Geometry- Advanced

[Aman verma]

Q:In the above figure a square of maximum possible area is circumscribed by the right angle triangle ABC in such a way that one of its side lies on the hypotenuse of the triangle. What is the area of the square ?

(A) [abc/(a^2+b^2+ab)]^2

(B) (a^2+b^2+c^2)/abc

(C) abc/(a^2+b^2+c^2)

(D) (a+b+c)/(a^2+b^2+c^2)

(E) (a^2+b^2+c^2)/3

[Brent@GMATPrepNow]

Q:In the above figure a square of maximum possible area is circumscribed by the right angle triangle ABC in such a way that one of its side lies on the hypotenuse of the triangle. What is the area of the square ?

(A) [abc/(a² + b² + ab)]²
(B) (a² + b² + c²)/abc
(C) abc/(a² + b² + c²)
(D) (a+b+c)/(a² + b² + c²)
(E) (a² + b² + c²)/3

We can solve this question using the Input-Output approach.
So, let's find a specific example where we can calculate the area of the square.
To do so, let's begin with a simple right triangle that we know a lot about: an isosceles right triangle (i.e., a 45-45-90 right triangle)


From here, notice that, since we've inserted a square with 90º angles, there are some additional 45º angles in this diagram.


Since the triangle in the bottom right corner has two equal angles, we know that the two opposite sides have equal length.


Next, since all sides in a square have equal length, we can see that we have many sides that are the same length.

At this point, we should recognize that the hypotenuse of the large right triangle is comprised of three equal lengths. So, let's choose some nice values for the length of the sides.

If we say that each leg has length 3, then the hypotenuse must have length 3√2


Since the hypotenuse of the large right triangle is comprised of three equal lengths, each line segment must have length √2


Great. We now have a diagram that meets the given conditions.
Here, we see that a = 3, b = 3 and c = 3√2
In this case, the area of the square = (√2)(√2) = 2

At this point, we'll plug a = 3, b = 3 and c = 3√2 into each answer choice and see which one yields an OUTPUT of 2

(A) [abc/(a² + b² + ab)]² = [27√2/(27)]² = [√2)]² = 2 BINGO!
(B) (a² + b² + c²)/abc = (36)/27√2 = something other than 2
(C) abc/(a² + b² + c²) = 27√2/(36) = something other than 2
(D) (a+b+c)/(a² + b² + c²) = (6 + 3√2)/(36) = something other than 2
(E) (a² + b² + c²)/3 = (36)/3 = 12

Answer: A

Cheers,
Brent

[Aman verma]

Q:In the above figure a square of maximum possible area is circumscribed by the right angle triangle ABC in such a way that one of its side lies on the hypotenuse of the triangle. What is the area of the square ?

(A) [abc/(a² + b² + c²)]²
(B) (a² + b² + c²)/abc
(C) abc/(a² + b² + c²)
(D) (a+b+c)/(a² + b² + c²)
(E) (a² + b² + c²)/3

We can solve this question using the Input-Output approach.
So, let's find a specific example where we can calculate the area of the square.
To do so, let's begin with a simple right triangle that we know a lot about: an isosceles right triangle (i.e., a 45-45-90 right triangle)


From here, notice that, since we've inserted a square with 90º angles, there are some additional 45º angles in this diagram.


Since the triangle in the bottom right corner has two equal angles, we know that the two opposite sides have equal length.


Next, since all sides in a square have equal length, we can see that we have many sides that are the same length.

At this point, we should recognize that the hypotenuse of the large right triangle is comprised of three equal lengths. So, let's choose some nice values for the length of the sides.

If we say that each leg has length 3, then the hypotenuse must have length 3√2


Since the hypotenuse of the large right triangle is comprised of three equal lengths, each line segment must have length √2


Great. We now have a diagram that meets the given conditions.
Here, we see that a = 3, b = 3 and c = 3√2
In this case, the area of the square = (√2)(√2) = 2

At this point, we'll plug a = 3, b = 3 and c = 3√2 into each answer choice and see which one yields an OUTPUT of 2

(A) [abc/(a² + b² + c²)]² = [27√2/(36)]² = something other than 2
(B) (a² + b² + c²)/abc = (36)/27√2 = something other than 2
(C) abc/(a² + b² + c²) = 27√2/(36) = something other than 2
(D) (a+b+c)/(a² + b² + c²) = (6 + 3√2)/(36) = something other than 2
(E) (a² + b² + c²)/3 = (36)/3 = 12

Hmmm, none of the answer choices yield the required output of 2. Are you sure you correctly transcribed the answer choices? (or perhaps I made an error somewhere??)

Cheers,
Brent

Hi Brent,

Yes, there has been an error in option(A) in the denominator. The last term of the denominator is ab instead of c^2.Regret the error. I have already corrected it in the actual post.I would like to know whether this problem can be solved algebraically. I don't have the detailed solution to this problem. I only have the net answer.

[Brent@GMATPrepNow]

Hi Brent,

Yes, there has been an error in option(A) in the denominator. The last term of the denominator is ab instead of c^2.Regret the error. I have already corrected it in the actual post.I would like to know whether this problem can be solved algebraically. I don't have the detailed solution to this problem. I only have the net answer.

Okay, I edited my answer accordingly.

I can't think of an algebraic approach. Of course, one may exist, but I don't see it.

Cheers,
Brent

[Mathsbuddy]

Hi Brent,

Yes, there has been an error in option(A) in the denominator. The last term of the denominator is ab instead of c^2.Regret the error. I have already corrected it in the actual post.I would like to know whether this problem can be solved algebraically. I don't have the detailed solution to this problem. I only have the net answer.

Okay, I edited my answer accordingly.

I can't think of an algebraic approach. Of course, one may exist, but I don't see it.

Cheers,
Brent

In combination with your great method, how about this to save some time:

By quickly by looking at dimensions, we can instantly eliminate B, C and D:

Using L=dimension of Length

(B) (a^2+b^2+c^2)/abc [eliminate because L^2/L^3 = 1/L not area L^2]

(C) abc/(a^2+b^2+c^2) [because L^3/L^2 = 1/L too, not L^2]

(D) (a+b+c)/(a^2+b^2+c^2) [because L/L^2 = 1/L , not L^2]


Perhaps some algebra method can come from this:

Let's call the small right angle triangle (with the hypotenuse being one side of the square) the "minor triangle". We will call the biggest triangle, the "major triangle".

Next we will draw a major square on the hypotenuse of the major triangle.

Now we have some relationships.
Rules of similar shapes mean that:
[minor square area:major square area] = [minor triangle area:major triangle area]
Also, from Pythagoras a^2 + b^2 = c^2 and p^2 + q^2 = r^2 where p, q and r are the dimensions of the minor triangle.
r^2 is what we are trying to find out in terms of a and b.
We also know that a/p = b/q = c/r

Also, if we consider the special case when a=b and p=q, surely we have enough information to solve it. I don't have time now, but I'll try later. Maybe you'll beat me to it!

[Mathsbuddy]

As discussed above, dimensions instantly remove answers b, c and d. This leaves just a and e.


Consider the case of a right-angle isosceles triangle with 45 degree base angles.
Imagine 4 of these triangles arranged so their right angles all meet a centre point. Now we have a large square of area c^2. Inside it we have 9 of the smaller squares. Hence, each edge contains exactly 3 squares.

If the side of a minor square is a third the side of the major square, then the minor square area cannot be a 2/3 of the area of the major square. Instead the area is c^2/9. Therefore answer E must be eliminated too.

E = (a^2+b^2+c^2)/3 = 2c^2/3 (by Pythagoras Theorem).

Hence answer A is the only candidate.