PLEASE PLEASE PLEASE HELP!!!!!!
In the following diagram, the radius of the outer circle is 8 cm and the radius of the inner circle is 4 cm. Find the following;

PLEASE PLEASE PLEASE HELP In the following diagram the radius of the outer circle is 8 cm and the radius of the inner circle is 4 cm Find the following class=
PLEASE PLEASE PLEASE HELP In the following diagram the radius of the outer circle is 8 cm and the radius of the inner circle is 4 cm Find the following class=

Respuesta :

Answer:

(See explanation for further details)

Step-by-step explanation:

1) The circumference of the outer circle is:

[tex]s_{EXT} = 2\pi\cdot (8\,cm)[/tex]

[tex]s_{EXT} = 16\pi\,cm[/tex]

2) The circumference of the inner circle is:

[tex]s_{INT} = 2\pi\cdot (4\,cm)[/tex]

[tex]s_{INT} = 8\pi\,cm[/tex]

3) The area of the outer circle is:

[tex]A_{EXT} = \pi\cdot (8\,cm)^{2}[/tex]

[tex]A_{EXT} = 64\pi\,cm^{2}[/tex]

4) The area of the inner circle is:

[tex]A_{INT} = \pi\cdot (4\,cm)^{2}[/tex]

[tex]A_{INT} = 16\pi\,cm^{2}[/tex]

5) The length of the arc BC is:

[tex]s_{BC} = \left(\frac{80^{\circ}}{360^{\circ}} \right)\cdot (16\pi\,cm)[/tex]

[tex]s_{BC} = \frac{32\pi}{9}\,cm[/tex]

6) The length of the arc ST is:

[tex]s_{ST} = \left(\frac{80^{\circ}}{360^{\circ}} \right)\cdot (8\pi\,cm)[/tex]

[tex]s_{ST} = \frac{16\pi}{9}\,cm[/tex]

7) The area of the sector BPC is:

[tex]A_{BPC} = \left(\frac{80^{\circ}}{360^{\circ}} \right)\cdot (64\pi\,cm^{2})[/tex]

[tex]A_{BPC} = \frac{128\pi}{9}\,cm^{2}[/tex]

8) The area of the sector SPT is:

[tex]A_{SPT} = \left(\frac{80^{\circ}}{360^{\circ}} \right)\cdot (16\pi\,cm^{2})[/tex]

[tex]A_{SPT} = \frac{32\pi}{9}\,cm^{2}[/tex]

9) The radian measure of the sector BPC is:

[tex]\theta = \left(\frac{80^{\circ}}{360^{\circ}}\right)\cdot (2\pi\,rad)[/tex]

[tex]\theta = \frac{4\pi}{9}\,rad[/tex]

10) The radian measure of the sector SPT is::

[tex]\theta = \left(\frac{80^{\circ}}{360^{\circ}}\right)\cdot (2\pi\,rad)[/tex]

[tex]\theta = \frac{4\pi}{9}\,rad[/tex]