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# help

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AB = 15, BC = 18, AC = 25, D is midpoint of BC, G is centroid.  Find DG. Nov 17, 2019

#1
+2 Based on one of the properties of a centroid. Segment GD is 1/3 the length of AD.

Sorry my method is pretty complicated so hang on.

Triangles ABD and ABC are similar. Since D is the midpoint, and they share AB as a side, their ratio is 1:2

We can find the area of ABC using Heron's formula. After one million years, we have a simplified radical of $$2\sqrt{4466}$$.

Since their ratio is 1:2, we know that the area of ABD is now $$\sqrt{4466}$$.

Side BD is 18 / 2 = 9

We know sides AB = 15 and BD = 9. So we can work backwards from Heron's formula to find the length of AD.

Working backwards from heron's formula: we calculate s. Since the length of side C is unknown, we can call it X.

We find that S  is 12 + 0.5x. We know the area is $$\sqrt{4466}$$. So we can make the following equation:

$$(0.5x+12)(0.5x+12-15)(0.5x+12-9)(0.5x+12-x)=4466$$

(0.5x+12)(0.5x-3)*(0.5x+3)(0.5x+12-x)=4466 = {x=-(2*sqrt(86)), x=2*sqrt(86), x=-(2*sqrt(67)), x=2*sqrt(67)}

So side C is either $$2\sqrt{86}$$, or $$2\sqrt{67}$$.

That means GD is either: $$\frac{2\sqrt{86}}{3}$$, or $$\frac{2\sqrt{67}}{3}$$....

Yup! THis is completely messed up! What a crazy method! I got two possible sides too, uh.. Help?

Nov 17, 2019
#2
+1

Better ways to do this but I think this can be solved by using the Law of Cosines twice

We have that

AC^2  = AB^2 + BC^2   - 2 ( AB) (BC) cos (ABC)

25^2 = 15^2 + 18^2  - 2 (15) (18) cos( ABC)

625 = 225 + 324 - [540 cos ABC]

-76/540 =  cos(ABC)  =  -19/135

So   applying this again to  find the length of AD  we have that

AD^2  =  AB^2 +( BC/2)^2 - 2 (AB) (BC) cos (ABC)

AD^2  = 15^2 + 9^2 - 2(15) (18) (-19/135)

So ....  since AD is a median, DG is (1/3) (AD) =  (1/3) sqrt(382)  = sqrt (382) / 3  ≈ 6.51 units

EDIT TO CORRECT AN ERROR....!!!!   Nov 18, 2019
edited by CPhill  Nov 18, 2019
#3
+1

yup I don't know how to use trig. thats why I screwed up my answer lol.

CalculatorUser  Nov 18, 2019