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# In the diagram above, <PQR is a right angle, and QS is

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Manager
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In the diagram above, <PQR is a right angle, and QS is [#permalink]  05 May 2012, 19:37
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Geometry.jpg [ 3.7 KiB | Viewed 973 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 May 2012, 02:25, edited 1 time in total.
Edited the question
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]  06 May 2012, 02:43
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Attachment:

Geometry.jpg [ 3.7 KiB | Viewed 960 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.

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

Hope it helps.
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Re: Geometry [#permalink]  06 May 2012, 02:16
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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
Manager
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Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]  06 May 2012, 15:58
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
Manager
Joined: 18 Oct 2010
Posts: 82
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Kudos [?]: 55 [0], given: 26

Re: Geometry [#permalink]  14 May 2012, 04:23
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 .
Re: Geometry   [#permalink] 14 May 2012, 04:23
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