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In the diagram above, <PQR is a right angle, and QS is : Problem Solving (PS)
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In the diagram above, <PQR is a right angle, and QS is

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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]
Bunuel 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.

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

Hope it helps.

ya that helps , great
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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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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]
Bunuel 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.

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]
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]
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]
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]
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]
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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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]
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]
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 ?

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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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 ?

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]
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]
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

PS/QS = QS/SR
25*4 = QS^2 = 100
QS = 10

Area PQR = 1/2 * PR * QS = 1/2 * 29*10 = 145

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

Let's label the three angles in ∆PQR as follows

This means we can label the angles in the red and blue triangles as follows

This means, the remaining angles in the red and blue triangles are as follows:

Since the red and blue triangles have the SAME three angles, they are SIMILAR triangles.
This means the ratios of their corresponding sides must be equal.

In the blue triangle, the side BETWEEN the circle and the square has length 25
In the red triangle, the side BETWEEN the circle and the square has length h
So one ratio is: 25/h

In the blue triangle, the side BETWEEN the star and the square has length h
In the red triangle, the side BETWEEN the star and the square has length 4
So another ratio is: h/4

Since the ratios of corresponding sides must be equal, we can write: 25/h = h/4
Cross multiply to get: h² = 100

Solve: h = 10 or h = -10
Since the height must be POSITIVE, we can be certain that h = 10

So the area of ∆PQR = (base)(height)/2 = (29)(10)/2 = 145