## Beat The GMAT Challenge Question – September 28, 2010

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The perimeter of a right triangle is . If the sum of the squares of the lengths of the three sides is 96, what is the area of the triangle?

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## 27 comments

Sunil on September 28th, 2010 at 7:38 am

Hi,

The area is 6*sqrt(3)

The solution for this question is as follows:

Let the three sides of the right triangle be a,b,c and lets suppose that c is the hyoptaneous of the right triangle.

So the area of the triangle is = (1/2)*a*b

Now perimeter = a+b+c = 6+6*sqrt(3)

And, a^2 + b^2 + c^2 = 96

As we assumed c to be the hypotaneous, so c = sqrt(a^2 + b^2)

so, a+b+sqrt(a^2 + b^2) = 6+6*sqrt(3) ------ (1)

and 2*(a^2 + b^2) = 96

so, a^2 + b^2 = 48

putting this value in the eqn 1, a+b = 6+6*sqrt(3) - 4*sqrt(3)

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

Squairing both sides of the eqn 2,

a^2 + b^2 + 2*a*b = 36+12+2*6*2*sqrt(3)

putting a^2 + b^2 = 48,

48 + 2*a*b = 48+2*6*2*sqrt(3)

so 2*a*b = 2*6*2*sqrt(3)

so, a*b = 6*2*sqrt(3)

Area = (1/2) *a*b = (1/2)*6*2*sqrt(3) = 6*sqrt(3)

Abhishek Pandey on September 28th, 2010 at 8:41 am

Let the sides of right angled triangle be ‘a’ and ‘b’, as shown in figure

I _

a I _ (a^2 + b^2)^1/2

I _

I_ _ _ _ _ _ _ _ _ _ _ _ _ __

b

Length = a

Height =b

So hypotenuse = (a^2 + b^2)^1/2

Given :

1) Perimeter of right angled triangle = 6 + 6(3)^1/2

=> a + b + (a^2 + b^2)^1/2= 6 + 6(3)^1/2

2) Sum of squares of the lengths of three sides = 96

=> a^2 + b^2 + ((a^2 + b^2)^1/2)^2 = 96

Solving equation 2:

=> a^2 + b^2 + ((a^2 + b^2)^1/2)^2 = 96

=> a^2 + b^2 + (a^2 + b^2 ) = 96

=> 2 (a^2 + b^2 ) = 96

=> a^2 + b^2 = 48

Putting the value of ‘a^2 + b^2 ‘ in equation 1:

=> a + b + (a^2 + b^2)^1/2= 6 + 6*sqrt(3)

=> a + b + (48)^1/2 = 6 + 6*sqrt(3)

=> a + b + 4(3)^1/2 = 6 + 6*sqrt(3)

=> a + b = 6 + 6*sqrt(3) - 4*sqrt(3)

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

From equation 1 : we have a + b = 6 + 2 *sqrt(3)

From equation 2 : we have a^2 + b^2 = 48

For the area of triangle we need to get value of ½ ab

Using first information :

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

( Squaring both sides )

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

=> a^2 + b^2 + 2ab = 36 + 12 + 24*sqrt(3)

=> a^2 + b^2 + 2ab = 48 + 24*sqrt(3)

(substituting second information - a^2 + b^2 = 48 )

=> 48 + 2ab = 48 + 24*sqrt(3)

=> 2ab = 24* sqrt(3)

=> ab = 12* sqrt(3)

So area of triangle = ½ ab

= ½ (12 * sqrt(3))

= 6 * sqrt(3)

shibbirahamed on September 28th, 2010 at 9:02 am

Let, the three sides of the triangle be ‘a’,’ b’ and ‘c’ where ‘c’ is the hypotenuse of the right triangle. So the area of the triangle is (1/2) a*b.

Given-

a+b+c = 6+6√3 ----------------------(1)

a^2+b^2+c^2 = 96 ----------------------(2)

Now, According to Pythagorean Theorem; a^2+b^2=c^2

Putting the value of a^2+b^2 into equation (2) we have

c^2+c^2 = 96

=>2c^2 = 96

=>c^2 = 48

=> c = √48

=> c = 4√3

Putting the value of 'c' into equation (1) we have

a+b+4√3=6+6√3

a+b =6+2√3

Now, 2*a*b =(a+b)^2-(a^2+b^2)

=(a+b)^2-c^2; [since a^2+b^2=c^2]

=(6+2√3)^2-48; [since c^2=48]

=24√3

therefore,

(1/2)a*b =6√3

So the area of the triangle is 6√3.

