In quadrilateral ABCD above, what is the length of AB ?
This topic has expert replies

 Master  Next Rank: 500 Posts
 Posts: 394
 Joined: 02 Jul 2017
 Thanked: 1 times
 Followed by:4 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
GMAT/MBA Expert
 Brent@GMATPrepNow
 GMAT Instructor
 Posts: 14498
 Joined: 08 Dec 2008
 Location: Vancouver, BC
 Thanked: 5254 times
 Followed by:1262 members
 GMAT Score:770
[spoiler=] [/spoiler][/quote]
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 345 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: xÂ² + 1Â² = 5Â²
Simplify: xÂ² + 1 = 25
So: xÂ² = 24
So: x = âˆš24 = âˆš[(4)(6)] = (âˆš4)(âˆš6) = 2âˆš6
Answer: C
Cheers,
Brent
Brent Hanneson  Creator of GMATPrepNow.com
Use my video course along with Beat The GMAT's free 60Day Study Guide
Watch these video reviews of my course
And check out these free resources
Use my video course along with Beat The GMAT's free 60Day Study Guide
Watch these video reviews of my course
And check out these free resources
If we draw diagonal \(BD\), we've created two right triangles: \(BCD\) and \(BAD\). We see that triangle \(BCD\) is a \(345\) 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 \(n\).
$$
\begin{align*}
1^2 + n^2 &= 5^2\\
1 + n^2 &= 25\\
n^2 &= 24\\
n &= \sqrt{24}\\
&= \sqrt{4} \cdot \sqrt{6} \\
&= 2\sqrt{6}
\end{align*}
$$
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 \(n\).
$$
\begin{align*}
1^2 + n^2 &= 5^2\\
1 + n^2 &= 25\\
n^2 &= 24\\
n &= \sqrt{24}\\
&= \sqrt{4} \cdot \sqrt{6} \\
&= 2\sqrt{6}
\end{align*}
$$
GMAT/MBA Expert
 Rich.C@EMPOWERgmat.com
 Elite Legendary Member
 Posts: 10346
 Joined: 23 Jun 2013
 Location: Palo Alto, CA
 Thanked: 2867 times
 Followed by:503 members
 GMAT Score:800
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 squareroot both sides, we'll have...
B = âˆš24
B = 2âˆš6
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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 squareroot both sides, we'll have...
B = âˆš24
B = 2âˆš6
Final Answer: C
GMAT assassins aren't born, they're made,
Rich