In the figure above, if triangles ABC, ACD, and ADE are isosceles righ

Question

https://gmatclub.com/forum/download/file.php?id=47603 In the figure above, if triangles ABC, ACD, and ADE are isosceles right triangles and the area of ΔABC is 6, then the area of ΔADE is A. 18 B. 24 C. 36 D. 12\sqrt{2} E. 24\sqrt{2} PS73602.01 Quantitative Review 2020 NEW QUESTION 2019-04-26_1753.png

BrentGMATPrepNow · Accepted Answer

In general, isosceles right triangles have the following properties. 
 https://i.imgur.com/eFd3Pve.png 
 Notice that the hypotenuse = (√2)(length of one leg)

So, let's label the sides of the BLUE triangle as follows:
 https://i.imgur.com/6MDqfaZ.png

This means each leg of the RED triangle has length (√2)(x) 
 https://i.imgur.com/4N6WEsW.png

The hypotenuse of an isosceles right triangle = (√2)(length of one leg) . . . 
 https://i.imgur.com/gXKWlGn.png 
. . . so the length of the hypotenuse = (√2)(√2)(x) =   2x

This means each leg of the GREEN triangle has length 2x 
 https://i.imgur.com/QLqDOXJ.png

What is the area of ΔADE?

Area of triangle = (base)(height)/2 
So, area of ΔADE = (  2x  )(  2x  )/2 =   2x²

So, what is the value of   2x²  ?

GIVEN: the area of ΔABC is 6
ΔABC is the BLUE triangle we started with. 
We can write: (x)(x)/2 = 6
Simplify: x²/2 = 6, which means  x² = 12

This means the area of ΔADE =   2x²   = 2( 12 ) = 24

Answer: B

Cheers,
Brent

IanStewart · Answer

An isosceles right triangle is a 45-45-90 triangle, and in a 45-45-90 triangle, the hypotenuse is √2 times as long as a leg. So the legs of the middle triangle are √2 times those of the smallest one, and the legs of the largest triangle are (√2)(√2) = 2 times as long as the legs of the smallest one. If both legs of the largest triangle are twice as long as those of the smallest, the area of the largest triangle will be 2*2 = 4 times the area of the smallest, so will be 24.

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Archit3110 · Answer

solve step by step for each ∆ ACB side ; 2√3
then ∆ side AC = 4√6
then∆AED ; side AD= 4√3
area of ∆AED = 1/2 *4√3*4√3
IMO B ; 24

EgmatQuantExpert · Answer

Solution 
  Given  
In this question, we are given
 • A diagram consists of three triangles ABC, ACD, and ADE – all of them are individually isosceles right triangles.
• The area of ΔABC is 6.

To Find  
We need to determine
 • The area of ΔADE.

Approach & Working  
Let us assume the length of AB = n
 • As ΔABC is an isosceles right angle triangle, AB = BC = n
• Therefore, AC = √ (n^2 + n^2)  = √2n

We also know that ΔACD is an isosceles right angle triangle.
 • Hence, AC = CD = √2n
• Therefore, AD = √ (2n^2 + 2n^2)  = 2n

Finally, ΔADE is an isosceles right-angle triangle.
 • Hence, AD = DE = 2n
• Thus, the area of ΔADE = ½ * 2n * 2n =  2n^2

We are also given that area of ΔABC is 6.
 • Hence, we can write = ½ * n * n = ½  n^2  = 6
 o Or,  n^2  = 12 
• Therefore, area of ΔADE =  2n^2  = 2 * 12 = 24

Hence, the correct answer is option B.

https://e-gmat.com/blogs/wp-content/uploads/2016/12/Geometry-Ability-Quiz-4-e1480581478957.png

ScottTargetTestPrep · Answer

Since the area of ABC is 6, AB and BC are both √12 (since ½(√12)^2 = ½(12) = 6). Furthermore, AC is √12 x √2 = √24 and hence, AD is √24 x √2 = √48. Since DE is also √48, the area of triangle ADE is:

½(√48)^2 = ½(48) = 24

Answer: B

altairahmad · Answer

The area of triangle ADC is twice the area of triangle ACB and the area of triangle AED is twice the area of triangle ADC.

Shrey08 · Answer

In the figure notice that:

1) ABCF forms a square with diagonal AC dividing it into equal halves.
Area of ABC = Area of ACF = 6

2) Area ADC = 2* Area of ACF = 12

3) ACDG forms a square with diagonal DA dividing it into equal halves.

4) Area of ADG = Area of ACD = 12

5) Area of ADE = 2* Area f ADG = 24

Basshead · Answer

\frac{base * height }{ 2 }= area

We're told the area of triangle ABC = 6. We can find:

AB = BC =  \sqrt{12}

CA = DC =  \sqrt{24}

DA = DE =  \sqrt{48}

\frac{\sqrt{48} *\sqrt{48} }{ 2} = 24

Answer is B.

MHIKER · Answer

In triangle ABC, AB=BC= X 
 \frac{1}{2}  *X*X=6 
 X^2 =12
X=  2\sqrt{3}

Now,  AC^2  =  (2\sqrt{3})^2 + (2\sqrt{3})^2 
  AC  =  \sqrt{24} =  2\sqrt{6}

So,  AC=DC=   2\sqrt{6}

Then  AD^2   =   (2\sqrt{6})^2 + (2\sqrt{6})^2 =  48 
 AD = \sqrt{48}  =  2\sqrt{12}

The are of the Triangle ADE =  \frac{1}{2}*   2\sqrt{12} * 2\sqrt{12} 
 =  \frac{1}{2}*12*2*2  =24

The answer is B

ThatDudeKnows · Answer

This problem is rated by GMAC as Medium Difficulty. I showed it to my 8-year-old daughter, Riley. She got it right. She knows diddly squat about right triangles.

So many geometry questions are doable without knowing anything about geometry. Be like Riley: use your eyes and ballpark!!

Just LOOK at ABC compared to ACD. ACD is twice as big as ABC. So ACD has an area of 12.
Now just LOOK at ACD compared to ADE. ADE is twice as big as ACD. So ADE has an area of 24.

Answer choice B.

Aaaaaand, on to the next question!

ThatDudeKnowsBallparking

Rainman91 · Answer

Although estimating is a good strategy, but is it not risky in this instance? To assume that it is twice as large without concrete data?

Kimberly77 · Answer

Great explanation always  BrentGMATPrepNow . One question that how will I know to assign x and not using just 1 here in 45-45-90 triangle. As if I use 1 then in final triangle ADE area calculation, I got bh/2= (2*2)/2 = 2? Thanks Brent

BrentGMATPrepNow · Answer

If you assign those values, the area of ΔABC won't equal 6 (we are told that the area of ΔABC is 6)

Kimberly77 · Answer

Make sense now thanks  BrentGMATPrepNow

woohoo921 · Answer

@brentgmatclubnow

Thank you for this helpful explanation. To confirm, for finding the hypotenuse of ABC, you can also just use the Pythagorean theorem, correct?
(xroot(2))^2 + (xroot(2))^2=c^2
c^2=(4x^.5)^.5
=2x

adityasuresh · Answer

Isocèles right triangles (45,45,90) is in the ratio X:X:root2X

In the figure above, if triangles ABC, ACD, and ADE are isosceles righ

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