AB

[goyalsau]

Tough one for me,

[GMATGuruNY]

Each of the answer choices contains √3. This means that we should look for a 30:60:90 triangle.

In a 30:60:90 triangle, the sides are proportioned x : x√3 : 2x. Since the side opposite the 60 degree angle is x√3, and the answer choices represent CB, the answer choices indicate the following: that angle CDO = 60 and angle COD = 30. See the drawing above.

This means that triangle CDO and triangle BOE are each 30:60:90 triangles.
Looking at triangle CDO, since CD = 6, CO = 6√3.
Looking at triangle BOE, since BE=2, BO = 2√3.
Thus, CB = 6√3 - 2√3 = 4√3.

The correct answer is B.

To confirm that angle COD=30, let's examine triangle BAO. Since OD bisects angle COD = 30, angle BOA = 60 and angle BAO = 30. So triangle BAO also is a 30:60:90 triangle, as indicated in the drawing above.
The shorter leg in this triangle is BO = 2√3.
The longer leg is AB = AE+EB = 4+2 = 6.
Does BO*√3 = AB? Yes, because 2√3*√3 = 6.

[Anurag@Gurome]

Here is another approach without assuming anything.



Refer to the image above.

A perpendicular EM on AO is drawn from E.
Now triangle OEB is congruent to triangle OEM.
Thus, EM = EB = 2 cm
and, OM = OB = x xm (say)

In the right-angled triangle AEM, say AM = y cm, thus

• .... (EM)² + (AM)² = (AE)²
  => (2)² + (y)² = (4)²
  => y² = 4² - 2² = 12
  => y = √12 = 2√3

Thus, AM = 2√3 cm

In the right-angled triangle ABO,

• .... (AB)² + (OB)² = (OA)²
  => (AE + EB)² + (OB)² = (OM + AM)²
  => (4 + 2)² + (x)² = (x + 2√3)²
  => 36 + x² = x² + (4√3)x + 12
  => (4√3)x = 36 - 12 = 24
  => x = 2√3

Thus, OB = 2√3 cm

Now triangle ODC and triangle OEB are similar. Thus,

• .... OB/OC = BE/CD = 2/6 = 1/3
  => OC = 3*(OB)
  => (OB + BC) = 3*(OB)
  => BC = 2*(OB) = 2*(2√3) = 4√3

The correct answer is B.

[goyalsau]

Hats off to your approach Sir, But i would like to ask one thing.

Here is another approach without assuming anything.



Refer to the image above.

A perpendicular EM on AO is drawn from E.
Now triangle OEB is congruent to triangle OEM.
Thus, EM = EB = 2 cm
and, OM = OB = x xm (say)

I am not able to understand why EM will be equal to EB, why it will be equal I can draw a different triangle with some different dimensions. But i don't think its compulsory that every time EM will be equal to EB, I am sure there must be some reasoning behind this.
Please Clear my doubt sir...

In the right-angled triangle AEM, say AM = y cm, thus

• .... (EM)² + (AM)² = (AE)²
  => (2)² + (y)² = (4)²
  => y² = 4² - 2² = 12
  => y = √12 = 2√3

Thus, AM = 2√3 cm

In the right-angled triangle ABO,

• .... (AB)² + (OB)² = (OA)²
  => (AE + EB)² + (OB)² = (OM + AM)²
  => (4 + 2)² + (x)² = (x + 2√3)²
  => 36 + x² = x² + (4√3)x + 12
  => (4√3)x = 36 - 12 = 24
  => x = 2√3

Thus, OB = 2√3 cm

Now triangle ODC and triangle OEB are similar. Thus,

• .... OB/OC = BE/CD = 2/6 = 1/3
  => OC = 3*(OB)
  => (OB + BC) = 3*(OB)
  => BC = 2*(OB) = 2*(2√3) = 4√3

The correct answer is B.

[goyalsau]

I think i got it,

It like drawing a Circle with Center E, and Radius of the Circle is EM or EB , & OA and OB are tangents on the circle from the same point. In that case there length will also be equal...............
When we join a tangent and the radius of the circle, Angle is always and Always 90 degree.

Thanks for the thoughtful explanation.................

[Anurag@Gurome]

Hats off to your approach Sir, But i would like to ask one thing.

I am not able to understand why EM will be equal to EB, why it will be equal I can draw a different triangle with some different dimensions. But i don't think its compulsory that every time EM will be equal to EB, I am sure there must be some reasoning behind this.
Please Clear my doubt sir...

