Question
What should come in place of both the question marks in the equation ?
$$\frac{?}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{?}$$
$$\frac{?}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{?}$$
Answer: Option A
$$\eqalign{
& {\text{Let,}} \cr
& {\text{ }}\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr
& {\text{Then,}} \cr
& \Leftrightarrow {x^2} = \sqrt {128 \times 162} \cr
& \Leftrightarrow {x^2} = \sqrt {64 \times 2 \times 18 \times 9} \cr
& \Leftrightarrow {x^2} = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr
& \Leftrightarrow {x^2} = 8 \times 6 \times 3 \cr
& \Leftrightarrow {x^2} = 144 \cr
& \Leftrightarrow x = \sqrt {144} \cr
& \Leftrightarrow x = 12 \cr} $$
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$$\eqalign{
& {\text{Let,}} \cr
& {\text{ }}\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr
& {\text{Then,}} \cr
& \Leftrightarrow {x^2} = \sqrt {128 \times 162} \cr
& \Leftrightarrow {x^2} = \sqrt {64 \times 2 \times 18 \times 9} \cr
& \Leftrightarrow {x^2} = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr
& \Leftrightarrow {x^2} = 8 \times 6 \times 3 \cr
& \Leftrightarrow {x^2} = 144 \cr
& \Leftrightarrow x = \sqrt {144} \cr
& \Leftrightarrow x = 12 \cr} $$
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