Question

I need help on those 3 problems!

Answer 1

Answer:

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

Step-by-step 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

$AG=\frac{2}{3}AD$ ---> the centroid divides each median in a ratio of 2:1

we have

$AD=15\ units$

substitute

$AG=\frac{2}{3}15=10\ units$

Part b) Find the length of the segment GD

we know that

$GD=\frac{1}{3}AD$ ---> the centroid divides each median in a ratio of 2:1

we have

$AD=15\ units$

substitute

$GD=\frac{1}{3}15=5\ units$

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

$CG=\frac{2}{3}CE$ ---> the centroid divides each median in a ratio of 2:1

we have

$CG=13\ units$

substitute

$13=\frac{2}{3}CE$

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

Find the length of the segment GE

$GE=\frac{1}{3}CE$

substitute

$GE=\frac{1}{3}19.5\\GE=6.5\ units$

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

$OL=\frac{2}{3}OQ$ ---> the centroid divides each median in a ratio of 2:1

substitute the given values

$8x=\frac{2}{3}(9x+6)$

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

$NL=\frac{2}{3}NP$ ---> the centroid divides each median in a ratio of 2:1

substitute the given values

$(x+4)=\frac{2}{3}(3x+3)$

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

$ML=\frac{2}{3}MR$ ---> the centroid divides each median in a ratio of 2:1

substitute the given values

$(10x-4)=\frac{2}{3}(12x+18)$

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

$DG=\frac{2}{3}DM$ --> the centroid divides each median in a ratio of 2:1

substitute the value of DM

$DG=\frac{2}{3}12=8\ units$