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3 years ago

Given △ABC, AD and BE are medians, AD ∩BE=G. Prove: GD/AG = DE/AB

Mathematics

1 answer:

saw5 [17]3 years ago

3 0

Answer:

See explanantion

Step-by-step explanation:

Since AD and BE are medians, then points D and E are midpoints of the sides BC and AC, respectively. Thus, segment DE is triangle's midline. By the Triangle midline theorem, DE is parallel to the side AB. Thus,

$\angle ABE\cong \angle BED;$
$\angle BAD\cong \angle EDA$

as alternate interior angles.

Moreover, $\angle BGA\cong \angle EGD$ as vertical angles.

This gives us that triangles ABG and DEG are similar by AAA postulate.

Hence,

$\dfrac{AB}{ED}=\dfrac{AG}{GD}$

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I need to find the surface area and volume of all three figures. If you could provide what equations you used to I will be grate

Margarita [4]

Answer:

Part 1) Sphere The surface area is equal to $SA=196\pi\ m^{2}$ and the volume is equal to $V=\frac{1,372}{3}\pi\ m^{3}$

Part 2) Cone The surface area is equal to $SA=(16+4\sqrt{65})\pi\ units^{2}$ and the volume is equal to $V=\frac{112}{3}\pi\ units^{3}$

Part 3) Triangular Prism The surface area is equal to $SA=51.57\ mm^{2}$ and the volume is equal to $V=17.388\ mm^{3}$

Step-by-step explanation:

Part 1) The figure is a sphere

a) Find the surface area

The surface area of the sphere is equal to

$SA=4\pi r^{2}$

we have

$r=14/2=7\ m$ ----> the radius is half the diameter

substitute

$SA=4\pi (7)^{2}$

$SA=196\pi\ m^{2}$

b) Find the volume

The volume of the sphere is equal to

$V=\frac{4}{3}\pi r^{3}$

we have

$r=14/2=7\ m$ ----> the radius is half the diameter

substitute

$V=\frac{4}{3}\pi (7)^{3}$

$V=\frac{1,372}{3}\pi\ m^{3}$

Part 2) The figure is a cone

a) Find the surface area

The surface area of a cone is equal to

$SA=\pi r^{2} +\pi rl$

we have

$r=4\ units$

$h=7\ units$

Applying Pythagoras Theorem find the value of l (slant height)

$l^{2}=r^{2} +h^{2}$

substitute the values

$l^{2}=4^{2} +7^{2}$

$l^{2}=65$

$l=\sqrt{65}\ units$

$SA=\pi (4)^{2} +\pi (4)(\sqrt{65})$

$SA=16\pi +4\sqrt{65}\pi$

$SA=(16+4\sqrt{65})\pi\ units^{2}$

b) Find the volume

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2}h$

we have

$r=4\ units$

$h=7\ units$

substitute

$V=\frac{1}{3}\pi (4)^{2}(7)$

$V=\frac{112}{3}\pi\ units^{3}$

Part 3) The figure is a triangular prism

a) The surface area of the triangular prism is equal to

$SA=2B+PL$

where

B is the area of the triangular base

P is the perimeter of the triangular base

L is the length of the prism

Find the area of the base B

$B=\frac{1}{2} (2.7)(2.3)=3.105\ mm^{2}$

Find the perimeter of the base P

$P=2.7*3=8.1\ mm$

we have

$L=5.6\ mm$

substitute the values

$SA=2(3.105)+(8.1)(5.6)=51.57\ mm^{2}$

b) Find the volume

The volume of the triangular prism is equal to

$V=BL$

where

B is the area of the triangular base

L is the length of the prism

we have

$B=3.105\ mm^{2}$

$L=5.6\ mm$

substitute

$V=(3.105)(5.6)=17.388\ mm^{3}$

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