angle is the same. it means that Q1 angle is equal to R angle.
and Q2 angle is equal to P angle.

now we have 2 right triangles..similar to each other.
leg 25 and height x for example
leg 4 and height x.

so:
25 corresponds to the x side of the smaller triangle, scale factor thus must be 25/x
x corresponds to the side equal to 4. so scale factor is x/4
of course the scale factor is the same...so we can set these two equal: 25/x = x/4 = cross multiply -> x^2 = 100, x=10.
base=29, height =10. 29x10/2 = 145.
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 18 Mar 2017, 14:26
1
mehdi2012 wrote:
is it true?

25^2 + qs^2 = qp^2
4^2 + qs^2 = qr^2
and then (by sum them) you have:
625+16+2(qs^2)=qp^2+qr^2 *
and you know from the main triangle:
pq^2+qr^2=(pr)^2=(25+4)^2=841 **

from * and ** we have:

pq^2 +qr^2 = 841 = 625+16+2qs^2
2qs^2=841-641= 200
qs=10
then the area of PQR is:
10* (25+4)/2 = 145


correct
I fell into a trap with area calculation :) it's important not to forget to divide 29*10 by 2... :) sometimes when I'm happy to find solution I make stupid very trivial mistakes :)
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 10 Apr 2019, 22:27
I agree that these are similar triangle( AAA theorem). I am struggling to validate the correct corresponding sides.
For instance in two smaller triangle,
(1) PS/QS=QS/SR
(2) PS/QS=SR/QS

All mentioned sides are bases of a right angled triangle. We are not even given the angles that could have indicated the correct corresponding sides. Thus the could be either
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In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 25 Jun 2019, 18:07
VeritasKarishma in the naming part for 2nd triangle, the common angle is R for both the triangles (PQR and QSR) shouldn't it be the first letter in the naming of triangles ?

just as you did it for the above triangles PQR and PSQ ?



VeritasKarishma wrote:
healthjunkie wrote:
Why do we write the proportion inversely as PS/QS=QS/QR? I wrote it as QS/SR=QS/PS and then got stuck from there but wondering why we choose to write the proportion that way.


What made you think that QS/SR=QS/PS is the correct relation?
I assume you found that the two small triangles are similar to each other. The point is how are they similar to each other? They are similar because they are both similar to the big triangle PQR.
Angle PQR = Angle QSP = angle QSR = 90 degrees
Ange P is common to PQR and PSQ so by AA, triangle PQR is similar to triangle PSQ - note the naming of the triangles. The angles which are equal are placed in corresponding positions. Angle P is common so it is the first vertex of each triangle. Then angle Q = angle S so we have Q and S as second vertices and the leftover as third vertices to get PQR and PSQ.

Similarly, angle R is common to triangle PQR and triangle QSR so by AA, triangle PQR is similar to triangle QSR - the naming of the triangles should be in order to ensure that you get the corresponding sides correctly.

So triangle PQR is similar to triangles PSQ and QSR. Now you know the corresponding sides:

PS/QS (Sides made by first two underlined letters)= SQ/SR (sides made by next two letters) = PQ/QR (sides made by first and third letters)
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 25 Jun 2019, 22:14
1
anmolgmat14 wrote:
VeritasKarishma in the naming part for 2nd triangle, the common angle is R for both the triangles (PQR and QSR) shouldn't it be the first letter in the naming of triangles ?

just as you did it for the above triangles PQR and PSQ ?



VeritasKarishma wrote:
healthjunkie wrote:
Why do we write the proportion inversely as PS/QS=QS/QR? I wrote it as QS/SR=QS/PS and then got stuck from there but wondering why we choose to write the proportion that way.


What made you think that QS/SR=QS/PS is the correct relation?
I assume you found that the two small triangles are similar to each other. The point is how are they similar to each other? They are similar because they are both similar to the big triangle PQR.
Angle PQR = Angle QSP = angle QSR = 90 degrees
Ange P is common to PQR and PSQ so by AA, triangle PQR is similar to triangle PSQ - note the naming of the triangles. The angles which are equal are placed in corresponding positions. Angle P is common so it is the first vertex of each triangle. Then angle Q = angle S so we have Q and S as second vertices and the leftover as third vertices to get PQR and PSQ.

Similarly, angle R is common to triangle PQR and triangle QSR so by AA, triangle PQR is similar to triangle QSR - the naming of the triangles should be in order to ensure that you get the corresponding sides correctly.

So triangle PQR is similar to triangles PSQ and QSR. Now you know the corresponding sides:

PS/QS (Sides made by first two underlined letters)= SQ/SR (sides made by next two letters) = PQ/QR (sides made by first and third letters)


The triangles can be named in any way. They can even be PQR and SRQ are similar. But, for our ease of finding corresponding sides, we name them as per corresponding angles
So PQR => angle P, then 90 degrees angle, then angle R
and QSR => angle Q, then 90 degrees angle, then angle R

You switch to RQP and RSQ if you wish but it makes no difference.
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Re: In the diagram above, <PQR is a right angle, and QS is  [#permalink]

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New post 02 Jul 2019, 06:53
For the triangle PQR would'nt the height be PQ base be QR and hypotenuse be PR. Bunuel could you explain this
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Re: In the diagram above, <PQR is a right angle, and QS is   [#permalink] 02 Jul 2019, 06:53
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