If 3a = 4b = 6c and a + b + c = $$27\sqrt {29} $$ then $$\sqrt {{a^2} + {b^2} + {c^2}} $$ is equal to
A. 87
B. $$3\sqrt {29} $$
C. 82
D. 83
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & 3a = 4b = 6c \cr & \Rightarrow \frac{{3a}}{{12}} = \frac{{4b}}{{12}} = \frac{{6c}}{{12}} \Rightarrow \frac{a}{4} = \frac{b}{3} = \frac{c}{2} = k \cr & \Rightarrow a = 4k,\,b = 3k,\,c = 2k \cr & a + b + c = 27\sqrt {29} \cr & 9k = 27\sqrt {29} \cr & k = 3\sqrt {29} \cr & a = 4 \times 3\sqrt {29} ,\,b = 3 \times 3\sqrt {29} ,\,c = 2 \times 3\sqrt {29} \cr & \sqrt {{a^2} + {b^2} + {c^2}} \cr & = \sqrt {29\left( {144 + 81 + 36} \right)} \cr & = \sqrt {29 \times 261} \cr & = \sqrt {29 \times 29 \times 9} \cr & = 29 \times 3 \cr & = 87 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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