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

Answer: C

Cheers,
Brent
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Join D and B
DCB is an right angle triangle

\(DB^2= CD^2 +BC^2\)

\(DB^2=9 +16=25\)

\(or DB=5\)

Also, \(DB^2= AB^2 + AD^2\)

\(25=AB^2 + 1\)

\(AB^2 = 24\)

\(AB= 2\sqrt{6}\)

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

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

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

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.



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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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
_________________

If we draw diagonal BD, There are 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. Now use the Pythagorean theorem to determine AB.

1^2 + AB^2 = 5^2

1 + AB^2 = 25

AB^2 = 24

AB = √24

AB = √4 x √6 = 2√6

Answer: C

HI,
can we also use the 30 : 60 : 90 triangle property to solve for the required side here ?

I tried using, but got a weird answer..

Hi Shrey9,

Neither of these triangles is a 30/60/90 right triangle - so that property does not apply. To use 30/60/90 rules, you need to either have the 3 angles or know that you have a right triangle and 2 of the sides that fit the relationship of 2 of the 3 sides of that type of triangle (re. X : X√3 : 2X)

GMAT assassins aren't born, they're made,
Rich
_________________

so what you're saying is Pythagoras can be applied to any right triangle (even if its just one angle, which is 90deg) and 30 :60 : 90 or 45 : 45 : 90 can't just be applied when we don't know other angles apart from the 90 deg one ? (coz it could be 90 : 46 : 44 etc..)

Hi Shrey9,

That is exactly correct! The Pythagorean Theorem applies to ANY Right Triangle; but the special Right Triangles only occur under specific conditions.

GMAT assassins aren't born, they're made,
Rich
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