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The figure shows the quadratic function y = x2 and the points D and G

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MathRevolution
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The figure shows the quadratic function y = x2 and the points D and G [#permalink] New post  03 Jun 2020, 04:20
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[GMAT math practice question]

The figure shows the quadratic function \(y = x^2\) and the points \(D\) and \(G\) on the graph. The coordinate of point \(A\) is \((0, 9)\), and \(□ABCD\) and \(□BEFG\) are squares. What is the difference between the area of \(□ABCD\) and that of \(□BEFG\)?

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A. \(1\)

B. \(2\)

C. \(3\)

D. \(4\)

E. \(5\)
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nick1816
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Re: The figure shows the quadratic function y = x2 and the points D and G [#permalink] New post  03 Jun 2020, 04:36
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Area of the ABCD = \(AD^2\) = [x-coordinate of D\(]^2\) = y-coordinate of D = 9

AD = AB = \(\sqrt{9}\) = 3

y-coordinate of B = (y-coordinate of D) - (AB) = 9-3 = 6 = y-coordinate of G

Area of the BEFG = \(BG^2 \)= [x-coordinate of G\(]^2 \)= y-coordinate of G = 6

Difference in the area = 9-6=3


MathRevolution wrote:
[GMAT math practice question]

The figure shows the quadratic function \(y = x^2\) and the points \(D\) and \(G\) on the graph. The coordinate of point \(A\) is \((0, 9)\), and \(□ABCD\) and \(□BEFG\) are squares. What is the difference between the area of \(□ABCD\) and that of \(□BEFG\)?

Attachment:
6.3ps.png


A. \(1\)

B. \(2\)

C. \(3\)

D. \(4\)

E. \(5\)
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dimri10
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Re: The figure shows the quadratic function y = x2 and the points D and G [#permalink] New post  03 Jun 2020, 04:41
Using the equation y= x^2
AD is 3 so square 1 area is 9
BE is √6 so square 2 area is 6
9-6 =3

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yashikaaggarwal
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Re: The figure shows the quadratic function y = x2 and the points D and G [#permalink] New post  03 Jun 2020, 05:52
Distance between Origin to A is 9 (given)
Since D is perpendicular to A
Also, Height of D = Height of A = 9 (in context to Y axis)
D lies on parabola y=x^2
9=x^2 => X=3.
Thee coordinate of D = (3,9)
The distance between point A and D is √(y2-y1)^2+(x2-x1)^2
=> √(9-9)^2+(0-3)^2 => √9 => 3
Area of square ABCD = S^2 = 3^2 => 9

Now distance of B from Orgin is 6 (OA-AB=BO)
Height of G from X-axis will be 6 as well. Since G is perpendicular on B
G lies on parabola y=x^2
=> 6=x^2 => x=√6
Therefore the area of BEFG = S2^2 => √6^2 => 6
The difference between area of square ABCD and BEFG is 9-6=3

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Re: The figure shows the quadratic function y = x2 and the points D and G [#permalink] New post  05 Jun 2020, 01:37
=>

Since the coordinate of point A\(\) is \((0, 9)\), we have \(D(3, 9), B(6, 0) \)

and \(G(√6, 6).\)

Then the length of the sides of square \(ABCD\) is \(3\), and that of sides of the square \(BEFG\) is \(√6.\)

The difference between areas of the two squares is \(3^2 - (√6)^2 = 9 - 6 = 3.\)

Therefore, C is the answer.
Answer: C
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Re: The figure shows the quadratic function y = x2 and the points D and G [#permalink]
05 Jun 2020, 01:37
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