Yuridias wrote:

Given a triangle ABC, where G is the intersection of point of medians, what fraction of the triangle ABC is shaded?

A. \(\frac{1}{4}\)

B. \(\frac{1}{12}\)

C. \(\frac{1}{6}\)

D. \(\frac{1}{8}\)

E. \(\frac{1}{9}\)

Since the question does not mention anything about the kind of triangle we have (isosceles, scalene, etc), we can assume that we can use ANY triangle.

So, let's use a nice equilateral triangle with sides of length 2

Area of an equilateral triangle = (√3)(side²/4)

So, area = (√3)(2²/4)

= (√3)(4/4)

=

√3 Now let's draw our medians

We can also add some angles

We need to find the area of the RED triangle in the middle

So, let's flip that red triangle like so...

If we draw an altitude from the top vertex, the line divides the triangle into two 30-60-90 special right triangles.

When we compare the RED 30-60-90 triangles with the base 30-60-90 triangle we can calculate the height (h)

Since we're comparing SIMILAR triangles, we can write:

0.5/

√3 =

h/

1Cross multiply to get: (0.5)(1) = (√3)(h)

Simplify: 1/2 = (√3)(h)

Solve: h = 1/(2√3)

We have the following

Area of triangle = (base)(height)/2

= (1)[1/(2√3)]/2

=

1/(4√3)What fraction of the triangle ABC is shaded?Fraction =

1/(4√3)/

√3=

[1/(4√3)](

1/√3)

= 1/12

Answer: B

Cheers,

Brent

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