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

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Bunuel
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In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   26 Apr 2019, 06:26
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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. \(24\sqrt{2}\)


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BrentGMATPrepNow
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   26 Apr 2019, 09:24
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Bunuel wrote:

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. \(24\sqrt{2}\)


In general, isosceles right triangles have the following properties.

Notice that the hypotenuse = (√2)(length of one leg)

So, let's label the sides of the BLUE triangle as follows:



This means each leg of the RED triangle has length (√2)(x)



The hypotenuse of an isosceles right triangle = (√2)(length of one leg) . . .

. . . so the length of the hypotenuse = (√2)(√2)(x) = 2x


This means each leg of the GREEN triangle has length 2x



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
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   26 Apr 2019, 09:04
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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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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   28 Apr 2019, 03:38
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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


Bunuel wrote:

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
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   23 May 2019, 05:10
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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.

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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   28 May 2019, 06:17
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Bunuel wrote:

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
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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
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   13 Jun 2020, 02:51
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.
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In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   26 Aug 2020, 21:28
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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
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   06 Nov 2020, 05:54
\(\frac{base * height }{ 2 }= area\)

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.
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In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   19 Nov 2020, 02:24
Top Contributor
Bunuel wrote:

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. \(24\sqrt{2}\)


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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
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In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   13 May 2022, 11:39
Expert Reply
Bunuel wrote:

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


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Attachment:
2019-04-26_1753.png


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
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   23 Oct 2022, 03:23
ThatDudeKnows wrote:
Bunuel wrote:

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


Show Spoiler
Attachment:
2019-04-26_1753.png


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


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?
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In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   18 Nov 2022, 02:43
BrentGMATPrepNow wrote:
Bunuel wrote:

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. \(24\sqrt{2}\)


In general, isosceles right triangles have the following properties.

Notice that the hypotenuse = (√2)(length of one leg)

So, let's label the sides of the BLUE triangle as follows:



This means each leg of the RED triangle has length (√2)(x)



The hypotenuse of an isosceles right triangle = (√2)(length of one leg) . . .

. . . so the length of the hypotenuse = (√2)(√2)(x) = 2x


This means each leg of the GREEN triangle has length 2x



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


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
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   18 Nov 2022, 08:14
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Kimberly77 wrote:
BrentGMATPrepNow wrote:
Bunuel wrote:

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. \(24\sqrt{2}\)


In general, isosceles right triangles have the following properties.

Notice that the hypotenuse = (√2)(length of one leg)

So, let's label the sides of the BLUE triangle as follows:



This means each leg of the RED triangle has length (√2)(x)



The hypotenuse of an isosceles right triangle = (√2)(length of one leg) . . .

. . . so the length of the hypotenuse = (√2)(√2)(x) = 2x


This means each leg of the GREEN triangle has length 2x



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


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


If you assign those values, the area of ΔABC won't equal 6 (we are told that the area of ΔABC is 6)
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   18 Nov 2022, 11:53
Make sense now thanks BrentGMATPrepNow
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   17 Dec 2022, 13:45
BrentGMATPrepNow wrote:
Bunuel wrote:

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. \(24\sqrt{2}\)


In general, isosceles right triangles have the following properties.

Notice that the hypotenuse = (√2)(length of one leg)

So, let's label the sides of the BLUE triangle as follows:



This means each leg of the RED triangle has length (√2)(x)



The hypotenuse of an isosceles right triangle = (√2)(length of one leg) . . .

. . . so the length of the hypotenuse = (√2)(√2)(x) = 2x


This means each leg of the GREEN triangle has length 2x



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


@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
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink] New post   09 Mar 2023, 01:08
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Isocèles right triangles (45,45,90) is in the ratio X:X:root2X
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Re: In the figure above, if triangles ABC, ACD, and ADE are isosceles righ [#permalink]
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