Looking at dimensions, we can quickly eliminate some answers upon inspection:

L denotes the dimension of length.

(A)(b+c)/bc -> L/L^2 = 1/L which is not L (Eliminate)
(B)(bc)/(b+c) -> L^2/L = L (OK)
(C)Square root(b^2+c^2) -> L (OK)
(D)[(b+c)^2]/bc -> L^2/L^2 = 1 which is not L (Eliminate)
(E)(bc)^2 -> (L^2)^2 = L^4 which is not L (Eliminate)

This leaves 2 options:
(bc)/(b+c) OR Square root(b^2+c^2)

Answer (C) cannot be correct because it would be the hypotenuse of a right angle triangle; and this hypotenuse would be larger than b or c, but AD is shorter. Hence answer (C) can be eliminated.

This just leaves one option.

Answer = [spoiler](B)[/spoiler]

Alternatively, consider the case of an isosceles with AD perpendicular to the base.

Answer (B) could simplify to AD = (bc)/(b+c) = b^2/2b = b/2

Also, from the diagram, AD = b * cos 60 = b/2

Therefore answer B is correct.

Mathsbuddy wrote:Looking at dimensions, we can quickly eliminate some answers upon inspection:

L denotes the dimension of length.

(A)(b+c)/bc -> L/L^2 = 1/L which is not L (Eliminate)
(B)(bc)/(b+c) -> L^2/L = L (OK)
(C)Square root(b^2+c^2) -> L (OK)
(D)[(b+c)^2]/bc -> L^2/L^2 = 1 which is not L (Eliminate)
(E)(bc)^2 -> (L^2)^2 = L^4 which is not L (Eliminate)

This leaves 2 options:
(bc)/(b+c) OR Square root(b^2+c^2)

Answer (C) cannot be correct because it would be the hypotenuse of a right angle triangle; and this hypotenuse would be larger than b or c, but AD is shorter. Hence answer (C) can be eliminated.

This just leaves one option.

Answer = [spoiler](B)[/spoiler]

Hi Mathsbuddy,

That's a great way to solve the problem, but I don't understand it. What's L and how you arrive at b+c = L and bc = L^2. Please elaborate and explain the method you used. Also please mention if this problem can be solved algebraically without using options or trigonometric ratios.

Aman verma wrote:

Mathsbuddy wrote:Looking at dimensions, we can quickly eliminate some answers upon inspection:

L denotes the dimension of length.

(A)(b+c)/bc -> L/L^2 = 1/L which is not L (Eliminate)
(B)(bc)/(b+c) -> L^2/L = L (OK)
(C)Square root(b^2+c^2) -> L (OK)
(D)[(b+c)^2]/bc -> L^2/L^2 = 1 which is not L (Eliminate)
(E)(bc)^2 -> (L^2)^2 = L^4 which is not L (Eliminate)

This leaves 2 options:
(bc)/(b+c) OR Square root(b^2+c^2)

Answer (C) cannot be correct because it would be the hypotenuse of a right angle triangle; and this hypotenuse would be larger than b or c, but AD is shorter. Hence answer (C) can be eliminated.

This just leaves one option.

Answer = [spoiler](B)[/spoiler]

Hi Mathsbuddy,

That's a great way to solve the problem, but I don't understand it. What's L and how you arrive at b+c = L and bc = L^2. Please elaborate and explain the method you used. Also please mention if this problem can be solved algebraically without using options or trigonometric ratios.

Hi Aman verma,

Thanks for the question.

L stands for the dimension of Length.

(Strictly speaking [L] = dimension of length, but let's just skip the parentheses!)

Other dimensions are: Mass [M] and Time [T]. Any physics equations or units can be reduced to M,L,T (and temperature sometimes) to quickly check if they are wrong.

length * length = area
So L^2 is area, not length

Similarly distance/time [=L/T] gives speed, so speed cannot be T/L, or M/T, or T^2, etc

Okay to answer your question:

b is a measure of length
c is a measure of length too

Therefore, if we ignore any values they might be, we can replace them each with L

So b+c becomes L + L = 2L which is still length, therefore this can be replaced with L (we throw away numbers as they have no dimension of their own)

and bc becomes L*L = L^2 which is area, not length.

After a little practice, it is easy and quick to spot rogue formulae that are inconsistent dimensionally.

It's the same as looking at units:

Inches + inches = inches (L+L = L)
Inches * inches = square inches (area)

A length is never an area, or a time, or a mass, or vice-versa.

Let me know if that helps.

