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In the diagram to the right, triangle PQR has a right angle
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Updated on: 29 Mar 2014, 02:25
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83% (02:31) correct 17% (02:48) wrong based on 202 sessions
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In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?
Re: In the diagram to the right, triangle PQR has a right angle
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29 Mar 2014, 02:39
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sudeeptasahu29 wrote:
In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?
According to the above, triangles PQR, PSQ and QSR must be similar.
Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR --> QS^2 = PS*SR = 16*9 --> QS = 4*3 = 12.
The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150.
Re: In the figure to the right, triangle PQR has a right angle
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29 Mar 2014, 00:54
1
2
sudeeptasahu29 wrote:
Triangle PQR has a right angle at Q and the line segment QS is perpendicular to PR. If line segment PS = 16 and SR has a length of 9, what is the area of triangle PQR?
a. 72 b. 96 c. 108 d. 150 e. 200
In right triangle PQR right angled at Q, QS^2 = PS * SR --------> qs^2 = 16*9 ------> qs = 4*3 = 12
Area = \(\frac{1}{2} QS*PR\) -----> \(\frac{1}{2} 12*25\) ------> 150
_________________
Re: In the diagram to the right, triangle PQR has a right angle
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15 Jun 2014, 23:35
Bunuel wrote:
sudeeptasahu29 wrote:
In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?
According to the above, triangles PQR, PSQ and QSR must be similar.
Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR --> QS^2 = PS*SR = 16*9 --> QS = 4*3 = 12.
The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150.
Answer: D.
Hope it's clear.
Hello Bunnel, how can we ascertain that PSQ and QSR but not PSQ and RSQ that are similar?
Re: In the diagram to the right, triangle PQR has a right angle
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15 Jun 2014, 23:38
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Kconfused wrote:
Bunuel wrote:
sudeeptasahu29 wrote:
In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?
According to the above, triangles PQR, PSQ and QSR must be similar.
Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR --> QS^2 = PS*SR = 16*9 --> QS = 4*3 = 12.
The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150.
Answer: D.
Hope it's clear.
Hello Bunnel, how can we ascertain that PSQ and QSR but not PSQ and RSQ that are similar?
Triangle QSR is the same as RSQ.
_________________
Re: In the diagram to the right, triangle PQR has a right angle
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16 Jun 2014, 04:24
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Thanks Bunnel! Wouldn't that mix up the corresponding sides? Obviously QSR and RSQ when matched with PSQ would result in different set of corresponding sides My question is, how can we find the right set of corresponding sides to match up in similar triangles? Thank you in advance! I've always had this problem with similar triangles.
Re: In the diagram to the right, triangle PQR has a right angle
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18 Jun 2014, 05:03
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Kconfused wrote:
Thanks Bunnel! Wouldn't that mix up the corresponding sides? Obviously QSR and RSQ when matched with PSQ would result in different set of corresponding sides My question is, how can we find the right set of corresponding sides to match up in similar triangles? Thank you in advance! I've always had this problem with similar triangles.
Re: In the diagram to the right, triangle PQR has a right angle
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06 Jul 2014, 09:30
Kconfused wrote:
Thanks Bunnel! Wouldn't that mix up the corresponding sides? Obviously QSR and RSQ when matched with PSQ would result in different set of corresponding sides My question is, how can we find the right set of corresponding sides to match up in similar triangles? Thank you in advance! I've always had this problem with similar triangles.
i had the same issue as in finding corresponding sides in similiar triangle. i suggest you watch some videos in khan academy relating to similiar triangles. there are few videos pertaining to it .watch them they are really helpful.
Re: In the diagram to the right, triangle PQR has a right angle
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02 Aug 2017, 20:33
Can we solve this by using the following approach?? the area of triangle PQR = area of triangle PQS+ area of QRS But the side QS must be equal to what ?? IS that equal to PS or SR Kindly help
Re: In the diagram to the right, triangle PQR has a right angle
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27 Oct 2017, 23:19
Bunuel wrote:
sudeeptasahu29 wrote:
In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?
A. 72 B. 96 C. 108 D. 150 E. 200
Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: http://gmatclub.com/forum/if-arc-pqr-ab ... l#p1154669
According to the above, triangles PQR, PSQ and QSR must be similar.
Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR --> QS^2 = PS*SR = 16*9 --> QS = 4*3 = 12.
The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150.
Re: In the diagram to the right, triangle PQR has a right angle
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28 Oct 2017, 00:07
zanaik89 wrote:
Bunuel wrote:
sudeeptasahu29 wrote:
In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?
A. 72 B. 96 C. 108 D. 150 E. 200
Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: http://gmatclub.com/forum/if-arc-pqr-ab ... l#p1154669
According to the above, triangles PQR, PSQ and QSR must be similar.
Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR --> QS^2 = PS*SR = 16*9 --> QS = 4*3 = 12.
The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150.
Answer: D.
Hope it's clear.
But 16+9 is the hypotenuse of the triangle PQR
Why are we using it as the height?
We are considering PR = 25 as the base of triangle PQR and QS as the height:
The area of triangle PQR = 1/2*base*height = 1/2*PR*QS = 1/2*(16+9)*12 = 150.
_________________
Now Area of the triangle is (\(\frac{1}{2}\))*base*hieght So, area of triangle PQR is (\(\frac{1}{2}\))*PR*QS = (\(\frac{1}{2}\))*(16+9)*12 = (\(\frac{1}{2}\))*25*12 = 150
Hence D
_________________
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Re: In the diagram to the right, triangle PQR has a right angle
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27 Mar 2018, 09:49
sudeeptasahu29 wrote:
Attachment:
Capture.PNG
In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?
A. 72 B. 96 C. 108 D. 150 E. 200
Standard type of right triangles are with sides 3 4 5 , 5 12 13, 8 15 17, and 7 24 25. 1) In this question we have hypotenuse as 25 (a multiple of 5) so we can choose 3 4 5 and 5 12 13. 2) Now hypotenuse is the largest side of a right triangle , this leaves only 3 4 5. In this case multiply each side of 3 4 5 with 5 (since hypotenuse is a multiple of 5) we get 15 20 25. Now calculate the area! (15*20)/2=150
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Re: In the diagram to the right, triangle PQR has a right angle &nbs
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27 Mar 2018, 09:49
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83% (02:31) correct 
