Question

when the denominator 1÷2√5 is rationalised it becomes,it becomes √k÷19,find k​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$\frac{1}{2 \sqrt{5} } \\ = \frac{ \sqrt{5} }{2 \sqrt{5} \times \sqrt{5} } \\ = \frac{ \sqrt{5} }{2 \times 5} \\ = \frac{ \sqrt{5} }{10}$

Now, according to the question

$\frac{ \sqrt{5} }{10} = \frac{ \sqrt{k} }{19} \\ = >10 \sqrt{k} = 19 \sqrt{5} \\ = > ({10 \sqrt{k} )}^{2} = ( {19 \sqrt{5} )}^{2} \\ = > 100k = 361 \times 5 \\ = > k = \frac{361 \times 5}{100} \\ = > k = \frac{1805}{100} \\ = > k = 18.05$