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04 May 2022, 03:45
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
57% (02:43) correct
43%
(02:03)
wrong
based on 21
sessions
History
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Time
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Not Attempted Yet
In triangle ABC, we have AC = BC = 7 and AB = 2. Suppose that D is a point on line AB such that B lies between A and D and CD = 8. What is BD ?
(A) 3
(B) \(2\sqrt{3}\)
(C) 4
(D) 5
(E) \(4\sqrt{2}\)
Are You Up For the Challenge: 700 Level Questions
(A) 3
(B) \(2\sqrt{3}\)
(C) 4
(D) 5
(E) \(4\sqrt{2}\)
Are You Up For the Challenge: 700 Level Questions
04 May 2022, 07:44
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The first step is to draw a triangle incorporating all the data given. The triangle is an isosceles triangle with AB as the unequal side and the vertex C containing the unequal angle.
4th May 2022 - Post 3.png [ 10.56 KiB | Viewed 4229 times ]
The angle bisector of the unequal angle of an isosceles triangle is also the perpendicular bisector of the unequal side.
Therefore, AE = EB = 1 unit
The length of the perpendicular bisector CE can be found using Pythagoras theorem since triangle BCE is a right angled triangle.
\(CE^2\) + \(EB^2\) = \(BC^2\).
Substituting and simplifying, we have CE = \(\sqrt{48}\).
Observe that triangle DCE is also a right angled triangle. Applying the Pythagoras theorem,
\(\\
CE^2\) + \(ED^2\) = \(CD^2\)
48 + \((EB + BD)^2\) = 64
Therefore,\( (1 + x)^2\) = 16
Expanding the LHS using standard identities, 1 + \(x^2\) + 2x = 16
Or, \(x^2\) + 2x = 15.
Of all the numbers given in the answer options, the only value that satisfies the above equation is 3.
The correct answer option is A.
Attachment:
4th May 2022 - Post 3.png [ 10.56 KiB | Viewed 4229 times ]
The angle bisector of the unequal angle of an isosceles triangle is also the perpendicular bisector of the unequal side.
Therefore, AE = EB = 1 unit
The length of the perpendicular bisector CE can be found using Pythagoras theorem since triangle BCE is a right angled triangle.
\(CE^2\) + \(EB^2\) = \(BC^2\).
Substituting and simplifying, we have CE = \(\sqrt{48}\).
Observe that triangle DCE is also a right angled triangle. Applying the Pythagoras theorem,
\(\\
CE^2\) + \(ED^2\) = \(CD^2\)
48 + \((EB + BD)^2\) = 64
Therefore,\( (1 + x)^2\) = 16
Expanding the LHS using standard identities, 1 + \(x^2\) + 2x = 16
Or, \(x^2\) + 2x = 15.
Of all the numbers given in the answer options, the only value that satisfies the above equation is 3.
The correct answer option is A.
21 Apr 2026, 20:07
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