Question

$\sqrt{1+(a-6{)}^2}=\sqrt{29}+\sqrt{36+(4-9{)}^2}$

Answer 1

Consider the given decimal number

$\Rightarrow \sqrt{1+{(a-6)}^2}=\sqrt{29}+\sqrt{36+(4-9)}$

$\Rightarrow \sqrt{1+{(a-6)}^2}=\sqrt{29}+\sqrt{36-5}$

$\Rightarrow \sqrt{1+{(a-6)}^2}=\sqrt{29}+\sqrt{29}$

$\Rightarrow \sqrt{1+{(a-6)}^2}=2\sqrt{29}$

$\Rightarrow 1+{(a-6)}^2=2\times 29$

$\Rightarrow {(a-6)}^2=58-1$

$\Rightarrow a=\sqrt{57}+6$

$\Rightarrow a=-\sqrt{57}+6$

Hence, this is the answer.