Question

If a, b, c are the lengths of a triangle so that a=xy, b=yz and c=zx, where x, y, z are natural numbers. Then x^2 +y^2+z^2 is:
A. ab/c+ bc/a+ca/b
B. a/bc+ b/can + c/ab
C. ab+bc+ac/a^2+b^2+c^2
D. a^2+b^2+c^2/ab+bc+ac​

Answer 1

Step-by-step explanation:

Option a is the correct answer

Answer 2

$xy = a=> xyz = az => z = \frac{xyz}{a}$

$x z = b=>xyz = by=> y = \frac{xyz}{b}$

$yz = c=>xyz = cx=> x = \frac{xyz}{c}$

also,  

       $abc = xy * xz*yz = (xyz)^2$  $= \frac{(b^2c^2 + c^2a^2 + a^2b^2)}{(abc)}$

now,

$x^2 + y^2 + z^2$

$= (\frac{xyz}{a})^2 + (\frac{xyz}{b})^2 + (\frac{xyz}{c})^2$

$= (xyz)^2 *( \frac{1}{a^2} + \frac{1}{b^2} +\frac{1}{c^2})$

$= \frac{(abc)(b^2c^2 + c^2a^2 + a^2b^2)}{ (abc)^2}$

$=\frac{b^2c^2+c^2a^2+a^2b^2}{ abc}$

$=\frac{b^2c^2}{abc} +\frac{c^2a^2}{abc} + \frac{a^2b^2}{abc}$

$= \frac{bc}{a} + \frac{ca}{b} + \frac{ab}{c}$

Correct option => A

My friend, If this comes in MCQ, instead of solving try putting value in the options, that will be faster than solving this.