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Agastya on September 28th, 2010 at 9:13 am

The answer is 6*sqrt(3)

Here is how we can solve it:

Let the base and height of the right triangle be b and h respectively. The hypotenuse will be sqrt(b^2+h^2). Using the perimeter clause we have:

b + h + sqrt(b^2+h^2) = 6 + 6*sqrt(3) --------- (1)

Since the sum of the square of sides is 96, we have:
b^2 + h^2 + (b^2+h^2) = 96
This gives us:

b^2 + h^2 = 48 -------------------------------(2)

Substituting Eqn. (2) in (1), we have:

b + h = 6 + 2*sqrt(3) ----------------------------(3)

If we square both sides of Eqn. (3) we get:

b^2 + h^2 + 2bh = 48 + (24*sqrt(3))

Substituting Eqn (2) in the above equation, we get:

48 + 2bh = 48 + 24*sqrt(3)

From this we can solve for bh:

bh = 12*sqrt(3)

Since the area of the right triangle is 0.5bh, we find that the area is:

Area of the right triangle = 0.5bh = 6*sqrt(3)

Saptarshi Roy on September 28th, 2010 at 9:40 am

let us take the sides as a,b & c with C being the hypotenuse.

==> a^2 + b^2 = c^2. ............................................ Statement 1
as per inputs in the question:

i.) a + b + c = 6 + 6√3.............................................Statement 2
ii.) a^2 + b^2 + c^2 = 96......................................... Statement 3

As perm Statements 1 and 3 replacing a^2 + b^2 in statement 3,

c^2 + c^2 = 96
==> c^2 = 48
==> c = 4√3 ......................................................... Statement 4

From 2 and 4 we get,

a + b = 6 + 2√3

Now squaring both sides of the above equation,
a^2 + b^2 + 2ab = 36 + 12 + 24√3
==> a^2 + b^2 + 2ab = 48 + 24√3

Now a^2 + b^2 = c^2 = 48

==> 48 + 2ab = 48 + 24√3
==> 2ab = 24√3
==> ab = 12√3

Now area of a right triangle = 1/2( base * altitude) = 1/2 (ab)

==> area = 1/2(ab) = 1/2 (12√3) = 6√3

Thus the answer is achieved........................

ASKI@1st on September 28th, 2010 at 10:05 am

Let’s call x, y, z the sides of the right triangle with z being the hypotenuse.

(1) Perimeter = P = x+y+z= 6+6*sqrt(3)
(2) Sum of squares of the lengths of the three sides = S = x^2+y^2+z^2=96

Area = A= 0.5*x*y =?

2 - We all know that for a right triangle : x^2+y^2=z^2, therefore
S= z^2+z^2=96 => 2*z^2=96 => so z=4*sqrt(3)

1 - P = (x+y+z) => (P-z)^2= (x+y)^2 = x^2+y^2+2*x*y = z^2+4*A
(P-z)^2 – z^2 = 4*A => (P-2*z)*P=4*A
4*A = (6+6*sqrt(3)-8*sqrt(3))*( 6+6*sqrt(3))=12*sqrt(3) => A=6*sqrt(3)

Well, well, well… (Cyndi Lauper – We’re the world)

Ashok Kadam on September 28th, 2010 at 10:10 am

Let the sides of the triangle be a,b and c where c is hypotenuse.
Since we are given that the perimeter of a right triangle is 6 + 6 √3.
=> a + b + c = 6 + 6 √3. ------------ Equation (1)

We are also given that the sum of the squares of the lengths of the three sides is 96.
=> a^2 + b^2 + c^2 = 96 ------------ Equation (2)

As per Pythagoras theorem,
a^2 + b^2 = c^2 ------------ Equation (3)

Combining (2) and (3), we get
2 * c^2 = 96.
c^2 = 48 = 16 * 3.
c = 4 √3.

Substituting the value of c in (1), we get
a + b + 4 √3 = 6 + 6 √3
a +b = 6 + 2 * √3

Now,
(a + b)^2 = a^2 + b^2 + 2*a*b
=> a*b = (6 + 2*√3)^2 - 48
=> a*b = 12 * √3

Thus the are of the triangle is
= 1/2 * a * b = 1/2 * 12 * √3.
= 6*√3.

