Question

If sides of a triangle ABC satisfy a relation 4a2 + 9b2 + c2 = 2a(a + 3b + c), then - qbq-xvvm-rnh​

Answer 1

Answer:

Correct option is

A

3

Given  a+b−c=2 and 2ab−c2=4

Using these two 2ab−4=c2=(a+b−2)2

⇒2ab−4=a2+b2+2ab−4a−4b+4⇒a2+b2−4a−4b+8=0

⇒(a−2)2+(b−2)2=0⇒a=b=2⇒c=2

Thus triangle is equilateral. Hence Δ2=(43.(2)2)2=3

Answer 2

Answer:

$a,b,c$ are in HP.

Step-by-step explanation:

Given:

$4x^{2} + 9b^{2 }+ 16c^{2} = 2(3ab + 6bc + 4ca)$

To find:

The Series

Solution:

$2(4a^{2} + 9b^{2} + 16c^{2} = 2*2(3ab + 6bc + 4ca)$

$4a^{2}+4a^{2}+9b^{2}+9b^{2}+16c^{2}+16c^{2}-12ab-24bc-16ca = 0$

$(2a-3b)^{2}+(3b-4c)^{2}+(4c-2a)^{2} = 0$

$2a-3b = 0, \\3b-4c = 0, \\4c-2a = 0$

$2a=3b=4c= x$

$a =\frac{x}{2} , b =\frac{x}{3}, c =\frac{x}{4}$

$2,3,4$ are in AP

$\frac{1}{c} + \frac{1}{a} =\frac{6}{x}$

$\frac{1}{c} + \frac{1}{a} =\frac{2}{b}$

So $a,b,c$ are in HP.