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In the diagram above, <PQR is a right angle, and QS is [#permalink]
05 May 2012, 18:37

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

64% (02:08) correct
36% (01:28) wrong based on 55 sessions

Attachment:

Geometry.jpg [ 3.7 KiB | Viewed 2282 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

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

Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
06 May 2012, 01:43

2

This post received KUDOS

Expert's post

BhaskarPaul wrote:

Attachment:

Geometry.jpg [ 3.7 KiB | Viewed 2265 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.

Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
06 May 2012, 14:58

Bunuel wrote:

BhaskarPaul 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.

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 .

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

Re: In the diagram above, <PQR is a right angle, and QS is [#permalink]
01 Aug 2013, 19:37

Bunuel wrote:

BhaskarPaul 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.