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## Kaplan: Similar right angle Triangles

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shubhamkumar Rising GMAT Star
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Kaplan: Similar right angle Triangles Sun Apr 15, 2012 10:47 am
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• Lap #[LAPCOUNT] ([LAPTIME])
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 Traingle PQR ?
A.125
B.145
C.240
D.290
E.It cannot be determined

OA: B
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Neo Anderson Rising GMAT Star
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Sun Apr 15, 2012 11:06 am
from similarity of triangles PQS & QRS

we get QS / 25 = 4 / QS

thus QS^2 = 100 => QS = 10

Area of triangle PQR = 1/2* QS* PR = 1/2*10*(25+4) = 290/2= 145 hence B.

shubham_k Just gettin' started!
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Sun Apr 15, 2012 11:48 am
First of all we can easily prove the similarity of triangles PQS & QRS by AAA test of similarity (/_qps = /_rqs). Now for similar triangles we have ratio proportional so QS / 25 = 4 / QS. Thus qs= 10. So area of triangle is 1/2*base*height = 1/2* qs*pr = 145. Thus B.

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 9:07 pm
shubhamkumar wrote:
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 Traingle PQR ?
A.125
B.145
C.240
D.290
E.It cannot be determined

OA: B
Using the Pythagoras Theorem:
In triangle PQR, PR² = PQ² + QR²
(25 + 4)² = PQ² + QR²
841 = PQ² + QR² ...Equation (1)

In triangle QSR, QR² = SQ² + SR²
QR² = SQ² + 4² ...Equation (2)

In triangle PQS, PQ² = SQ² + SP²
PQ² = SQ² + 25² ...Equation (3)

From equations (2) and (3), put the value of QR² and PQ² in equation 1,
841 = SQ² + 25² + SQ² + 4²
2SQ² = 841 - 625 - 16
2SQ² = 200
SQ² = 100
SQ = 10
Area of triangle PQR = (1/2) * 29 * 10 = 29 * 5 = 145 sq units

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spider Just gettin' started!
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Sun Apr 15, 2012 9:44 pm
Neo Anderson wrote:
from similarity of triangles PQS & QRS

we get QS / 25 = 4 / QS

thus QS^2 = 100 => QS = 10

Area of triangle PQR = 1/2* QS* PR = 1/2*10*(25+4) = 290/2= 145 hence B.
I did not get How come triangles PQS & QRS are similar?

Neo Anderson Rising GMAT Star
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Mon Apr 16, 2012 1:18 am
spider wrote:
Neo Anderson wrote:
from similarity of triangles PQS & QRS

we get QS / 25 = 4 / QS

thus QS^2 = 100 => QS = 10

Area of triangle PQR = 1/2* QS* PR = 1/2*10*(25+4) = 290/2= 145 hence B.
I did not get How come triangles PQS & QRS are similar?
from the figure /_ RSQ = /_ SPQ + /_ SQP = 90 (external /_ of a triangle) ---- (A)

also we know, /_ RQP = /_RQS + /_ SQP = 90 -----(B)

from (A) and (B) we get /_ RQS = /_ SPQ

also we know both triangles are right angled triangles

thus all three angles of both the traingle are equal!

from AAA similarity we can say that

triangle PQS and QRS are similar!

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