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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 \(□BEFG\)?
6.3ps.png [ 16.14 KiB | Viewed 2345 times ]
A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)
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 [ 16.14 KiB | Viewed 2345 times ]
A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)
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
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
03 Jun 2020, 04:41
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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
Posted from my mobile device
AD is 3 so square 1 area is 9
BE is √6 so square 2 area is 6
9-6 =3
Posted from my mobile device
