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

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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 \((\frac{1}{2})*12*6 = 36\) 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

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 -> \(72 - [1/2*6*1]\)\(= 72 - 3 = 69\)
2. 2 -> \(72 - [1/2*6*2]\)\(= 72 - 6 = 66\)
3. 3 -> \(72 - [1/2*6*3]\)\(= 72 - 9 = 63\)
4. 4 -> \(72 - [1/2*6*4]\)\(= 72 - 12 = 60\)
5. 5 -> \(72 - [1/2*6*5]\)\(= 72 - 15 = 57\)

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

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

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.

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

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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