Question

PLEASE PLEASE PLEASE HELP ME

Given: △CDE
m∠D=90°, DK⊥ CE
CD:DE=3:5, KE=CK+8
Find: CE

Answer 1

CE = 17

Explanation:

∵ m∠D = 90

∵ DK ⊥ CE

∴ m∠KDE = m∠KCD⇒Complement angles to angle CDK

In the two Δ KDE and KCD:

∵ m∠KDE = m∠KCD

∵ m∠DKE = m∠CKD

∵ DK is a common side

∴ Δ KDE is similar to ΔKCD

∴
`(KD)/(KC)=(DE)/(CD)=(KE)/(KD)`

∵ DE : CD = 5 : 3

∴
`(KD)/(KC)=(5)/(3)`

∴ KD = 5/3 KC

∵ KE = KC + 8

∵
`(KE)/(KD)=(5)/(3)`

∴
`(KC+8)/((5)/(3)KC )=(5)/(3)`

∴
`KC + 8 = (25)/(9)KC`

∴
`(25)/(9)KC - KC=8`

∴
`(16)/(9)KC=8`

∴ KC = (8 × 9) ÷ 16 = 4.5

∴ KE = 8 + 4.5 = 12.5

∴ CE = 12.5 + 4.5 = 17