ABCD is a quadrilateral

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pesfunk: Master | Next Rank: 500 Posts; Posts: 286; Joined: Tue Sep 21, 2010 5:36 pmFollowed by: 5 members; Location: Kolkata, India; Thanked: 11 times

by pesfunk » Fri Apr 15, 2011 5:41 pm

Could someone help me with this one please ?

ABCD is a quadrilateral. A rhombus is a quadrilateral whose sides are all congruent. BCEF is a
rhombus and shares a common side with the quadrilateral ABCD. The area of which one is greater:
ABCD or BCEF ?
(1) ABCD is a square.
(2) BCEF is not a square.

OA C

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Source: Beat The GMAT — Data Sufficiency | Fri Apr 15, 2011 5:41 pm

vineeshp: Legendary Member; Posts: 965; Joined: Thu Jan 28, 2010 12:52 amFollowed by: 34 members; Thanked: 156 times; GMAT Score:720

by vineeshp » Fri Apr 15, 2011 9:53 pm

TOugh one to explain.

A square that shares the same side length with a rhombus will be larger than the rhombus.

Just cannot figure out how to explain this.

Vineesh,
Just telling you what I know and think. I am not the expert.

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Sat Apr 16, 2011 1:20 am

hi guys let me try
(1) if abcd is square that shares the the side with rhombus bcef then they have equal sides say-x
and the area of square=x^2,
the area of rombus will=x^2*sin(b) where sinb is the value of angle formed by the sides od rombus
if sin(b)=1 (sin90=1) then their areas are equal x^2*1=x^2
sin can`t be greater that 1, so if angle between sides of rombus is less than 90. sin will be less then 1 and product of x^2*(value less that 1)will be <then simple x^2
but we are not given any property of rombus
and 1 st insuff
(2)rombus is not a square, it happens that the area of rombus=x^2*sinb, and sinb<1
but we have no info about abcd
both
area of rombus=x^2*sinb, where sinb<1
area of square=x^2
clearly that area of square is greater then that of rombus
both suff
i agree this is hard one problem

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rohu27: Legendary Member; Posts: 586; Joined: Tue Jan 19, 2010 4:38 amFollowed by: 5 members; Thanked: 31 times; GMAT Score:730

by rohu27 » Sat Apr 16, 2011 2:46 am

Let me try to explain this without heavy math involved.

now ABCD is a quadrilateral. BCEF is a rhombus. both share common side BC. ABCD can be anythng-sqare,rhombus,rectangle.
now stat1: abcd is sqauare. so area=x^2 (assuming x is the side) so the rhombus too has all sides as x. area of rhombus can alo be expresses as b*h where b is one side (As all sides are equal) h is the height.here b=x
now we need to knw whether x^2 is greater than or less than x*h. how to find the height of rhombus?
u will need to draw the pic to get this better. the opp sides in a rhombus are parallel.
once u draw a perpendicular from one side to the other, you can see a right triangle with one of the sides as hypotenuse.
so whtevr the measurement of the height it WILL be less than the hypotenuse. so h is always less than b. this i enough to say that area of square is >area of rhombus
BUT
stat1 just says that BCEF is a rhombus. square is also a type of rhombu, in which case both areas will eb equal.
so we need stat 2 also which confirms it is not the case.

using 1 & 2 its clear than area of ABCD is more.
main point here is to establish for a given side area of square is > area of rhombus

hope im not missing anythgn here in my explanation

Have time? https://www.beatthegmat.com/730-q-50-v40 ... 88058.html

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conceptd: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Sat Dec 17, 2011 9:54 pm

by conceptd » Wed Jul 04, 2012 5:57 am

This question can be very easily solved with a property of a quadilateral:-

A square has a larger area than any other quadrilateral with the same perimeter

Now for (1) ABCD is a square, but we know nothing about BCEF. Another property is

If the diagonals of a rhombus are equal, then that rhombus must be a square. Now as we see,

we only know that BCEF is a Rhombus, we know nothing about the angle. Hence (1) insufficient.

(2) Insufficient

(1) + (2), here we get that BCEF is not a square, hence clearly Area ABCD > Area BCEF Sufficient

Hence C

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