Sanitha Alam on September 28th, 2010 at 9:48 am

The answer is 6 √3.

Solution:

Let the sides of the triangle be a,b,c with a as the altitude, b as the base and c as the hypotenuse.

Given:

a+b+c=6+6√3---(1)

a^2 + b^2 + c^2 = 96---(2)

For a right triangle, a^2 + b^2= c^2---(3)

(2) & (3) => 2 c^2=96 => c^2=48

=> c= 4√3 ---(4)

(1) & (4) => a+b=6+2√3

We know that:

(a+b)^2=a^2+b^2+2a.b

(6+2√3)^2=48+2a.b

Area=1/2(ab)=6 √3

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)

Kumaran on September 28th, 2010 at 9:32 am

Let the three sides of the right angled triangle be a, b and c where c is the hypotenuse. We know that c^2 = a^2 + b^2.

Given that a^2+b^2+c^2 = 96 and a+b+c (perimeter) = 6+6√3

Substituting c^2 = a^2 + b^2 in a^2+b^2+c^2 = 96.

2c^2 = 96 => c^2 = 48 => c = 4√3.

Substituting value of c in a+b+c = 6+6√3 => a+b = 6+2√3.

We know that c^2 = (a^2+b^2) = 48 and (a+b) = 6+2√3

(a+b)^2 = a^2 +b^2 + 2ab

(6+2√3)^2 = 48 + 2ab.

Simplifying we get 48 + 24√3 = 48 +2ab => 24√3 = 2ab => ab=12√3

Now area of the triangle is (1/2)*ab => (1/2)*12√3 = 6√3.

Answer is 6√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........................

s.sharma on September 28th, 2010 at 9:41 am

The answer is 6 multiplied by the square root of 3 ( 6* sqrt3 ).

Let the 3 sides be p, b, and h. p stands for perpendicular side, b stands for the base side and the h stands for the hypotenuse.

we are given two things

1st: p + b + h = 6+ 6 sqrt(3)

2nd: p^2 + b^2 + h^2 = 96

and we already know that p^2 + b^2 = h^2

from these 3 equations the product pb can be deduced.

pb = 12* sqrt(3)

therefore, the area is 6*sqrt(3).

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 10:07 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. ------------ (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 ------------ (2)

As per Pythagoras theorem,

a^2 + b^2 = c^2 ------------ (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.

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.

Anand on September 28th, 2010 at 10:24 am

Clearly it is an 30:60:90 triangle and sides are in ratio 1:sqrt(3):2

a+a.sqrt(3)+2a=6+6.sqrt(3)-->Eqn 1

a^2+(a^2)3+4a^2=96 or a=2sqrt(3)

Sub a in Eqn1. 3 sides are 2sqrt(3),6,4sqrt(3)

Area=1/2*(Product of 2 legs containing 90)=1/2*2sqrt(3)*6

=6sqrt(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.

Anahatha on September 28th, 2010 at 10:31 am

This should be a 30-60-90 triangle because

perimeter= 6+6√3=6(1+√3)

The perimeter of such a triangle should be a multiple of (1+2+√3)

or 3+√3 or 3(1+√3)

a^2+b^2+c^2=96

taking out the common ratio

ratio^2 * (1+4+3)=96

ratio=2√3

Now we get the base and height as 2√3 and 6 ( by applying ratio 1:√3:2) and hence area 6√3

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.

pnk on September 28th, 2010 at 10:58 am

perimeter = 6 + 6 Sqrt (3)

since it contains sqrt => triplets will not be the sides

trial and error to find possible sides. Since perimeter contains sqrt implies one of the sides is a factor of sqrt of a number.

trial 1: perimeter = 6 + sqrt (3) + 5 sqrt (3). but in this case sum of squares of sides ie 6, sqrt (3) and 5 sqrt (3) is will not be 96

trial 2: perimeter = 6 + 2 sqrt (3) + 4 sqrt (3). Square of sides ie 6, 2sqrt (3) and 4 sqrt (3) is 96.

