Question

Can someone explain this to me please

Answer 1

c. 36·x

Explanation:

Part A

The details of the circle are;

The area of the circle, A = 12·π cm²

The diameter of the circle, d =
`\overline {AB}`

Given that
`\overline {AB}` is the diameter of the circle, we have;

The length of the arc AB = Half the the length of the circumference of the circle

Therefore, we have;

A = 12·π = π·d²/4 = π·
`\overline {AB}`²/4

Therefore;

12 =
`\overline {AB}`²/4

4 × 12 =
`\overline {AB}`²

`\overline {AB}`² = 48

`\overline {AB}` = √48 = 4·√3

`\overline {AB}` = 4·√3

The circumference of the circle, C = π·d = π·
`\overline {AB}`

Arc AB = Half the the length of the circumference of the circle = C/2

Arc AB = C/2 = π·
`\overline {AB}`/2

`\overline {AB}` = 4·√3

∴ C/2 = π·4·√3/2 = 2·√3·π

The length of arc AB = 2·√3·π cm

Part B

The given parameters are;

The length of
`\overline {OF}` = The length of
`\overline {FB}`

Angle D = angle B

The radius of the circle = 6·x

The measure of arc EF = 60°

The required information = The perimeter of triangle DOB

We have;

Given that the base angles of the triangles DOB are equal, we have that ΔDOB is an isosceles triangle, therefore;

The length of
`\overline {OD}` = The length of
`\overline {OB}`

The length of
`\overline {OB}` =
`\overline {OF}` +
`\overline {FB}` =
`\overline {OF}` +
`\overline {OF}` = 2 ×
`\overline {OF}`

∴ The length of
`\overline {OD}` = 2 ×
`\overline {OF}` = The length of
`\overline {OB}`

Given that arc EF = 60°, and the point 'O' is the center of the circle, we have;

∠EOF = The measure of arc EF = 60° = ∠DOB

Therefore, in ΔDOB, we have;

∠D + ∠B = 180° - ∠DOB = 180° - 60° = 120°

∵ ∠D = ∠B, we have;

∠D + ∠B = ∠D + ∠D = 2 × ∠D = 120°

∠D = ∠B = 120°/2 = 60°

All three interior angles of ΔDOB = 60°

∴ ΔDOB is an equilateral triangle and all sides of ΔDOB are equal

Therefore;

The length of
`\overline {OD}` = The length of
`\overline {OB}` = The length of
`\overline {DB}` = 2 ×
`\overline {OF}`

The perimeter of ΔDOB = The length of
`\overline {OD}` + The length of
`\overline {OB}` + The length of
`\overline {DB}` = 2 ×
`\overline {OF}` + 2 ×
`\overline {OF}` + 2 ×
`\overline {OF}` = 6 ×
`\overline {OF}`

∴ The perimeter of ΔDOB = 6 ×
`\overline {OF}`

The radius of the circle =
`\overline {OF}` = 6·x

∴ The perimeter of ΔDOB = 6 × 6·x = 36·x