Math Problem Statement

If 3a = 4b = 6c and a + b + c = 27sqrt(29) find sqrt(a ^ 2 + b ^ 2 + c ^ 2)

Solution

We are given the system of equations:

3a=4b=6c3a = 4b = 6c

and

a+b+c=2729a + b + c = 27\sqrt{29}

Step 1: Express aa, bb, and cc in terms of one variable

From the equation 3a=4b=6c3a = 4b = 6c, we can introduce a variable kk such that:

3a=4b=6c=k3a = 4b = 6c = k

This gives us the following expressions:

a=k3,b=k4,c=k6a = \frac{k}{3}, \quad b = \frac{k}{4}, \quad c = \frac{k}{6}

Step 2: Use the equation a+b+c=2729a + b + c = 27\sqrt{29}

Substitute a=k3a = \frac{k}{3}, b=k4b = \frac{k}{4}, and c=k6c = \frac{k}{6} into the equation a+b+c=2729a + b + c = 27\sqrt{29}:

k3+k4+k6=2729\frac{k}{3} + \frac{k}{4} + \frac{k}{6} = 27\sqrt{29}

To add these fractions, find the least common denominator (LCD) of 3, 4, and 6, which is 12:

k3=4k12,k4=3k12,k6=2k12\frac{k}{3} = \frac{4k}{12}, \quad \frac{k}{4} = \frac{3k}{12}, \quad \frac{k}{6} = \frac{2k}{12}

Now add the fractions:

4k12+3k12+2k12=9k12=3k4\frac{4k}{12} + \frac{3k}{12} + \frac{2k}{12} = \frac{9k}{12} = \frac{3k}{4}

So, we have:

3k4=2729\frac{3k}{4} = 27\sqrt{29}

Step 3: Solve for kk

Multiply both sides by 4:

3k=108293k = 108\sqrt{29}

Now divide by 3:

k=3629k = 36\sqrt{29}

Step 4: Find a2+b2+c2\sqrt{a^2 + b^2 + c^2}

Now, we need to compute a2+b2+c2\sqrt{a^2 + b^2 + c^2}. Using the expressions for aa, bb, and cc:

a=k3=36293=1229a = \frac{k}{3} = \frac{36\sqrt{29}}{3} = 12\sqrt{29} b=k4=36294=929b = \frac{k}{4} = \frac{36\sqrt{29}}{4} = 9\sqrt{29} c=k6=36296=629c = \frac{k}{6} = \frac{36\sqrt{29}}{6} = 6\sqrt{29}

Now compute a2+b2+c2a^2 + b^2 + c^2:

a2=(1229)2=144×29=4176a^2 = (12\sqrt{29})^2 = 144 \times 29 = 4176 b2=(929)2=81×29=2349b^2 = (9\sqrt{29})^2 = 81 \times 29 = 2349 c2=(629)2=36×29=1044c^2 = (6\sqrt{29})^2 = 36 \times 29 = 1044

Now add these:

a2+b2+c2=4176+2349+1044=7569a^2 + b^2 + c^2 = 4176 + 2349 + 1044 = 7569

Step 5: Take the square root

a2+b2+c2=7569=87\sqrt{a^2 + b^2 + c^2} = \sqrt{7569} = 87

Thus, the value of a2+b2+c2\sqrt{a^2 + b^2 + c^2} is 8787.


Would you like any further details or clarifications on the steps?

Related questions:

  1. How would the solution change if the values of a+b+ca + b + c were different?
  2. How can this method be applied to solve other systems of proportional equations?
  3. What is the significance of introducing the variable kk in solving this type of equation?
  4. Can this approach be generalized for ma=nb=pcma = nb = pc type of systems?
  5. How do different numerical values of aa, bb, and cc affect the value of a2+b2+c2\sqrt{a^2 + b^2 + c^2}?

Tip:

When dealing with proportional equations, using a common variable like kk can simplify the process of solving for multiple unknowns at once.

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Math Problem Analysis

Mathematical Concepts

Algebra
Proportional Equations
Radicals

Formulas

a = k/3, b = k/4, c = k/6
a^2 + b^2 + c^2
√(a^2 + b^2 + c^2)

Theorems

Proportional Relationships
Pythagorean Theorem

Suitable Grade Level

Grades 10-12