# Please solve q 56 ?

May 12, 2018

option (4) is acceptable

#### Explanation:

$a + b - c$

$= {\left(\sqrt{a} + \sqrt{b}\right)}^{2} - {\left(\sqrt{c}\right)}^{2} - 2 \sqrt{a b}$

$= \left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right) \left(\sqrt{a} + \sqrt{b} - \sqrt{c}\right) - 2 \sqrt{a b}$

$= \left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right) \left(\sqrt{c} - \sqrt{c}\right) - 2 \sqrt{a b}$

$= \left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right) \times 0 - 2 \sqrt{a b}$

$= - 2 \sqrt{a b} < 0$

So $a + b - c < 0 \implies a + b < c$
This means sum of the lengths of two sides is less than the third side . This is not possible for any triangle.
Hence formation of triangle is not possible i.e option (4) is acceptable

May 24, 2018

Option (4) is correct.

#### Explanation:

Given,

$\rightarrow \sqrt{a} + \sqrt{b} = \sqrt{c}$

$\rightarrow {\left(\sqrt{a} + \sqrt{b}\right)}^{2} = {\left(\sqrt{c}\right)}^{2}$

$\rightarrow a + 2 \sqrt{a b} + b = c$

$\rightarrow a + b - c = - 2 \sqrt{a b}$

$\rightarrow a + b - c < 0$

$\rightarrow a + b <$$c$

So, no formation of triangle is possible.