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

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

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Updated on: 29 Mar 2014, 02:25
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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?

A. 72
B. 96
C. 108
D. 150
E. 200

Originally posted by sudeeptasahu29 on 28 Mar 2014, 23:17.
Last edited by Bunuel on 29 Mar 2014, 02:25, edited 1 time in total.
Renamed the topic and edited the question.
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Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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29 Mar 2014, 02:39
3
7
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: if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html#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.

Hope it's clear.
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Re: In the figure to the right, triangle PQR has a right angle  [#permalink]

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

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29 Mar 2014, 01:00
Dear Narenn,

Thank you for the reply. Please explain how you got this relation QS^2 = PS * SR??
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Re: In the figure to the right, triangle PQR has a right angle  [#permalink]

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29 Mar 2014, 01:39
2
2
sudeeptasahu29 wrote:
Dear Narenn,

Thank you for the reply. Please explain how you got this relation QS^2 = PS * SR??

It's a standard relationship based on similar triangle theory.

We should note that triangle PQR, triangle PSQ, and triangle QSR are similar

In similar triangles, ratios of corresponding sides are equal.

We know that triangle PSQ is similar to triangle QSR

so, $$\frac{base of PSQ}{height of PSQ} = \frac{base of QSR}{height of QSR}$$

$$\frac{QS}{PS} = \frac{RS}{QS}$$ --------> $$QS^2$$ = PS*RS

Similarly you can also prove that $$PQ^2$$= PS*PR and $$QR^2$$ = SR*PR
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Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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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?

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: if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html#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.

Hope it's clear.

Hello Bunnel, how can we ascertain that PSQ and QSR but not PSQ and RSQ that are similar?
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Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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15 Jun 2014, 23:38
1
1
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?

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: if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html#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.

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

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16 Jun 2014, 04:24
1
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.
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Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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18 Jun 2014, 05:03
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Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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

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09 Dec 2016, 01:53
Hi Bunnel,

I tried solving this problem by using the 45-45-90 triangle approach for triangle QSR !! Why is this approach incorrect ?
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Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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09 Dec 2016, 01:59
anuj11 wrote:
Hi Bunnel,

I tried solving this problem by using the 45-45-90 triangle approach for triangle QSR !! Why is this approach incorrect ?

First of all, you don't show your work at all and asks what's wrong with it. Next, why do you assume that QSR is 45-45-90 triangle?
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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, 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
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Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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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.

Hope it's clear.

But 16+9 is the hypotenuse of the triangle PQR

Why are we using it as the height?
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Posts: 50730
Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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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.

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

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28 Oct 2017, 11:04
2
Here is the basic mathematical way of solving this question.

Attachment:
File comment: Image related to solution of this problem

Traingle.jpg [ 9.23 KiB | Viewed 12105 times ]

Let QR = x, PQ = y and QS = z
Given PS = 16 and SR = 9

Triangle PQR is right angle triangle
$$x^2 + y^2 = (16+9)^2 = 25^2$$ -----------------------------------------Equation 1

Triangle QSR is right angle triangle

$$z^2 + 9^2 = x^2$$ ------------------------------------------------Equation 2

Triangle QPS is right angle traingle

$$z^2 + 16^2 = y^2$$ -----------------------------------------------Equation 3

Putting the values of x^2 and y^2 from equation 2 and 3 into equation 1, we get

$$z^2 + 9^2 + z^2 + 16^2 = 25^2$$

$$2z^2 + 9^2 + 16^2 = 25^2$$ ---------------------------------Equation 4

Solving equation 4 for z, we get z=12

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

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20 Dec 2017, 00:12
Hi.

Can we solve this by using the property : The median from right angle in a triangle is half of the hypotenuse?

Implying QS = 1/2* PR and then using Pythagorean Theorem ? Is this method valid as well ?
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Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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20 Dec 2017, 00:44
sandysilva wrote:
Hi.

Can we solve this by using the property : The median from right angle in a triangle is half of the hypotenuse?

Implying QS = 1/2* PR and then using Pythagorean Theorem ? Is this method valid as well ?

QS is not the median, it's height.
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Re: In the diagram to the right, triangle PQR has a right angle  [#permalink]

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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 [#permalink] 27 Mar 2018, 09:49
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