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Geometery DS Q.. i jus dun get it

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by rohit_gmat » Sun Oct 23, 2011 3:30 am
Image

In the figure above (see attachment), segment BD has length 6. What is the length of segment AC?

1) The area of quadrilateral ABCD is 60

2) The length of segment AD is 10


OA A

Is there any concept to be used here? Is it a parallelogram or smth?? plz help.

thanks!
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by Amiable Scholar » Sun Oct 23, 2011 4:47 am
No this is not parallelogram but definitely a trapezium as angle ABD = angle BDC
so line AB || DC
Now here
AC^2= BD ^2 + (Projection of AC over DC ) ^2
NOW Project of AC over DC = DC + AB (its visible from figure)
so we have BD , We just need sum of AB and DC here

now looking over first statement
area of quadilateral = 60
or
1/2*AB*BD + 1/2*BD*DC = 60
BD*(AB + DC )=120
Which gives us value of AB + DC so 1 is alone sufficient

now looking over second statement
AD = 10
only can give us the value of AB (using pytha theorem )alone no info about DC
so statement 2 is not alone sufficient .
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by Cheese12 » Sun Oct 23, 2011 11:56 pm
rohit_gmat wrote:Image

In the figure above (see attachment), segment BD has length 6. What is the length of segment AC?

1) The area of quadrilateral ABCD is 60

2) The length of segment AD is 10


OA A

Is there any concept to be used here? Is it a parallelogram or smth?? plz help.

thanks!


I solved it this way....

from the question stem : we can derive that (AB+DC)^2 + 6^2 = AC^2 ....
( I got this eq by drawing an imaginary line down from A to say point Z. Then we can say that length ZC= AB+DC ... and we know that AZ=6=BD )

Statement 1 -> area = 60 = (1/2*DC*6)+ (1/2*AB*6) = 60 ----> AB+DC= 20 --> SUFF
Statement 2 -> AD =10 .. from this we can find out AB.. but still we dont know DC.. so INSUFF..

Answer is A.
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by GMATGuruNY » Mon Oct 24, 2011 10:43 am
rohit_gmat wrote:Image

In the figure above (see attachment), segment BD has length 6. What is the length of segment AC?

1) The area of quadrilateral ABCD is 60.

2) The length of segment AD is 10.


OA A

Is there any concept to be used here? Is it a parallelogram or smth?? plz help.

thanks!
Since both AB and CD are perpendicular to BD, AB is parallel to CD.
Thus, ABCD is a trapezoid whose bases are AB and CD and whose height is BD=6.
Area of a trapezoid = (b1 + b2)/2 * h.
Thus, area of ABCD = (AB+CD)/2 * 6 = 3(AB+CD).

Statement 1: The area of quadrilateral ABCD is 60.
Thus:
3(AB+CD) = 60
AB+CD = 20.
Image
In the figure above, since AE is perpendicular to CE, quadrilateral ABDE is a rectangle and ∆ACE is a right triangle:
In rectangle ABDE, AE=BD=6 and DE=AB.
Thus, in ∆ACE, CE = AB+CD = 20.
Since ∆ACE is a right triangle, 6² + 20² = AC².
Thus, the length of AC can be determined.
SUFFICIENT.

Statement 2: The length of segment AD is 10.
Image
The figures above illustrate that AC can be different lengths.
INSUFFICIENT.

The correct answer is A.
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by factor26 » Tue Oct 25, 2011 4:38 pm
Since both AB and CD are perpendicular to BD, AB is parallel to CD.
Thus, ABCD is a trapezoid whose bases are AB and CD and whose height is BD=6.
Area of a trapezoid = (b1 + b2)/2 * h.
Thus, area of ABCD = (AB+CD)/2 * 6 = 3(AB+CD)
how can we say this is a trapezoid?
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by GMATGuruNY » Wed Oct 26, 2011 4:56 am
factor26 wrote:
Since both AB and CD are perpendicular to BD, AB is parallel to CD.
Thus, ABCD is a trapezoid whose bases are AB and CD and whose height is BD=6.
Area of a trapezoid = (b1 + b2)/2 * h.
Thus, area of ABCD = (AB+CD)/2 * 6 = 3(AB+CD)
how can we say this is a trapezoid?
A trapezoid is a quadrilateral with at least 2 parallel sides.
Since AB || CD, ABCD is a trapezoid.
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