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In the figure above, the area of square ABCF is 25 and triangle CGD is

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In the figure above, the area of square ABCF is 25 and triangle CGD is  [#permalink]

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New post 15 Mar 2018, 23:04
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E

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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
2018-03-16_1012.png [ 8.33 KiB | Viewed 828 times ]

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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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New post 15 Mar 2018, 23:05
Bunuel wrote:
Image

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

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New post 15 Mar 2018, 23:37
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Image

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

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New post 16 Mar 2018, 02:49
1
Bunuel wrote:
Image

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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New post 19 Mar 2018, 15:13
Bunuel wrote:
Image

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.

Answer: E
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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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New post 25 Mar 2018, 11:02

Solution



Image

    •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 \(DEFG= FE*GF\)
    = \(3*2=6\)

Answer: E
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Re: In the figure above, the area of square ABCF is 25 and triangle CGD is &nbs [#permalink] 25 Mar 2018, 11:02
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