In the figure below, what is the area of the region shown?

Question

(A) \(8 \sqrt{3}\)

(B) 16

(C) \(12 + 4 \sqrt{3}\)

(D) \(8 + 8 \sqrt{3}\)

(E) \(16 + 8 \sqrt{3}\)

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2020-12-22_15-58-07.png [ 18.4 KiB | Viewed 2356 times ]

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Official Answer

E

imeanup · Answer

Consider  	riangle ABC ,  AB^2=AC^2-BC^2=8^2-4^2=48

AB=4\sqrt{3}

Area of  	riangle ABC =\frac{1}{2}*BC*AB=\frac{1}{2}*4\sqrt{3}*4=8\sqrt{3}

We can find the  \angle BCA = 30^{\circ}  and  \angle ACB=60^{\circ}  using  \frac{PPB}{HHB} . Therefore,  \angle CAD=60^{\circ}  and  \angle DCA=30^{\circ}  which makes both  	riangle ABC \approx 	riangle ADC  (similar).

Area of  	riangle ADC =8\sqrt{3}

But we don't have any such option  16\sqrt{3} .

So, the  	riangle ADC  is an isoceles  	riangle , therefore,  AD=DC

AC^2=AD^2+DC^2=2AD^2=>AD=4\sqrt{2}

Area of  	riangle ADC = \frac{1}{2}*4\sqrt{2}*4\sqrt{2}=16

Total-Area = 16+8\sqrt{3}

IMO  Ans E

TarunKumar1234 · Answer

Area of Triangle ABC = 1/2 *BC*AB = 1/2*4*4Sqt3= 8*Sqrt(3)

Now, we have Option D and E to pickup.

AC=8 so, AD+DC >AC=8, So, both sides AD and DC will be greater than 4.

Hence, Total Area = 8*Sqrt(3) + 1/2 * (>4)*(>4)

so, Option E.

ShankSouljaBoi · Answer

Please explain the highlighted. Flipping a triangle over the other just for juxtaposition does not prove similarity. Further, I think u have used RHS to prove similarity, which is not the case here either.

ShankSouljaBoi · Answer

Property used: Perpendicular from right angle divides the hypotenuse equally. Check the image. Its E

Bunuel , I think this is a 700+ question

ShankSouljaBoi · Answer

to add to ur solution for clarity , lets try to decipher how did you get to AD and DC each has to be greater than 4. Let AD be a and DC be b Then, for the triangle ADC to exist a-b < AC < a+b a-b < 8 < a+b .... condition 1 ::::::::> a+b/2 > 4 Now using AM >= GM (Something greater than 4) >= GM. The maximum GM can be when a = b So, (Something greater than 4) >= sqrt (a^2) ::::::::> (Something greater than 4) >= a .... condition 2 We will have to drop = in condition 2 because if a = b = 4 ... condition 1 is violated as 8 < a+b ====> 8 < 8 Therefore we can conclude that both a and b are greater than 4 .

imeanup · Answer

ShankSouljaBoi

You haven't read the complete explanation. There are two possibilities here. Both are being mentioned in the explanation. Moreover, the triangles are "similar" based on AAA triangle similarity properties. However, that is not the case here based on the answer choices the triangle is isosceles.

ShankSouljaBoi · Answer

The figure is not drawn to scale. I did read your explanation completely, and your contention that only two possibilities exist is false. What if I have 29 degrees and 61 degrees and 90 degrees. It's not mentioned that scaling has been considered. Further, how did you get to AAA, please show via image?

imeanup · Answer

Yes, you are correct. It is not similar;  ShankSouljaBoi

riangle ABC , 
 BC=4 
 AB=4\sqrt{3}

Using  AB^2+BC^2=AC^2  Pyth Theorem since the  \angle B =90^{\circ}  we can find the length of  AB

As the triangle is a right triangle, we can do the following trigonometry;
 \frac{BC}{AB}=tan θ^{\circ}

Still, we don't know the  θ^{\circ} . So, we can find that now:
 \frac{4}{4\sqrt{3} }= \frac{1}{\sqrt{3}}=tan 30^{\circ}

Therefore,  \angle CAB=30^{\circ}  and  \angle ACB=60^{\circ}

riangle ADC .  A+D=90^{\circ}  but it is not  30,60  or  60,30

ShankSouljaBoi · Answer

takeaway is two dimensions don't give u similarity 3 do. SSS AAA SAS RHS ASA are ok. ASS IS NOT

Posted from my mobile device

Fdambro294 · Answer

Just because the 2 triangles share a hypotenuse does not necessarily mean that these 2 triangles are similar.

The perpendicular height drawn to the hypotenuse does not bisect the triangle. It does so when the right triangle is Isoscles.

The height drawn from the 90 degree vertex to the hypotenuse will create 3 similar triangles within that same triangle.

But we do not have any other lengths besides the hypotenuse = 8

Anyone find a way to solve this problem?

In the figure below, what is the area of the region shown?

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GMAT Problem Solving (PS) Questions

In the figure below, what is the area of the region shown?

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