In the figure above, the area of the shaded region is

Question

https://gmatclub.com/forum/download/file.php?id=50467 In the figure above, the area of the shaded region is A. 8\sqrt{2} B. 4\sqrt{3} C. 4\sqrt{2} D. 8(\sqrt{3} - 1) E. 8(\sqrt{2} - 1) PS57302.01 2019-09-21_1806.png

BrentGMATPrepNow · Accepted Answer

Let's see what might happen if we add an auxiliary line to our diagram as follows:
 https://i.imgur.com/fvayv5G.png 
This tells us that:  area of shaded region =   ( area of ∆ABC ) - ( area of ∆ABD )

Let's start by finding the length of AB. 
To do this we can focus on the red triangle below.
Since we have a right triangle, we can apply the Pythagorean theorem to write: 4² + 4² = (side AB)²
Simplify: 32 = (side AB)²
Find the square root of both sides: √32 = side AB
Simplify:   4√2 = side AB  
When we add this to our diagram we get:
 https://i.imgur.com/3m6VJgK.png

At this point, we can see that  ∆ABC  is an  equilateral triangle

Area of equilateral triangle  = (\frac{\sqrt{3}}{4})(side^2)

So, the area of ∆ABC  = (\frac{\sqrt{3}}{4})(4\sqrt{2})^2= (\frac{\sqrt{3}}{4})(32)   =8\sqrt{3}   
 https://i.imgur.com/kserQj6.png

Area of triangle  = \frac{(base)(height)}{2} 
So, the area of ∆ABD  = \frac{(4)(4)}{2}    =8  
 https://i.imgur.com/qAGISWz.png

We're now ready to answer the question...
 area of shaded region =   ( area of ∆ABC ) - ( area of ∆ABD )

=   8\sqrt{3}   -   8

=  8(\sqrt{3}-1)

Answer: D

Cheers,
Brent

CareerGeek · Answer

Area of the square = 4^2 = 16
Diagonal of the square =  4\sqrt{2} 
So, the horizontal diagonal of the square forms an equilateral triangle of side  4\sqrt{2} 
--> Area of equilateral triangle =  \sqrt{3} /4* (4\sqrt{2})^2  =  8\sqrt{3}

--> Shaded Area = Area of equilateral triangle - 1/2(Area of square)
=  8\sqrt{3} - 8  
=  8(\sqrt{3} - 1)

chetan2u · Answer

Join BC as shown, BC= 4\sqrt{2} , as it is diagonal of square of side 4
Area of shaded region=Area of equilateral triangle ABC - Area of isosceles right angled triangle with equal sides as4 -BC =  (4\sqrt{2})^2*\sqrt{3}/4-\frac{1}{2}*4*4=8\sqrt{3}-8 = 8(\sqrt{3} - 1)

D

hemantbafna · Answer

Beautiful question!

Please find attached the explanation.

GMATWhizTeam · Answer

Solution:

Please refer to the image here:

Attachment:

triangle.png [ 51.65 KiB | Viewed 10942 times ]

Area of shaded region = Area of equilateral triangle \(ABC\) - Area of right isoceles triangle \(BCD\)

\(⇒ \frac{\sqrt3 }{4} \times (4\sqrt2)^2 - \frac{1}{2}\times 4\times 4\)

\(⇒ 8\sqrt3 - 8\)

\(⇒ 8(\sqrt3 -1)\)

Hence the right answer is Option D.

Neghineh · Answer

Could someone please explain to me how to calculate this step by step?  ⇒ 8\sqrt3 - 8

⇒ 8(\sqrt3 -1)

GMATWhizTeam

gmatt1476

Or anyone else?

DanTheGMATMan · Answer

Create an equilateral triangle by drawing a diagonal through the square. The diagonal will be 4 \sqrt{2}  because the sides of the square are 4.

The area of an equilateral triangle is  3^{2}  \sqrt{3} /4

so that will be 8 \sqrt{3} .

Subtract the area of the half of the square that's you've just included in the equilateral triangle, so

8 \sqrt{3}  - 8

In the figure above, the area of the shaded region is

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In the figure above, the area of the shaded region is

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