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If 3a = 4b = 6c and a + b + c = $$27\sqrt {29} $$ then $$\sqrt {{a^2} + {b^2} + {c^2}} $$ is equal to
Answer & Solution
Correct Answer:
Option
A
$$\eqalign{
& 3a = 4b = 6c \cr
& \Rightarrow \frac{{3a}}{{12}} = \frac{{4b}}{{12}} = \frac{{6c}}{{12}} \Rightarrow \frac{a}{4} = \frac{b}{3} = \frac{c}{2} = k \cr
& \Rightarrow a = 4k,\,b = 3k,\,c = 2k \cr
& a + b + c = 27\sqrt {29} \cr
& 9k = 27\sqrt {29} \cr
& k = 3\sqrt {29} \cr
& a = 4 \times 3\sqrt {29} ,\,b = 3 \times 3\sqrt {29} ,\,c = 2 \times 3\sqrt {29} \cr
& \sqrt {{a^2} + {b^2} + {c^2}} \cr
& = \sqrt {29\left( {144 + 81 + 36} \right)} \cr
& = \sqrt {29 \times 261} \cr
& = \sqrt {29 \times 29 \times 9} \cr
& = 29 \times 3 \cr
& = 87 \cr} $$
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