Gurpreet singh on September 28th, 2010 at 10:24 am

a^2 + b^2 + c^2 = 96 where a and b are legs and c is hypotenuse.

also a^2+b^2 = c^2 => 2c^2= 96 => c = 4* sqrt(3)

Perimeter = a+b+c = 6+6* sqrt(3) = a+b+4* sqrt(3)

=> a+b = 6+2* sqrt(3)

a = 6 and b = 2* sqrt(3) => area = 1/2 * 2* sqrt(3) *6

=6* sqrt(3)

PS: I verified it quickly by reconciling 6^2 + {2* sqrt(3) }^2 = 48 = c^2

So we don't have to deduce it.

Chris Lele on September 28th, 2010 at 10:56 am

We can assume it's a 30:60:90 triangle and plug in values. The obvious number of 3 doesn't quite work--we get a perimeter of 9 + 3√3. We need a number that is slightly bigger and can quickly see that plugging in 4 will give us too large a number. Since the perimeter is in the form of x +x√3, we need to stick to integers or some integer with a √3. 2√3 is between 3 and 4. Plugging it in will give you the perimeter of 6 + 6√3. Then we just multiply the base and height-- x = 2√3 and x√3 = 6--to get 12√3 and we divide by 2. Ans: 6√3.

Another note: Working this way, we don't need the extra info. that the sum of the squares of the three sides = 96. Were this a data sufficiency problem, knowing that the triangle was a right triangle with the perimeter of 6 + 6√3 would allow us to derive the area. That is, there is only one possible triangle that conforms to these parameters.

saurabh goyal on September 28th, 2010 at 11:29 am

Answer is same as your guys have already mentioned 6√3
But I did a bit differently,
We know the sum of three square of three sides is 96.
We know the sum of three sides is 6+6√3
Now one side must definitely be in the range of √3 to 6√3
If square all the terms from √3 to 6√3 then the square will be 3,12,27,48,75,108.
Now 108 is more than 96 that means one side can’t be 6√3 it should be lesser than that. It should be in the range of √3 to 5√3.
Square of 1 to 6 is 1,4,9,25,36
Now we have a set of 11 different values and we know 3 of them are sides of a triangle, then by hit and trial I got my answer in second attempt. As 4√3,6,2√3.

Ishaan Singhal on September 28th, 2010 at 11:46 am

We know:
Let the 3 sides be a,b,c where c is the hypotenuse of the right triangle.

Now,
a + b + c = 6 + 6√3.................(I)
a² + b² + c² = 96.....................(II)

we need to find area of triangle = ½*a*b
we also know from Pythagoras Theorem that a² + b² = c²..........(III)
from eqns (II) and (III)
c² + c² = 96
hence
c = ±4√3
we neglect the negative value as a side cannot be less than 0
hence we have c = 4√3..........(IV)
from eqn (I) and (IV)

a + b = 6 + 2√3

From here we can solve this in two ways:-
1)
Since a + b is equivalent to a rational number plus an irrational number hence either of the two variables is equal to 6 and the other one is equal to 2√3. In each case the product of the two is
a * b = 6 * 2√3 = 12√3
hence
Area of triangle = ½*a*b = 6√3

2)
As per the Involution we have the following algebraic expression
(a + b)² = a² + b² + 2*a*b
Substituting the known values we have
(6 + 2√3)² = c² + 2*a*b
36 + 12 + 24√3 = 48 + 2*a*b
solving for ½*a*b we will get
½*a*b = 6√3

Thanks for the question
Ishaan

Chandra Pallaka on September 28th, 2010 at 12:23 pm

Assume a & b are two legs of the triangle
1. a+b+sqrt(a^2+b^2)=6+6sqrt(3)
2. a^2+b^2+sqrt(a^2+b^2)=96 --> a^2+b^2=48
if you substitute 2 in 1 then a+b=6+2sqrt(3)
use this formula
(a+b)^2=a^2+b^2+2ab
(6+2sqrt(3))^2=48+2ab
2ab=36+24sqrt(3)+12-48
2ab/4=24sqrt(3)/4
area = 6sqrt(3)