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GMATPrep - Isoceles triangle, value of x?

Expert replies
by California4jx » Sun May 17, 2009 8:56 am
how do we find two isosceles triangles values ? OA: C
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Source: — Data Sufficiency |

by DanaJ » Sun May 17, 2009 9:48 am
This question has been answered before, but here goes:

1 tells you that QR = RS, which makes QRS an isosceles triangle. This means that angles RQS and RSQ share the same measurement. Since the sum of angles in a triangle is 180, then RQS + RSQ + QRS = 180.

Say RQS = RSQ = a. Then you get that a + a + QRS = 180, so a = (180 - a)/2.

This is not enough to solve the problem though.

2 tells you that ST - UT, making SUT an isosceles triangle. Again, you have that angles UST = SUT = b and UST + SUT + STU = 180. This means that 2b + STU = 180, or b = (180 - STU)/2.

Again, this alone is insufficient.

But put the two together to find x. Notice that RSQ + x + UST = 180, or in other words that a + b + x = 180. Replace a and b to get:

(180 - a)/2 + (180 - b)/2 + x = 180

180 - (a + b)/2 + x = 180

x = (a + b)/2

But notice that a and b are actually the non-right angles of triangle PRT. This means that their sum will be 180 - (measurement of right angle) = 180 - 90 = 90.

So x = (a + b)/2 = 90/2 = 45.

Answer C.
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by cramya » Sun May 17, 2009 9:59 am
Stmt I

RQ=RS

ANGLE RQS = ANGLE RSQ = lets say each = angle a

No way to find x

INSUFF

Stmt II

ST = TU

ANGLE TUS = ANGLE TSU = lest say angle b each

no way to find x

INSUFF

TOGETHER:

In isoceles triangle RQS

angle R = 180 - 2a (since a+a+angle R = 180 for any traingle)

In isolceles triangle SUT

angle T = 180-2b


However in traingle RPT

we know 90 + angle R + angle T = 180

90 + 180 - 2a + 180-2b = 180

a+b = 135

Also

x+a+b = 180 (angles in a striaght line on one side add up to 180)

x = 180-135
= 45


No need to solve once u know u can find x but just for FYI I have solved

Hope this helps.

C
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by aj5105 » Sun May 17, 2009 10:16 am
We get 2 angles (other than 90)of triangle RPT to be 67.5
x = 45

(C)
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by California4jx » Sun May 17, 2009 12:32 pm
Thanks to all !
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