If 3a = 4b = 6c and a + b + c = \(27\sqrt{29}\) , then \(\sqrt{a^2+

1.	If 3a = 4b = 6c and a + b + c = \(27\sqrt{29}\) , then \(\sqrt{a^2+b^2+c^2}\) is(a) \(3\sqrt{29}\) (b) 81(c) 87(d) 29
Answer» (c) 87 3a = 4b = 6c \(\Rightarrow\) 4b = 6c \(\Rightarrow\) b = \(\frac{3}{2}c\) and 3a = 4b \(\Rightarrow\) a = \(\frac{4}{3}b\) = \(\frac{4}{3}\times \frac{3}{2}c\) = 2c \(\therefore\) a + b + c = \(27\sqrt{29}\) \(\Rightarrow\) 2c + \(\frac{3}{2}c\) + c = \(27\sqrt{29}\) \(\Rightarrow\) \(\frac{9}{2}c\) = \(27\sqrt{29}\) \(\Rightarrow\) c = \(6\sqrt{29}\) Now, \(\sqrt{a^2+b^2+c^2}\) = \(\sqrt{(a+b+c)^2-2(ab+bc+ca)}\) = \(\sqrt{(27\sqrt{29})^2-2(2c\times\frac{3}{2}c+\frac{3}{2}c\times c+c\times 2c)}\) = \(\sqrt{729 \times29 - 2(3c^2+\frac{3}{2}c^2+2c^2})\) = \(\sqrt{729 \times29 - 2\times \frac{13c^2}{2}}\) = \(\sqrt{729 \times29 - 13\times(6\sqrt{29)^2}}\) = \(\sqrt{29(729-468)}\) = \(\sqrt{29\times261}\) = \(\sqrt{29\times29\times9}\) =29 x 3 = 87

If 3a = 4b = 6c and a + b + c = \(27\sqrt{29}\) , then \(\sqrt{a^2+b^2+c^2}\) is(a) \(3\sqrt{29}\) (b) 81(c) 87(d) 29

Answer»

(c) 87

3a = 4b = 6c \(\Rightarrow\) 4b = 6c \(\Rightarrow\) b = \(\frac{3}{2}c\)

and 3a = 4b \(\Rightarrow\) a = \(\frac{4}{3}b\) = \(\frac{4}{3}\times \frac{3}{2}c\) = 2c

\(\therefore\) a + b + c = \(27\sqrt{29}\)

\(\Rightarrow\) 2c + \(\frac{3}{2}c\) + c = \(27\sqrt{29}\)

\(\Rightarrow\) \(\frac{9}{2}c\) = \(27\sqrt{29}\) \(\Rightarrow\) c = \(6\sqrt{29}\)

Now, \(\sqrt{a^2+b^2+c^2}\)

= \(\sqrt{(a+b+c)^2-2(ab+bc+ca)}\)

= \(\sqrt{(27\sqrt{29})^2-2(2c\times\frac{3}{2}c+\frac{3}{2}c\times c+c\times 2c)}\)

= \(\sqrt{729 \times29 - 2(3c^2+\frac{3}{2}c^2+2c^2})\)

= \(\sqrt{729 \times29 - 2\times \frac{13c^2}{2}}\)

= \(\sqrt{729 \times29 - 13\times(6\sqrt{29)^2}}\)

= \(\sqrt{29(729-468)}\) = \(\sqrt{29\times261}\) = \(\sqrt{29\times29\times9}\)

=29 x 3 = 87