Question

I need help on those 3 problems!

Answer 1

Part 1)

a)
`AG=10\ units`

b)
`GD=5\ units`

c)
`CD=12\ units`

d)
`GE=6.5\ units`

e)
`GB=13\ units`

Part 2)

a)
`x=2`

b)
`x=2`

c)
`x=8`

Part 3)

a) The height of the truss is 12 units

b) The centroid of triangle DEF is 8 units down from D

Explanation:

Part 1)

we know that

A centroid of a triangle is the point where the three medians of the triangle meet. A median of a triangle is a line segment from one vertex to the mid point on the opposite side of the triangle

The centroid divides each median in a ratio of 2:1

Part a) Find the length of the segment AG

we know that

`AG=(2)/(3)AD` ---> the centroid divides each median in a ratio of 2:1

we have

`AD=15\ units`

substitute

`AG=(2)/(3)15=10\ units`

Part b) Find the length of the segment GD

we know that

`GD=(1)/(3)AD` ---> the centroid divides each median in a ratio of 2:1

we have

`AD=15\ units`

substitute

`GD=(1)/(3)15=5\ units`

Part c) Find the length of the segment CD

we know that

In the right triangle CGD

Applying the Pythagorean Theorem

`CG^2=GD^2+CD^2`

we have

`CG=13\ units\\GD=5\ units`

substitute

`13^2=5^2+CD^2`

`CD^2=144\\CD=12\ units`

Part d) Find the length of the segment GE

we know that

`CG=(2)/(3)CE` ---> the centroid divides each median in a ratio of 2:1

we have

`CG=13\ units`

substitute

`13=(2)/(3)CE`

`CE=13(3)/2\\CE=19.5\ units`

Find the length of the segment GE

`GE=(1)/(3)CE`

substitute

`GE=(1)/(3)19.5\\GE=6.5\ units`

Part e) Find the length of the segment GB

we know that

In the right triangle GBD

Applying the Pythagorean Theorem

`GB^2=GD^2+DB^2`

we have

`GD=5\ units`

`DB=CD=12\ units` ---> D is the midpoint segment CB

substitute

`GB^2=5^2+12^2`

`GB^2=169\\GB=13\ units`

Part 2) Point L is the centroid of triangle NOM

Find the value of x

Part a) we have

OL=8x and OQ=9x+6

we know that

`OL=(2)/(3)OQ` ---> the centroid divides each median in a ratio of 2:1

substitute the given values

`8x=(2)/(3)(9x+6)`

solve for x

`24x=18x+12\\24x-18x=12\\6x=12\\x=2`

Part b) we have

NL=x+4 and NP=3x+3

we know that

`NL=(2)/(3)NP` ---> the centroid divides each median in a ratio of 2:1

substitute the given values

`(x+4)=(2)/(3)(3x+3)`

solve for x

`3x+12=6x+6\\6x-3x=12-6\\3x=6\\x=2`

Part c) we have

ML=10x-4 and MR=12x+18

we know that

`ML=(2)/(3)MR` ---> the centroid divides each median in a ratio of 2:1

substitute the given values

`(10x-4)=(2)/(3)(12x+18)`

solve for x

`30x-12=24x+36\\30x-24x=36+12\\6x=48\\x=8`

Part 3)

Part a) Find the altitude of the truss

Let

M ----> the midpoint of segment FE

DM ---> the altitude of the truss

Applying Pythagorean Theorem in the right triangle FDM

`FD^2=FM^2+DM^2`

substitute the given values

`15^2=9^2+DM^2`

`DM^2=225-81\\DM^2=144\\DM=12\ units`

therefore

The height of the truss is 12 units

Part b) How far down from D is the centroid of triangle DEF?

we know that

`DG=(2)/(3)DM` --> the centroid divides each median in a ratio of 2:1

substitute the value of DM

`DG=(2)/(3)12=8\ units`