Trig?

Geometry Level 3

L, M and N are the midpoints of the sides AB, BC and CA of Δ \Delta ABC. AM, BN and CL intersect at G.

If AM = 5 x 3 = 5x - 3 units and GM = x + 2 = x + 2 units, then what is the length of AG?


The answer is 13.

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1 solution

Well, there is a theorem stating that a line joining a midpoint of one side of a triangle and the centriod of the triangle has a measurement of 1 3 \frac {1}{3} of the measurement of the line joining the midpoint of the same side of the triangle and it's opposite vertex.

So, we get:

x + 2 5 x 3 = 1 3 \frac {x + 2}{5x - 3} = \frac {1}{3}

3 x + 6 = 5 x 3 3x + 6 = 5x - 3

2 x = 9 2x = 9

x = 4.5 x = 4.5

Getting the difference between AM and Gm, we get:

5 x 3 ( x + 2 ) 4 x 5 4 ( 4.5 ) 5 18 5 = 13 5x - 3 - (x + 2) \Rightarrow 4x - 5 \Rightarrow 4(4.5) - 5 \Rightarrow 18 - 5 = \boxed {13}

There.

Nice souliton

Mardokay Mosazghi - 7 years, 2 months ago

AG*MG=AM THEN AG=5X-3-X-2, X=1.25 PUT 5X1.25-1.25-2IS EQUAL TO 6.25-1.25-2=3.25 =AG

amar nath - 7 years ago

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