Question

If 3a=4b=6c a+b+c=27√29 find (a^2+b^2+c^2)^(1/2) want to 2 marks answer

Answer 1

Let 3a = 4b = 6c = k, where k is a constant.
Therefore,
a = k/3
b = k/4
c = k/6
a + b + c = 27√29
=> (k/3) + (k/4) + (k/6) = 27√29
=> (3k/4) = 27√29
=> k = 36√29.
Therefore,
a = k/3 = 12√29
b = k/4 = 9√29
c = k/6 = 6√29
Therefore,
${( {a}^{2} \: + \: {b}^{2} \: + \: {c}^{2} ) }^{ \frac{1}{2} }$
[(12√29)^2 + (9√29)^2 + (6√29)^2]^(1/2)
=> 87.

Answer 2

√(a² + b² + c²) = 87  if 3a = 4b = 6c  & a + b + c = 27√29

Step-by-step explanation:

Missing information :

3a = 4b = 6c

Let say 3a = 4b = 6c  = 12k

=> a = 4k  

b = 3k  

c = 2k

a + b + c = 27√29

4k + 3k + 2k = 27√29

=> 9k = 27√29

=> k = 3√29

a = 4k = 12√29

b = 3k = 9√29

c = 2k = 6√29

√(a² + b² + c²)

= √ ( (12√29)² + (9√29)² + (6√29)²)

= √(29)( 144 + 81 + 36)

= √(29(261)

= √29 * 29 * 9

= 29 * 3

= 87

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