Triangle Question

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Triangle Question

by tpetel » Tue Feb 17, 2009 5:10 am
Please see attatchment

Given that:

(1) QR=RS
(2) ST=TU

x=?

Together they should satisfy a solution, but I don't see it... any help?
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Source: — Data Sufficiency |

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by DanaJ » Tue Feb 17, 2009 5:36 am
Well, it is obvious that taken separately, the stmts are not enough.
Taken together, things change.
1. tells us that triangle QRS is isosceles. Since the sum of its angles is 180 degrees (like with any other triangle) and it is isosceles (with measurement RQS = RSQ), this makes the measurement of RSQ = [180 - m(QRS)]/2
2. tells us that triangle UTS is isosceles as well. This means that we have m(UST) = [180 - m(UTS)]/2.
Now, x will be 180 - m(RSQ) - m(UST) = 180 - 180 + [m(QRS) + m(UTS)]/2 = [m(QRS) + m(UTS)]/2
Now here comes the magic: since triangle PRT is a right triangle and angle RPT is 90 degrees, this makes m(QRS) + m(UTS) = 180 - 90 = 90 degrees.
Thus you have measurement of x = 90/2 = 45.

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by tini » Tue Feb 17, 2009 6:45 am
we know <x+<qsr+<tsr=180..eqn 1

<qrs+<uts=90....eqn 2 as rpt is a right angled triangle

from isosceles trangle rqs we get 2<qsr+<qrs=180, or 2<qsr=180-<qrs...eqn 3

from isosceles triangle tur we get 2<tsu=180-<uts..eqn 4

adding eqn 3 & 4,
2(<qsr+<tsu)=2*180-(<qrs+<uts)=360-90=270 (substitute value from eqn 3)
<qsr+<tsu=135

from eqn 1 , <x=180-135=45

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by earth@work » Tue Feb 17, 2009 8:32 am
x=45
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