Triangle EFG is isosceles with EG= FG=16cm GH is an arc of a circle, centre F, with angle HFG=0.85 radians. Find a. the length of arc GH
b. the length of EF
c. the perimeter of the shaded region
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Answer:
∠F = ∠G where triangle EFG is isoscles. EF=EG and EH prep to FH
Step-by-step explanation:
in Δ EHF & ΔEHG
EF = EG ( given)
∠EHF = ∠EHG = 90° (EH ⊥ FG)
EH = EH ( common)
=> Δ EHF ≅ ΔEHG
=> ∠EFH = ∠EGH
=> ∠F = ∠G
QED proved
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Answer:
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Step-by-step explanation:
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