For the triangles OEB and OEM,, all the corresponding angles are equal in measure. Thus they are similar. Which means (EB/EM) = (OB/OM) = (OE/OE) = 1. Hence not only EB = EM, but all the corresponding sides are equal in length. For two triangle if their corresponding sides are equal in length and their corresponding angles are equal in size, we say the triangles are congruent in nature.

The explanation you've provided is more complicated than required. In fact the reasoning behind your explanation (i.e. why the lengths of tangents from a given point to a particular circle will be equal) is also what I've explained above.

[goyalsau]

For the triangles OEB and OEM,, all the corresponding angles are equal in measure. Thus they are similar. Which means (EB/EM) = (OB/OM) = (OE/OE) = 1.

Sir, As far as i know when two triangle are similar ratio of there sides is equal , It does not mean that there ratio is equal to 1, When its ratio is equal to one then we can say that they are congruent.

Take for example A right Angle Triangle with sides 3 , 4 , 5 & 6, 8 , 10

Both these triangles are similar but there ratio of sides are 1/2.

Hence not only EB = EM, but all the corresponding sides are equal in length. For two triangle if their corresponding sides are equal in length and their corresponding angles are equal in size, we say the triangles are congruent in nature.

The explanation you've provided is more complicated than required. In fact the reasoning behind your explanation (i.e. why the lengths of tangents from a given point to a particular circle will be equal) is also what I've explained above.

I hope i am making my point this time....

[Anurag@Gurome]

Sir, As far as i know when two triangle are similar ratio of there sides is equal , It does not mean that there ratio is equal to 1, When its ratio is equal to one then we can say that they are congruent.

Take for example A right Angle Triangle with sides 3 , 4 , 5 & 6, 8 , 10

Both these triangles are similar but there ratio of sides are 1/2

I hope i am making my point this time....

Obviously you can't take the ratio as 1.
But in this case note that the side OE is common two both the triangles. Thus the ratio of the lengths of their corresponding sides will be equal to 1 as (OE/OE) = 1.

Hope it is clear now.

[goyalsau]

Sir, As far as i know when two triangle are similar ratio of there sides is equal , It does not mean that there ratio is equal to 1, When its ratio is equal to one then we can say that they are congruent.

Take for example A right Angle Triangle with sides 3 , 4 , 5 & 6, 8 , 10

Both these triangles are similar but there ratio of sides are 1/2

I hope i am making my point this time....

Obviously you can't take the ratio as 1.
But in this case note that the side OE is common two both the triangles. Thus the ratio of the lengths of their corresponding sides will be equal to 1 as (OE/OE) = 1.

Hope it is clear now.

thanks sir, If i will be given the opportunity to choose the person for the next fields Medal , With no second thought You will be my fields medal winner....

[RACHVIK]

Is there a simpler way??

[GMATGuruNY]

Is there a simpler way??

I think that using the answer choices is the quickest way to solve, as in my post above. I've added a drawing to make the explanation easier to follow. The time to solve was less than 1 minute.

[anshumishra]

Tough one for me,

Please see the attached diagram.

tan z = 6/(x+y) = 2/y => 6y = 2x+2y => x= 2y

tan 2z = AB/BO = 6/Y

Since; tan 2z = 2tan z/(1 - tan^2 z)
=> 6/y = 2*(2/y)/(1-4/y^2)
=> 6y^2 -24 = 4y^2
=> y^2 = 12
=> y = 2*sqrt(3)
BC = x = 2y = 4*sqrt(3)

[goyalsau]

Since; tan 2z = 2tan z/(1 - tan^2 z)

Is it a formula.....

[anshumishra]

Since; tan 2z = 2tan z/(1 - tan^2 z)

Is it a formula.....

Yes. Sometimes having a math background, gives you multiple options.

[fskilnik@GMATH]

Hi there!

Please follow the steps in the figure.

Focus: CB = 2x = ?

01. OB = CB/2 = 2x/2 = x because triangles EAD and EBO are similar with ratio of similarity 2:1 in that order.

01. Triangles OBE and OEM are congruent (therefore "arrows for x and 2" are justified) ;

02. Triangle AEM is 30-60-90 (hypothenuse is twice one of the other sides) therefore AM = 2*sqrt(3)

03. Pythagoras on triangle OBA (remember we are looking for 2x):

(2*sqrt(3) + x)^2 = 6^2 + x^2
12+4x*sqrt(3) = 36
6+2x*sqrt(3) = 18
2x*sqrt(3) = 12 then multiplying by sqrt(3) we get 2x = 12*sqrt(3)/3 = 4*sqrt(3)

Regards,
Fabio.