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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] New post 05 May 2012, 18:37
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Attachment:
Geometry.jpg
Geometry.jpg [ 3.7 KiB | Viewed 4062 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 May 2012, 01:25, edited 1 time in total.
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Re: Geometry [#permalink] New post 06 May 2012, 01:16
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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] New post 06 May 2012, 01:43
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BhaskarPaul wrote:
Attachment:
Geometry.jpg
Geometry.jpg [ 3.7 KiB | Viewed 4044 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: In the diagram above, <PQR is a right angle, and QS is [#permalink] New post 06 May 2012, 14:58
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] New post 14 May 2012, 03:23
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] New post 31 Jul 2013, 06:00
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] New post 01 Aug 2013, 19:37
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] New post 28 Feb 2015, 12:50
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink] New post 24 May 2015, 19:27
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] New post 24 May 2015, 20:54
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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] New post 25 May 2015, 08:35
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] New post 02 Jun 2015, 19:32
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] New post 02 Jun 2015, 21:02
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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
Re: In the diagram above, <PQR is a right angle, and QS is   [#permalink] 02 Jun 2015, 21:02
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