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In the diagram above, <PQR is a right angle, and QS is
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Updated on: 06 May 2012, 01:25
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Question Stats:
64% (02:16) correct 36% (02:40) wrong based on 273 sessions
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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
Re: In the diagram above, <PQR is a right angle, and QS is
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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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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
Re: In the diagram above, <PQR is a right angle, and QS is
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06 May 2012, 01:43
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BhaskarPaul wrote:
Attachment:
Geometry.jpg [ 3.7 KiB | Viewed 21487 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.
Re: In the diagram above, <PQR is a right angle, and QS is
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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.
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 .
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
Re: In the diagram above, <PQR is a right angle, and QS is
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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.
Re: In the diagram above, <PQR is a right angle, and QS is
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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.
Re: In the diagram above, <PQR is a right angle, and QS is
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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!
Re: In the diagram above, <PQR is a right angle, and QS is
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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!
Re: In the diagram above, <PQR is a right angle, and QS is
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02 Jun 2015, 21:02
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
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11 Apr 2016, 18:09
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.
Re: In the diagram above, <PQR is a right angle, and QS is
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18 Mar 2017, 13:26
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
Re: In the diagram above, <PQR is a right angle, and QS is
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64% (02:16) correct 
it's important not to forget to divide 29*10 by 2... 
