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CD=AB=AD=4
Mnrt pitagoras
DE^2= AD^2+AE^2
25= 16+AE^2
AE^2=9
AE=3
Kliling trapesium CDEF adlh
CD+DE+EA+AB+BF+CF
=4+5+3+4+3+5
=24
CD = AD = 4 cm, ED = 5 cm
maka
AE² = ED² - AD²
⇒ AE² = 5² - 4²
⇒ AE² = 25 - 16
⇒ AE² = 9
⇒ AE = 3 cm
Sehingga
AE = BF = 3 cm
maka
K = CD + FC + BF + AB + AE + ED
= 4 + 5 + 3 + 4 + 3 + 5
= 24 cm
Jawab : D
Nomer 10
L EFCD = 240
⇒ CD · DE = 240
⇒ 16 · DE = 240
⇒ DE = 15 cm
maka
FB² = 17² - 15²
⇒ FB² = 289 - 225
⇒ FB² = 64
⇒ FB = 8 cm
dan
AE² = 25² - 15²
⇒ AE² = 625 - 225
⇒ AE² = 400
⇒ AE = 20 cm
amka
AB = 20 + 16 + 8
⇒ AB = 44 cm
Seingga
L = (1/2)(CD + AB)(DE)
= (1/2)(16 + 44)(15)
= (1/2)(60)(15)
= 450 cm²
Jawab : B
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