Answer:
400
Step-by-step explanation:
first work out the radius of priya's circle with the equation a =
we know the diameter of a circle is 2 times the radius so the diameter of priya's circle is 2x
this means that if the diameter of dan's circle is 4 times bigger it must be
8x
to find the radius of dan's circle we divide the diameter by to 2 to give
4x
then use the a= equation again
which equals 400
hope that helped
Answer:
∠EFD ≅ ∠EGD ⇒ A
Arc ED ≅ arc FD ⇒ C
m arc FD = 120° ⇒ E
Step-by-step explanation:
Let us revise some facts
In the quadrilateral CDGE
∵ m∠G = 60°
∵ m∠GDC = m∠GEC = 90°
- By using the 5th rule above
∴ m∠G + m∠GDC + m∠DCE + m∠GEC = 360°
∴ 60 + 90 + m∠DCE + 90 = 360
∴ 240 + m∠DCE = 360
- Subtract 240 from both sides
∵ m∠DCE = 120°
In circle C
∵ ∠DCE is a central angle subtended by arc DE
∵ ∠DFE is an inscribed angle subtended by arc DE
- By using the 2nd rule above
∴ m∠DFE = m∠∠DCE
∵ m∠DCE = 120°
∴ m∠DFE = (120)
∴ m∠DFE = 60°
- That means ∠EFD ≅ ∠EGD because their measure is 60°
∴ ∠EFD ≅ ∠EGD
In Δ EFD
∵ EF = FD
∵ m∠DFE = 60°
- By using the 4th rule above
∴ Δ EFD is an equilateral triangle
∴ ED = FD = FE
In circle C
∵ Side ED subtended by arc ED
∵ Side FD subtended by FD
∵ Side ED ≅ side FD ⇒ proved
- By using the 1st rule above
∴ Arc ED ≅ arc FD
∵ m∠ECD = 120°
∵ ∠ECD is a central angle subtended by arc ED
- By using the 3rd rule above
∴ m∠ECD = m arc ED
∴ m of arc ED = 120°
∵ Arc ED ≅ arc FD
∴ m arc ED = m arc FD
∴ m arc FD = 120°