β ππππππ = 35.76A line π·πΈ is drawn parallel to base π΅πΆ of β³ π΄π΅πΆ, meeting π΄π΅ in π· and π΄πΆ in at πΈ. If2 ππ, find the length of π΄πΈ.

# β ππππππ = 35.76A line π·πΈ is drawn parallel to base π΅πΆ of β³ π΄π΅πΆ, meeting π΄π΅ in π· and π΄πΆ in at πΈ. If2 ππ, find the length of π΄πΈ.

1. A
10ππ
2. B
2ππ
3. C
6ππ
4. D
12ππ

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### Solution:

π΄π΅
π΅π·

= 4 ππ and πΈ =
From the question we have,
π΄π΅ = 4 ππ and πΆπΈ = 2 cm
π΅π·
Also, π·πΈ β₯ π΅πΆ
By the property of similar triangles,

π΄π· π΄πΈ
π΄π΅ = π΄πΆ
β 1 β π΄π· = 1 β π΄πΈ
π΄π΅ π΄πΆ
π΄π΅ β π΄π·

π΄πΆ β π΄πΈ π΅π· πΆπΈ
β π΄π΅ =

π΄πΆ β π΄π΅ = π΄πΆ
β 1  = 2
4 π΄πΆ
β π΄πΆ = 8 cm

π΄πΈ = π΄πΆ β πΆπΈ = 8 β 2
π΄πΈ = 6ππ

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