tolong dijawab panjang keliling dan luas persegi panjang

pakai phitagoras
AF = akar(A0^2 - OF^2)
AF = akar(5^2 - 3^2
AF = 4cm

AB = 2AF
AB = 8cm

AC = 2AO
AC = 10
BC = akar(AC^2 - AB^2)
BC = akar(10^2 - 8^2)
BC = 6cm

Keliling = 2(AB + BC)
2(8+6)
28cm

Luas = AB . BC
6.8
48cm^2

(opsi C)

#CMIIW

Ao= 5; AC= 10
FO=3 ;EF =6
DIT: K DAN L
JWB :
AF^2 = AO ^2 - FO2
= 5^2-3^2
=25- 9
=16
AF = √16 = 4 ; AB= 8


BC^2=AC^2-AB^2
= 10^2- 8^2
=100- 64
= 36
BC = √36 =6

K= 2p + 2l
=2.8+2.6
=16+12
=28

L= p x l
=8×6
=48cm^2