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

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Tue Dec 11, 2018 12:28 am

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[Math Revolution GMAT math practice 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?

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

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Source: Beat The GMAT — Problem Solving | Tue Dec 11, 2018 12:28 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Dec 11, 2018 7:56 am

Max@Math Revolution wrote:[Math Revolution GMAT math practice 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?

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

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.

We have a free video on this topic as well as other assumptions we can make about diagrams on the GMAT: https://www.gmatprepnow.com/module/gmat-geometry?id=863

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.

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 . . .

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

Brent Hanneson - Creator of GMATPrepNow.com

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

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

by fskilnik@GMATH » Tue Dec 11, 2018 1:30 pm

Max@Math Revolution wrote:[Math Revolution GMAT math practice 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?

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

Very nice problem, Max. Congrats!

$$? = {S_{{\rm{shaded}}}} = {S_{\Delta {\rm{ADC}}}} - {S_{\Delta {\rm{AEF}}}}$$
$${S_{\Delta {\rm{ADC}}}} = {{{S_{{\rm{square}}}}} \over 2}\,\, = \,\,{{12 \cdot 12} \over 2}\,\, = \,\,72$$
$${S_{\Delta {\rm{AEF}}}}\,\, = \,\,\,{{AE \cdot {h_{AE}}} \over 2}\,\,\, = \,\,\,3 \cdot {h_{AE}}\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,12$$
$$\left( * \right)\,\,\left\{ {\,\,\,\left. \matrix{
\Delta {\rm{AEF}} \sim \Delta CBF\,\, \hfill \cr
{\rm{ratio}}\,\,{\rm{of}}\,\,{\rm{similarity}}\,\,AE:BC = 1:2\,\,\, \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{{{h_{AE}}} \over {{h_{BC}}}} = {1 \over 2}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{h_{AE}} = {1 \over 3}\left( {AB} \right) = 4} \right.$$
$$?\,\, = \,\,72 - 12\,\, = \,\,60$$

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

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Thu Dec 13, 2018 4:56 am

=>

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.

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 (1/2)*12*6 = 36 and the area of triangle BCF is (1/2)*12*8 = 48.
The area of the shaded quadrilateral CDEF is 144 - 36 - 48 = 60.

Therefore, the answer is C.
Answer: C

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