Therefore sides are 2 sqrt (3) and 6.

Area = 1/2*2sqrt(3)*6 = 6 sqrt(3)

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.

saurabh goyal on September 28th, 2010 at 11:30 am

I am sorry i forget to mention 16 in while i was mentioning the squares of 1 to 6.

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 11:59 am

Assume a & b are two legs of the triangle. I can draw following equations from given information.

1. a+b+sqrt(a^2+b^2) = 6+6sqrt(3)

2. 2(a^2+b^2) = 96 --> a^2+b^2 = 48=4sqrt(3)

If I substitute 2 in 1 then I get a+b=6+2sqrt(3)

If I use below formula

(a+b)^2 = a^2+b^2+2ab

(6+2sqrt(3))^2=48+2ab

2ab = 36+24sqrt(3)+12-48

2ab=24sqrt(3)

2ab/4=24sqrt(3)/4

1/2 ab = 6sqrt(3)

Ans: area is 6sqrt(3)

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)

anj on September 28th, 2010 at 3:21 pm

The Answer is 6√3.

Since perimeter is given as 6+6√3 that means either diagonal is 6√3 (largest length) or there are 2 sides whose total is 6√3.

and there is √3 in the perimeter that means it is a 30-60-90 triangle and not 45-45-90 triangle(since no √2).

If we assume diagonal as 6√3 the we get the perimeter as 9+6√3....and therefore the assumption is wrong ....

Now it is clear that there are 2 sides that have √3 multiple, whose addition will be 6√3....

Therefore we assume that diagonal is 4√3 because it is next lowest number and divisible by 2 to give us side in front of angle 60 as 6, also side in front of angle 30 as 2√3

That means we have got 3 sides as 4√3(diagonal), 2√3 and 6.

If we add all these measurements then we get the perimeter.

Put these values in the second given condition of addition of squares and it satisfies the condition.

Now if we calculate the area with fomula 1/2 (product of perpendicular sides) we get answer as 6√3

Harish on September 28th, 2010 at 6:57 pm

Let the sides be s1,s2 and s3(hypotenuse). We have to find (s1*s2)/2 - area of triangle

The three equations are -

1)s1+s2+s3=6+(6*3^(1/2))

2)s1^2+s2^2+s3^2 = 96

3)s1^2+s2^2=s3^2

Solving equation 2) and 3) gives S3 =4*(3^(1/2))

=>s1+s2 = 6+2*(3^(1/2))

Square the above equation

(s1+s2)^2 = (6+2*(3^(1/2)))^(2)

=>2s1s2 = 24*(3^(1/2))

=>Area = (s1s2)/2 = 6*(3^(1/2))

Answer is 6*(3^(1/2))

GMAT Delhi on September 28th, 2010 at 9:18 pm

Right triangle with sides a,b,c

Given:

(i) a+b+c = 6+6√3

(ii) a²+b²+c² = 96

(iii) a²+b² = c² (pythogaras theorem)

To find: Area of triangle = 1/2*(a*b) [assuming c is the hypotenuse]

Solution:

Putting (iii) in (ii),

c²+c² = 96

2c² = 96

c² = 48

c = 4√3

Putting value of c in (i),

a+b+4√3 = 6+6√3

a+b = 6+6√3 - 4√3

a+b = 6+2√3

Squaring both sides,

(a+b)² = (6+2√3)²

(a²+b²)+2ab = 36+12+24√3

(c²)+2ab = 36+12+24√3 [using (iii)]

48+2ab = 48+24√3 [substituting value of c]

ab = 12√3

Therefore,

Area of triangle = 1/2*(ab) = 1/2*12√3 = 6√3.

Eric Bahn on September 28th, 2010 at 10:08 pm

Congrats to Sanitha Alam for being selected as this week's BTG Challenge Question winner! Sanitha: we are going to contact you shortly to give you your prize!

All the best!

Ashok Kadam on September 28th, 2010 at 11:55 pm

Congrats Sanitha!! well done!! you're the luckiest

Sanitha Alam on September 29th, 2010 at 9:03 am

Thanks a lot