What is the area of the triangle in the figure above?

Question

https://gmatclub.com/forum/download/file.php?id=28755 What is the area of the triangle in the figure above? A. 2+2\sqrt{3} B. 4\sqrt{3} C. 8 D. 8\sqrt{3} E. 12 2015-12-27_2141.png

chetan2u · Accepted Answer

Hi,
the Q may be 700+, since you require to know to play around with Geometrical figures..
 the best way is to look how we can draw a 90 degree angle , which will help us to break the figure down to simpler terms 
  https://gmatclub.com/forum/download/file.php?mode=view&id=28776  ..
here we see the third angle C shouold be 45, so we draw a perpendicular from B to AC..
now we have two triangles ABD and BCD..
lets see the first triangle BCD..
angle BCD and CBD are 45 and CDB is 90...
 Hyp = BC= 2\sqrt{2} ...
so the sides BD and CD=2..
area of BCD=1/2 * 2*2=2...

lets see the second triangle ABD..
it is 30-60-90 triangle...
and therefore the sides BD:AD:AB are in ratio  1:\sqrt{3}:2 ..
BD is 2, so AD =  2\sqrt{3} ..
so  area = 1/2 * 2\sqrt{3}*2=2\sqrt{3} ..

total Area=2+ 2\sqrt{3} 
ans A

EMPOWERgmatRichC · Answer

Hi All,

Since the GMAT is built around existing patterns/rules/logic, the Geometry questions that you face on Test Day will almost always be based on standard shapes and formulas that you already know (or should know). Any time a question involves a shape that is 'weird', that shape can almost always be broken-down into smaller standard shapes - you can then use your knowledge of those shapes to answer the given question (although the math involved might be 'step-heavy').

Here, the weird shape is a 30/45/105 triangle. As chetan2u pointed out in his explanation, this triangle can be split into a 30/60/90 and a 45/45/90. From there, you can use the given side length, along with the rules for those 2 right triangles, to figure out ALL of the other side lengths and the total area of the shape.

GMAT assassins aren't born, they're made,
Rich

ScottTargetTestPrep · Answer

If we sketch the altitude (i.e., height) from the 105-degree angle to the opposite side, we have created two special right triangles: a 0-60-90 right triangle at the top and a 45-45-90 right triangle at the bottom (and the only labeled side will be the hypotenuse of the 45-45-90 right triangle) .

Since the 45-45-90 right triangle at the bottom has hypotenuse 2√2, each of its legs is 2. Since one of these legs is the height of the overall triangle, it’s also the side opposite the 30-degree angle of the 30-60-90 right triangle at the top. So the side opposite the 60-degree angle is 2√3.

Now, using the fact that the area of a right triangle is ½ the product of its two legs, the area of the 45-45-90 right triangle is ½ x 2 x 2 = 2, and the area of 30-60-90 right triangle is ½ x 2 x 2√3 = 2√3. Therefore, the area of the overall triangle is 2 + 2√3.

Answer: A

cadeepakbohra · Answer

Can someone suggest mistake in below solution.

Suppose Triangle is ABC with Angle A1 = 30, B = 105 and C = 45. Now, to calculate height, we draw one perpendicular line AD from A on BC (Extended). So now Triangle ADC is a right angle (45-90-45) and other triangle is ADB (15-90-75).
Now come to line DC which is having B as one point on it. Angle DAB = 15 and BAC = 30, So DB : BC = 15 : 30 = 1:2. BC is given = 2 * sq. root 2. So total DC = BC *3/2 = 3 * Sq. root 3.

As ADC is a 90-45-45 triangle, so AD will also be 3 * Sq. root3.

Now Triangle ABC. Base = BC = 2 * sq. root 2 and height = AD = 3 * Sq. root3

Area = (2 * sq. root 2) * (3 * Sq. root3) /2 = 6.

Can someone explain, what principle is wrong in above calculation.

Deepak

Argp · Answer

Here's a solution using the trigonometry table:

Qaisarjutt · Answer

in question 4 why base 2 under root 2 is used rather than 2

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What is the area of the triangle in the figure above?

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What is the area of the triangle in the figure above?

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