The quadrilateral ABCD is a square with sides of length 12.

Question

The quadrilateral ABCD is a square with sides of length 12. E is the mid-point of AD. F is the point of intersection of AC and BE. What is the area of the shaded quadrilateral CDEF?

Attachment:

12.11.png [ 10.38 KiB | Viewed 14904 times ]

A. 30
B. 45
C. 60
D. 75
E. 90

Show Answer

Official Answer

C

fskilnik · Accepted Answer

Very nice problem, Max. Congrats! (Kudos!)

https://i.ibb.co/hZmP0Lc/11-Dec18-5r.gif

? = {S_{{\rm{shaded}}}} = {S_{\Delta {\rm{ADC}}}} - {S_{\Delta {\rm{AEF}}}}

{S_{\Delta {\rm{ADC}}}} = {{{S_{{\rm{square}}}}} \over 2}\,\, = \,\,{{12 \cdot 12} \over 2}\,\, = \,\,72

{S_{\Delta {\rm{AEF}}}}\,\, = \,\,\,{{AE \cdot {h_{AE}}} \over 2}\,\,\, = \,\,\,3 \cdot {h_{AE}}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,12

\left( * \right)\,\,\left\{ {\,\,\,\left. \matrix{
 \Delta {\rm{AEF}} \sim \Delta CBF\,\, \hfill \cr 
 {\rm{ratio}}\,\,{\rm{of}}\,\,{\rm{similarity}}\,\,AE:BC = 1:2\,\,\, \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{{{h_{AE}}} \over {{h_{BC}}}} = {1 \over 2}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{h_{AE}} = {1 \over 3}\left( {AB} \right) = 4} \right.

?\,\, = \,\,72 - 12\,\, = \,\,60

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

BrentGMATPrepNow · Answer

This question could take a while to answer (especially if you don't know where to begin!

. Fortunately, we can take advantage of an IMPORTANT feature about geometric diagrams on the GMAT: The diagrams in GMAT problem solving questions are DRAWN TO SCALE unless stated otherwise. Since we are NOT told that the diagram is not drawn to scale, we can safely assume that it IS drawn to scale. So, we can use this fact to solve the question by simply "eyeballing" the diagram. Notice that if we draw a line straight down from point E, the square is cut into two EQUAL pieces. https://i.imgur.com/kt3I2Fl.png We know this because E is the mid-point of AD Since the area of the square is 144, the area of blue shaded region is 72 Notice that the area of the original gray shaded region is LESS THAN the area of the blue shaded region above. So, we know that the area of the original gray shaded region is LESS THAN 72 ELIMINATE D and E The next step will require some mental agility. If we mentally move the gray portion (highlighted in red) to the unshaded region below . . . https://i.imgur.com/mTrCNlS.png We can see that there will still be small region that is NOT shaded. So, the area of the original gray shaded region is just a little but LESS THAN 72. So, the correct answer must be C RELATED VIDEO https://www.youtube.com/watch?v=MaPuJh3m3Pk

MathRevolution · Answer

=>

To find the area of the shaded quadrilateral, we need to subtract the areas of triangles ABE and BCF from the area of square ABCD.

We can position the square in the xy-plane as shown in the diagram below.

Attachment:

12.13.png [ 23.05 KiB | Viewed 13652 times ]

The equations of AC and BE are \(y = -x+12\) and \(y = 2x\), respectively.
When we solve the system of equations: \(y = -x+12\) and \(y = 2x\) simultaneously, we obtain
\(–x + 12 = 2x, 3x = 12\) and \(x = 4\). Thus, the point F is \((4, 2*4) = (4,8).\)
The area of the square ABCD is \(12*12 = 144.\)
The area of triangle ABE is \((\frac{1}{2})*12*6 = 36\) and the area of triangle BCF is \((\frac{1}{2})*12*8 = 48.\)
The area of the shaded quadrilateral CDEF is \(144 – 36 – 48 = 60.\)

Therefore, the answer is C.
Answer: C

Pritishd · Answer

Sharing an alternate approach. The  area of the shaded portion  would be equal to the area of  ADC  - area of  AEF

1. Area of Triangle  ADC  =  1/2*AD*CD 
 ABCD  is a square with all sides of equal measure i.e.  12 , hence  AD = CD = 12 . Using this information, we can find area of triangle  ADC :
 1/2*AD*CD   =   1/2*12*12   =   72

2. Area of Triangle  AEF  =  1/2*AE*  perpendicular height from point  F  to base  AE 
We are being told that E is the midpoint of side AD hence  AE = ED = 6 . Now we only need to find the height of the perpendicular dropped on the base from point F.
In PS geometry the figures are drawn to scale unless explicitly called out. Hence as the point F is below the midpoint of AB, the perpendicular from F to the base will be have a length that would be less than 6 and greater than 0 i.e. 1, 2, 3, 4, 5.

Lets now plug in each of the 5 numbers in the below equation:

area of the shaded portion  = area of  ADC  - area of  AEF

1. 1 ->  72 -    = 72 - 3 = 69 
2. 2 ->  72 -    = 72 - 6 = 66 
3. 3 ->  72 -    = 72 - 9 = 63 
4. 4 ->  72 -    = 72 - 12 = 60 
5. 5 ->  72 -    = 72 - 15 = 57

Only 4 as the perpendicular length matches one of the answer choices hence answer is 60 (Ans.  C )

Krish728 · Answer

Hi,

I didn't understand the part where you are calculating the height of AE. How come height of AE/ height of BC becomes 1/3 (AB) from 1/2?

Can you please explain?

Thanks

fskilnik · Answer

Hi  Krish728  ,

Thank you for your interest in our approach/method.

We are dealing with two similar triangles, therefore not only their corresponding sides (as commonly used), but also their corresponding heights are related to the ratio of similarity of the triangles involved (in this case, 1:2)!

Much more on that is explored (and explained) in our preparation.

Regards and success in your studies,
Fabio.

anubhavshankar81 · Answer

Hi,

Can someone explain as to how options A and B can be eliminated in the event that the figure is not drawn to scale? It can be especially useful as this'll help in having a common approach to solve such problems regardless of the figure being drawn to scale or not.

Thank You.

Regards,
Anubhav

Rainman91 · Answer

PS questions are generally drawn to scale unless it is stated otherwise.

Also, even if you drew a 12 x 12 Square. The mid point E would be 6. You can then further divide the square into 4ths so the completely shaded region is 36. And since A to C is a diagonal, we know that the bottom right quadrant is half of 36 = 18.

So 36 eliminates A. 
36 + 18 = 54. Eliminates B.

Half of the square is not shaded. So it MUST be less than 72. D and E eliminated.

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The quadrilateral ABCD is a square with sides of length 12.

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GMAT Problem Solving (PS) Questions

The quadrilateral ABCD is a square with sides of length 12.

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