Question

See the attachment below.

Answer 1

Answers:

a)
`L=208\pi`

b)
`A=0.18 cm^(2)`

Explanation:

a) The area of the sector of a circle
`A_(c)` is given by:

`A_(c)=(rL)/(2)=(r^(2)\theta)/(2)` (1)

Where:

`r=144` is the radius

`L` is the length of arc

`\theta=260\°=(13)/(9)\pi` is the angle in radians (knowing
`360\°=2\pi` to make the conversion)

Isolating
`L` from (1) :

`L=r\theta` (2)

`L=144((13)/(9)\pi)` (3)

`L=208\pi` (4) This is the length of arc

b) If we want to find the shaded area
`A`, we have to find the area of the sector of a circle
`A_(c)` and substract the area of the triangle
`A_(t)`.

We already know
`A_(c)=(r^(2)\theta)/(2)` (5)

Where:

`r=4cm`

`\theta=30\°=(\pi)/(6)` (remembering
`360\°=2\pi`)

Hence:

`A_(c)=((4cm)^(2)((\pi)/(6)))/(2)` (6)

`A_(c)=(4)/(3)\pi cm^(2)` (7) area of the sector of the circle

On the other hand, the area of a triangle is given by:

`A_(t)=(bh)/(2)` (8)

Where:

`b` is the base of the triangle

`h` is the height of the triangle

If we divide this triangle in half, we will have two right triangles, each one with a height
`h` and a base
`(b)/(2)`, and hypotenuse=4cm.

In addition, each triangle will have the following angles (in degrees):
`90\°`,
`15\°` (the half of
`30\°`),
`75\°` (knowing the three inner angles of a triangle sum to
`180\°`).

Having this clear, let's use the trigonometric functions sine and cosine to find the values of
`h` and
`b`:

`sin\alpha=(Oppositeside)/(Hypotenuse)` (9)

`cos\alpha=(Adjacentside)/(Hypotenuse)` (10)

`sin(15\°)=((b)/(2))/(4cm)` (11)

`b=8sin(15\°)=2.070cm` (12) This is the base of the triangle

`cos(15\°)=(h)/(4cm)` (13)

`h=4cos(15\°)=3.86cm` (14) This is the height of the triangle

Substituting (12) and (14) in (8):

`A_(t)=((2.070cm)(3.86cm))/(2)` (15)

`A_(t)=3.99cm^(2) \approx 4cm^(2)` (16) This is the area of the triangle

Substracting the area of the triangle from the area of the sector of the circle:

`A=A_(c)-A_(t)` (17)

`A=(4)/(3)\pi cm^(2)- 4cm^(2)` (18)

Finally we have the area of the shaded portion:

`A=0.18 cm^(2)`