Zar8yahid
Answered

Refer to the circle.

If AE = 41, what must be the length of cord EF so that the perimeter of ∆EFS is 50?

If CF=50, what must be the length of chord AC so that the perimeter of ∆ACS is 60?

Wala pong sumasagot :(​

Refer To The CircleIf AE 41 What Must Be The Length Of Cord EF So That The Perimeter Of EFS Is 50 If CF50 What Must Be The Length Of Chord AC So That The Perime class=

Sagot :

KallesEliasid

Problem:

If AE = 41, what must be the length of cord EF so that the perimeter of ∆EFS is 50?

Solution:

AE = 41

ES = AE/2

ES = 41/2

ES = 20.5

ES = AS = FS = CS = 20.5

∆EFS = 50

∆EFS = ES + FS + EF

50 = 20.5 + 20.5 + EF

EF = 50 - 20.5 - 20.5

EF = 9

Answer:

Chord EF = 9

Problem:

If CF = 50, what must be the length of chord AC so that the perimeter of ∆ACS is 60?

Solution:

CF = 50

CS = CF/2

CS = 50/2

CS = 25

CS = FS = ES = AS = 25

∆ACS = 60

∆ACS = CS + AS + AC

60 = 25 + 25 + AC

AC = 60 - 25 - 25

AC = 10

Answer:

Chord AC = 10

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