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In the diagram to the right, triangle PQR has a right angle [#permalink]

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29 Mar 2014, 00:17

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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?

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
_________________

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 diagram to the right, triangle PQR has a right angle [#permalink]

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16 Jun 2014, 00:35

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?

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 [#permalink]

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16 Jun 2014, 05:24

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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.

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 [#permalink]

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06 Jul 2014, 10:30

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 [#permalink]

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17 Jul 2015, 14:09

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the diagram to the right, triangle PQR has a right angle [#permalink]

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04 Aug 2016, 22:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the diagram to the right, triangle PQR has a right angle [#permalink]

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02 Aug 2017, 21:33

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

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