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Math Expert V
Joined: 02 Sep 2009
Posts: 58472
In the figure above, the area of square ABCF is 25 and triangle CGD is  [#permalink]

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Difficulty:   15% (low)

Question Stats: 95% (01:25) correct 5% (01:51) wrong based on 60 sessions

HideShow timer Statistics In the figure above, the area of square ABCF is 25 and triangle CGD is an isosceles right triangle. What is the area of rectangle DEFG?

A. 15
B. 12
C. 9
D. $$6\sqrt{2}$$
E. 6

Attachment: 2018-03-16_1012.png [ 8.33 KiB | Viewed 915 times ]

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Math Expert V
Joined: 02 Sep 2009
Posts: 58472
Re: In the figure above, the area of square ABCF is 25 and triangle CGD is  [#permalink]

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Bunuel wrote: In the figure above, the area of square ABCF is 25 and triangle CGD is an isosceles right triangle. What is the area of rectangle DEFG?

A. 15
B. 12
C. 9
D. $$6\sqrt{2}$$
E. 6

Attachment:
2018-03-16_1012.png

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Senior PS Moderator V
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Posts: 3332
Location: India
GPA: 3.12
Re: In the figure above, the area of square ABCF is 25 and triangle CGD is  [#permalink]

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1 Since the area of the square ABCF is 25, the side of the square is 5.

Since the triangle GCD is an isosceles right triangle, the sides are in the ratio $$1:1:\sqrt{2}$$
We are given that the hypotenuse of the triangle is $$3\sqrt{2}$$, the other sides are equal to 3.

In rectangle GFDE, GF = 5 - 3 = 2. Also, GD = 3.

Therefore, the area of the rectangle GFDE is 6(Option E)
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Retired Moderator V
Joined: 28 Mar 2017
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Re: In the figure above, the area of square ABCF is 25 and triangle CGD is  [#permalink]

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1
Bunuel wrote: In the figure above, the area of square ABCF is 25 and triangle CGD is an isosceles right triangle. What is the area of rectangle DEFG?

A. 15
B. 12
C. 9
D. $$6\sqrt{2}$$
E. 6

Attachment:
2018-03-16_1012.png

10 second question.

Straight "E".

Isosceles right triangle has sides in the ratio 1(short):1(shhort):root2(hypo)
In the figure hypot=3*root2
So short side = 3
Now for rectabgle:
l=3 b=2
Area=6
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Re: In the figure above, the area of square ABCF is 25 and triangle CGD is  [#permalink]

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Bunuel wrote: In the figure above, the area of square ABCF is 25 and triangle CGD is an isosceles right triangle. What is the area of rectangle DEFG?

A. 15
B. 12
C. 9
D. $$6\sqrt{2}$$
E. 6

Attachment:
2018-03-16_1012.png

Since triangle CGD is an isosceles right triangle (i.e., 45-45-90), both sides CG and GD = 3√2/√2 = 3. Since the area of square ABCF is 25, side CF = √25 = 5. Thus side GF is 5 - 3 = 2.

Finally, the area of rectangle DEFG is GF x GD = 2 x 3 = 6.

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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3092
Re: In the figure above, the area of square ABCF is 25 and triangle CGD is  [#permalink]

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Solution •Area of square $$ABCF= 25$$

Let us assume the length of the sides of the square is ‘$$a$$’, then:

•$$a^2=25$$
•$$a=5$$ (Distance cannot be negative so $$a$$ cannot be $$-5$$)

•Triangle $$GCD$$ is an isosceles right-angled triangle.
•GCD is $$45- 45- 90$$ triangle.
•$$GC:GD: DC$$= $$1:1:\sqrt{2}$$
GC= GD= $$3$$

•Since $$GD=FE$$, hence $$FE=3$$.
•Since $$CF= AB=CG + GF$$ = $$5$$, hence GF=$$2$$.

Area of rectangle $$DEFG= FE*GF$$
= $$3*2=6$$

_________________ Re: In the figure above, the area of square ABCF is 25 and triangle CGD is   [#permalink] 25 Mar 2018, 12:02
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