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In quadrilateral ABCD above, what is the length of AB ?

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In quadrilateral ABCD above, what is the length of AB ?  [#permalink]

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New post 26 Apr 2019, 01:38
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

84% (01:09) correct 16% (01:34) wrong based on 146 sessions

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Re: In quadrilateral ABCD above, what is the length of AB ?  [#permalink]

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New post 26 Apr 2019, 02:11
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1
Join D and B
DCB is an right angle triangle
DB^2= CD^2 +BC^2
=9 +16=25
or DB=5
Also, DB^2= AB^2 + AD^2
25=AB^2 + 1
AB^2 = 24
AB= 2*(6^0.5)
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Re: In quadrilateral ABCD above, what is the length of AB ?  [#permalink]

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New post 27 Apr 2019, 05:28
Bunuel wrote:
Image
In quadrilateral ABCD above, what is the length of AB ?


A. \(\sqrt{26}\)

B. \(2\sqrt{5}\)

C. \(2\sqrt{6}\)

D. \(3\sqrt{2}\)

E. \(3\sqrt{3}\)


PS58502.01
OG2020 NEW QUESTION

Attachment:
2019-04-26_1235.png


join BD we get two ∆ right angled BCD and ABD ; BCD ; 3:4:5 and BD = 5 so 25-1 = AB^2
AB - \(2\sqrt{6}\)
IMO C
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Re: In quadrilateral ABCD above, what is the length of AB ?  [#permalink]

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New post 27 Apr 2019, 17:03
Hi All,

We're asked for the length of AB in quadrilateral ABCD.

When dealing with 'weird' shapes, it often helps to break the shape down into 'pieces' that are easier to deal with. Here, if you draw a line from B to D, you will from 2 RIGHT TRIANGLES.

Triangle BCD has legs of 3 and 4, so it's a 3/4/5 right triangle.
Triangle BAD then has a leg of 1 and a hypotenuse of 5. We can use the Pythagorean Formula to find the missing leg...

1^2 + B^2 = 5^2
1 + B^2 = 25
B^2 = 24

From here, if we square-root both sides, we'll have...
B = √24
B = 2√6

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Re: In quadrilateral ABCD above, what is the length of AB ?  [#permalink]

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New post 29 Apr 2019, 10:49
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Top Contributor
Bunuel wrote:
Image
In quadrilateral ABCD above, what is the length of AB ?


A. \(\sqrt{26}\)

B. \(2\sqrt{5}\)

C. \(2\sqrt{6}\)

D. \(3\sqrt{2}\)

E. \(3\sqrt{3}\)


PS58502.01
OG2020 NEW QUESTION

Attachment:
2019-04-26_1235.png

First add a line to join B and D:
Image


If we focus on the blue right triangle, we can EITHER recognize that legs of length 3 and 4 are part of the 3-4-5 Pythagorean triplet, OR we can apply the Pythagorean Theorem.
Image
Either way, we'll see that the triangle's hypotenuse (BD) must have length 5

Now, when we focus on the red right triangle, we can . . .
Image
. . . apply the Pythagorean Theorem to write: x² + 1² = 5²
Simplify: x² + 1 = 25
So: x² = 24
So: x = √24 = √[(4)(6)] = (√4)(√6) = 2√6

Answer: C

Cheers,
Brent

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Re: In quadrilateral ABCD above, what is the length of AB ?  [#permalink]

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New post 30 Apr 2019, 23:50

Solution



Given:

    • AD= 1
    • CD=1
    • BC=1

To Find:
    • The length of AB.

Approach and Working:

Let us join the points B and D.

Image

Since ∆ BCD is a right -angled triangle, we can apply Pythagoras theorem in ∆ BCD.
    • Thus, \(BD^2\) = \(CD^2\) +\(BC^2\)
      o \(BD^2\) = \(4^2\) +\(3^2\) = 16+9 = 25
      o \(BD^2\) = 25
         BD=5

Now, we can apply Pythagoras theorem in triangle ABD.
    • Hence, \(BD^2\) =\(AD^2\) + \(AB^2\)
      o 25 = 1 + \(AB^2\)
         \(AB^2\) = 24
         AB = √24 = 2√6.

Hence, option C is the correct answer.

Correct Answer: Option C
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Re: In quadrilateral ABCD above, what is the length of AB ?  [#permalink]

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New post 02 May 2019, 17:27
Bunuel wrote:
Image
In quadrilateral ABCD above, what is the length of AB ?


A. \(\sqrt{26}\)

B. \(2\sqrt{5}\)

C. \(2\sqrt{6}\)

D. \(3\sqrt{2}\)

E. \(3\sqrt{3}\)


PS58502.01
OG2020 NEW QUESTION

Attachment:
2019-04-26_1235.png


If we draw diagonal BD, we’ve created two right triangles: BCD and BAD. We see that triangle BCD is a 3-4-5 right triangle. So we see that side BD = 5.

Therefore, triangle BAD is a right triangle with a leg of 1 and a hypotenuse of 5. We can let side AB = n and use the Pythagorean theorem to determine n.

1^2 + n^2 = 5^2

1 + n^2 = 25

n^2 = 24

n = √24

n = √4 x √6 = 2√6

Answer: C
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Re: In quadrilateral ABCD above, what is the length of AB ?   [#permalink] 02 May 2019, 17:27
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