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Luda [366]
2 years ago
7

Triangle A B C has centroid G. Lines are drawn from each point through the centroid to the midpoint of the opposite side to form

line segments A F, B D, and C E. The length of line segment A G is 19 x + 14 and the length of line segment D G is 9 x + 2.
G is the centroid of triangle ABC.

What is the length of GF?

units
Mathematics
1 answer:
Helga [31]2 years ago
3 0

Based on the calculations, the length of GF is equal to 26 units.

<h3>How to find the length of GF?</h3>

In Geometry, the centroid of any triangle always divides the lines in a ratio of 2:1. Thus, G divides line B F as follows:

2DG = BD

2(9x + 2) = 40

18x + 4 = 40

18x = 40 - 4

x = 36/18

x = 2.

Similarly, G divides line A F as follows:

2GF = A F

GF = A F/2

GF = (19x + 14)/2

GF = (19(2) + 14)/2

GF = 52/2

GF = 26 units.

Read more on centroid here: brainly.com/question/15015349

#SPJ1

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Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel
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Answer with Step-by-step explanation:

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F=<0,-8>=0i-8j=-8j

\theta=\frac{\pi}{3}

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be=r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j

By using cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}

Force along the plane will be=\mid F_x\mid=F\cdot r

Force along the plane will be =\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3N

By using i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0

Therefore, force along the plane=\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)

2.The vector perpendicular to the plane=r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j

The force perpendicular to the plane=\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

The force perpendicular to the plane=4N

Therefore, F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

Sum of two component of force=F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)

Sum of two component of force=2\sqrt 3i-6j-2\sqrt3 i-2j=-8j

Hence,sum of two component of forces=Total force.

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