Problem Solving

soumya_joy wrote:Let a, b and c are the sides of a triangle ABC. Given (a + b + c) (b + c - a) = kbc, then k
will lie between.
A. -1 and 1 B. -4 and 4 C. 0 and 4 D. 4 and 6

The proper mathematical approach for solving this problem is well beyond the reach of GMAT. So I'm going to solve this problem using picking number approach.

(a + b + c) is always positive and as the sum of two sides of a triangle is always greater than the third side, (b + c - a) is also positive. Hence, (a + b + c)(b + c - a) is always positive.

Hence, option A and B cannot be the answer.

Now, if we take a = b = c = 1, (a + b + c)(b + c - a) = 3

Hence, C is the only possible answer.

If anyone is looking for the mathematical approach see my next post.

Mathematical Approach:
--> (a + b + c)(b + c - a) = kbc
--> (b² + c² - a² + 2bc) = kbc
--> (b² + c² - a²) = (kbc - 2bc)

Now, cosine of angle A in a triangle is given by cos A = (b² + c² - a²)/(2bc)

Hence, cos A = (b² + c² - a²)/(2bc) = (kbc - 2bc)/(2bc) = (k - 2)/2

Now, -1 ≤ cos A ≤ 1
---> -1 ≤ (k - 2)/2 ≤ 1
---> -2 ≤ (k - 2) ≤ 2
---> 0 ≤ k ≤ 4

The correct answer is D.