Mathsbuddy wrote:
Hi Aman verma,

Thanks for the question.

L stands for the dimension of Length.

(Strictly speaking [L] = dimension of length, but let's just skip the parentheses!)

Other dimensions are: Mass [M] and Time [T]. Any physics equations or units can be reduced to M,L,T (and temperature sometimes) to quickly check if they are wrong.

length * length = area
So L^2 is area, not length

Similarly distance/time [=L/T] gives speed, so speed cannot be T/L, or M/T, or T^2, etc

Okay to answer your question:

b is a measure of length
c is a measure of length too

Therefore, if we ignore any values they might be, we can replace them each with L

So b+c becomes L + L = 2L which is still length, therefore this can be replaced with L (we throw away numbers as they have no dimension of their own)

and bc becomes L*L = L^2 which is area, not length.

After a little practice, it is easy and quick to spot rogue formulae that are inconsistent dimensionally.

It's the same as looking at units:

Inches + inches = inches (L+L = L)
Inches * inches = square inches (area)

A length is never an area, or a time, or a mass, or vice-versa.

Let me know if that helps.

Simply amazing!

I think that took my comprehension to a different dimension. I haven't done any maths beyond my 12th grade so I couldn't have possibly thought of that kind of method. Now I wonder can that method be used to check the solution for any geometry problem. Cause any geometry problem is basically a problem on length or mass dimension in space. I want to improve my spatial reasoning. Can you suggest me any book that can help me understand the use of this method on maths problem ? Is there any short handbook that can help me learn to use this kind of concept on quant problems; something that I can complete in a couple of days ? Can I download it from the net ?

Aman verma wrote:

Q: In a triangle ABC Angle BAD = Angle CAD = 60 degrees. What is the length of AD ?

(A)(b+c)/bc

(B)(bc)/(b+c)

(C)Square root(b^2+c^2)

(D)[(b+c)^2]/bc

(E)(bc)^2

The constraint that ∠BAD=∠CAD=60 will be satisfied if ∆ABD and ∆ACD are each a 30-60-90 triangle.
In a 30-60-90 triangle, the sides are in the following ratio:
x : x√3 : 2x.

Plug in the following case:

Here, a=4√3, b=4, c=4, and AD=2.

The question stem asks for the value of AD=2. This is our target.
Now plug a=4√3, b=4, and c=4 into the answers to see which yields our target of 2.
Only B works:
(bc)/(b+c) = (4*4)/(4+4) = 16/8 = 2.

The correct answer is B.

Aman verma wrote:

Mathsbuddy wrote:
Hi Aman verma,

Thanks for the question.

L stands for the dimension of Length.

(Strictly speaking [L] = dimension of length, but let's just skip the parentheses!)

Other dimensions are: Mass [M] and Time [T]. Any physics equations or units can be reduced to M,L,T (and temperature sometimes) to quickly check if they are wrong.

length * length = area
So L^2 is area, not length

Similarly distance/time [=L/T] gives speed, so speed cannot be T/L, or M/T, or T^2, etc

Okay to answer your question:

b is a measure of length
c is a measure of length too

Therefore, if we ignore any values they might be, we can replace them each with L

So b+c becomes L + L = 2L which is still length, therefore this can be replaced with L (we throw away numbers as they have no dimension of their own)

and bc becomes L*L = L^2 which is area, not length.

After a little practice, it is easy and quick to spot rogue formulae that are inconsistent dimensionally.

It's the same as looking at units:

Inches + inches = inches (L+L = L)
Inches * inches = square inches (area)

A length is never an area, or a time, or a mass, or vice-versa.

Let me know if that helps.

Simply amazing!

I think that took my comprehension to a different dimension. I haven't done any maths beyond my 12th grade so I couldn't have possibly thought of that kind of method. Now I wonder can that method be used to check the solution for any geometry problem. Cause any geometry problem is basically a length or mass dimension in space. I want to improve my spatial reasoning. Can you suggest me any book that can help me understand the use of this method on maths problem ? Is there any short handbook that can help me learn to use this kind of concept on quant problems; something that I can complete in a couple of days ? Can I download it from the net ?

In this situation, it's only useful for eliminating answers that are dimensionally incorrect. I wouldn't advise using it for anything else. As with all problem solving, care must be taken not to make a mistake too! Remember that loose numbers are ignored, so you can't depend on it to find the formula. Wikepedia has some stuff on dimensional analysis, but it goes too deep. https://www.allmathwords.org/en/d/dimens ... lysis.html uses units as dimensions. Just keep it simple and it may be useful. Good luck!