a, b, and c are sides of a right triangle. What is the area of the tri

Hi,

We know its a right angle triangle..


(1) a = 4.
this gives us just one side, clearly insuff

(2) a + b + c = 12
this gives the perimeter. again clearly insuff.

Combined..
we know one side is 4 and sum of two other sides is 12-4=8...
and these two sides are related by pythagorean theorem...
so it should be suff..

but lest find teh answer
the most important info we can gather from this is a is not the biggest/hypotenuse
so let b be the hypotenuse, so b= \(\sqrt{a^2+c^2}\)..
b= \(\sqrt{4^2+c^2}\)...
\(\sqrt{16+c^2}\)...
substitute in the sum of two other sides is 12-4=8...
c+\(\sqrt{16+c^2}\)=8..
\(\sqrt{16+c^2}\)=8-c..
16+c^2=(8-c)^2..
16+c^2=64+c^2-16c...
16c=48 or c=3...
area =3*4/2=6..
suff..

[Alter to algebraic way
a+b+c = 12
c – hypotenuse

1.case
If C= 4 , hypotenuse cant be four ,since a or b will become higher than four to make a+b+c = 12
So 4 should a or b

2. case
Consider 4 as a or b
A + 4 + C = 12 from here we know 3,4,5 will fit
3+4+5 = 12 (area will be 6 units)
2+4+ 6 = 12 (not a triangle itself , a+b>c here a+b = c)
So only possible triangle here is 3,4,5
But if numbers get tough this approach will not be easy, algebraic way will be better
By using this method we will able to save time

Statement 1 allows for creating any right triangle one side of which is 4 long. The area can be anything.

Insufficient.

Statement 2 allows for the creation of various right triangles with various areas.

Insufficient.

Combining the statements we know the following.

4 cannot be the hypotenuse as the hypotenuse must be the longest side, and the remaining 8 cannot be divided into 4 and 4 as doing that would not create a right triangle.

As soon as 4 is used for one of the sides, we have 8 left.

3, 4 and 5 work for the two sides and the hypotenuse.

If we increase the length of the 3 side and reduce the length of the hypotenuse, then a² + b² will be greater than c².

If we decrease the length of the 3 side and increase the length of the hypotenuse, then a² + b² will be less than c².

So once the perimeter is defined and the length of one side is defined, the lengths of the other side and the hypotenuse are defined.

The correct answer is C.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.