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What is the perimeter of quadrangle ABCD in which a circle [#permalink]
 21 Mar 2008, 22:16
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What is the perimeter of quadrangle ABCD in which a circle is inscribed? 1) AB+DC=8 2) BC=5
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Re: DS: perimeter of quadrangle [#permalink]
21 Mar 2008, 23:28
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A1) AB+DC=8 LB=BM, MC=CN, ND=DP, PA=AL P=AB+BC+DC+DA=(AB+DC)+(BC+DA)=8+(BM+MC+DP+DA)=8+(BL+NC+DN+AL)=8+(AB+DC)=16 Sufficient 2) insufficient By the way, we have a typical pattern of GMAT DS trap (A,B vs C): the fist condition is complex and seems to be insufficient the second condition is very simple and insufficient. It would help to support fist condition.
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Re: DS: perimeter of quadrangle [#permalink]
22 Mar 2008, 06:54
walker wrote: A
1) AB+DC=8
LB=BM, MC=CN, ND=DP, PA=AL
P=AB+BC+DC+DA=(AB+DC)+(BC+DA)=8+(BM+MC+DP+DA)=8+(BL+NC+DN+AL)=8+(AB+DC)=16
Sufficient
2) insufficient
By the way, we have a typical pattern of GMAT DS trap (A,B vs C):
the fist condition is complex and seems to be insufficient
the second condition is very simple and insufficient. It would help to support fist condition. Thanks walker. excellent explanation!
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Re: DS: perimeter of quadrangle [#permalink]
02 Apr 2008, 11:41
Nice explanation..i get why its A..
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Re: DS: perimeter of quadrangle [#permalink]
03 Apr 2008, 06:14
pls explain why
LB=BM, MC=CN, ND=DP, PA=AL
what is the formula for this ?
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Re: DS: perimeter of quadrangle [#permalink]
03 Apr 2008, 06:38
shobuj wrote: pls explain why
LB=BM, MC=CN, ND=DP, PA=AL
what is the formula for this ? Let O is the center of the circle. OL=OM - the radius of the circle OLB=90, OMB=90 - properties of a tangent line OB is common for triangles OLB and OMB So, triangles OLB and OMB are equal, and LB=BM.
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Re: DS: perimeter of quadrangle [#permalink]
09 Apr 2008, 12:51
thanks walker nice explaination u make me greatful to u by ur several explaination but here is i wanna know though OB is common and OL=OM why LB=LM is this another properties if both <OLB=<OMB AND OL=OM therefore LB=LM
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Re: DS: perimeter of quadrangle [#permalink]
09 Apr 2008, 14:18
shobuj wrote: thanks walker nice explaination u make me greatful to u by ur several explaination but here is i wanna know though OB is common and OL=OM why LB=LM is this another properties if both <OLB=<OMB AND OL=OM therefore LB=LM I think you mean LB=BM, BO is the common side with B being the middle point. 
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Re: DS: perimeter of quadrangle [#permalink]
09 Apr 2008, 22:53
shobuj wrote: though OB is common and OL=OM why LB=LM is this another properties if both <OLB=<OMB AND OL=OM therefore LB=LM We have here two right triangles with the same hypotenuse and the same leg. So, we can express LB and MB as sqrt(OB^2-OL^2) and sqrt(OB^2-OM^2). As OL=OM --> LB=MB
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Re: DS: perimeter of quadrangle [#permalink]
10 Apr 2008, 01:11
very nice explanation Walker!
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Re: DS: perimeter of quadrangle
[#permalink]
10 Apr 2008, 01:11
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