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

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Math Expert
Joined: 02 Sep 2009
Posts: 55271
In quadrilateral ABCD above, what is the length of AB ?  [#permalink]

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26 Apr 2019, 01:38
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5% (low)

Question Stats:

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

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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 [ 8.99 KiB | Viewed 1209 times ]

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

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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
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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27 Apr 2019, 05:28
Bunuel wrote:

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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27 Apr 2019, 17:03
Hi All,

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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29 Apr 2019, 10:49
1
Top Contributor
Bunuel wrote:

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:

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.

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 . . .

. . . 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

Cheers,
Brent

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

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30 Apr 2019, 23:50

Solution

Given:

• CD=1
• BC=1

To Find:
• The length of AB.

Approach and Working:

Let us join the points B and D.

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.

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

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02 May 2019, 17:27
Bunuel wrote:

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

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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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