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What is the perimeter of quadrangle ABCD in which a circle

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Senior Manager
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What is the perimeter of quadrangle ABCD in which a circle [#permalink] New post 21 Mar 2008, 21:16
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A
B
C
D
E

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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] New post 21 Mar 2008, 22:28
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Expert's post
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.
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Re: DS: perimeter of quadrangle [#permalink] New post 22 Mar 2008, 05:54
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] New post 02 Apr 2008, 10:41
Nice explanation..i get why its A..
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Re: DS: perimeter of quadrangle [#permalink] New post 03 Apr 2008, 05:14
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] New post 03 Apr 2008, 05:38
Expert's post
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] New post 09 Apr 2008, 11:51
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] New post 09 Apr 2008, 13:18
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. :-D
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Re: DS: perimeter of quadrangle [#permalink] New post 09 Apr 2008, 21:53
Expert's post
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] New post 10 Apr 2008, 00:11
very nice explanation Walker!
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Re: DS: perimeter of quadrangle   [#permalink] 10 Apr 2008, 00:11
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