Answer:
c. 36·x
Step-by-step explanation:
Part A
The details of the circle are;
The area of the circle, A = 12·π cm²
The diameter of the circle, d = [tex]\overline {AB}[/tex]
Given that [tex]\overline {AB}[/tex] 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·[tex]\overline {AB}[/tex]²/4
Therefore;
12 = [tex]\overline {AB}[/tex]²/4
4 × 12 = [tex]\overline {AB}[/tex]²
[tex]\overline {AB}[/tex]² = 48
[tex]\overline {AB}[/tex] = √48 = 4·√3
[tex]\overline {AB}[/tex] = 4·√3
The circumference of the circle, C = π·d = π·[tex]\overline {AB}[/tex]
Arc AB = Half the the length of the circumference of the circle = C/2
Arc AB = C/2 = π·[tex]\overline {AB}[/tex]/2
[tex]\overline {AB}[/tex] = 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 [tex]\overline {OF}[/tex] = The length of [tex]\overline {FB}[/tex]
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 [tex]\overline {OD}[/tex] = The length of [tex]\overline {OB}[/tex]
The length of [tex]\overline {OB}[/tex] = [tex]\overline {OF}[/tex] + [tex]\overline {FB}[/tex] = [tex]\overline {OF}[/tex] + [tex]\overline {OF}[/tex] = 2 × [tex]\overline {OF}[/tex]
∴ The length of [tex]\overline {OD}[/tex] = 2 × [tex]\overline {OF}[/tex] = The length of [tex]\overline {OB}[/tex]
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 [tex]\overline {OD}[/tex] = The length of [tex]\overline {OB}[/tex] = The length of [tex]\overline {DB}[/tex] = 2 × [tex]\overline {OF}[/tex]
The perimeter of ΔDOB = The length of [tex]\overline {OD}[/tex] + The length of [tex]\overline {OB}[/tex] + The length of [tex]\overline {DB}[/tex] = 2 × [tex]\overline {OF}[/tex] + 2 × [tex]\overline {OF}[/tex] + 2 × [tex]\overline {OF}[/tex] = 6 × [tex]\overline {OF}[/tex]
∴ The perimeter of ΔDOB = 6 × [tex]\overline {OF}[/tex]
The radius of the circle = [tex]\overline {OF}[/tex] = 6·x
∴ The perimeter of ΔDOB = 6 × 6·x = 36·x
