In isosceles △abc (ac = bc) with base angle 30° cd is a median. how long is the leg of △abc, if sum of the perimeters of △acd and △bcd is 20 cm more than the perimeter of △abc?

Given that, In isosceles triangle:

Perimeter of Δ ACD +Δ BCD +20cm = Perimeter of Δ ABC

(Length of AC + Length of AD + Length of CD) + (Length of BC + Length of BD+ Length of CD) +20 cm = (Length of AC + Length of BC + Length of AB)

Since, Length of AD= Length of BD

Therefore,

Length of BD + Length of CD + 10cm = 1/2 of Length of AB

1/2 of Length of AB = Length of BD

So, Length of CD= 10 cm

tan 30° = Length of BD/Length of CD

Length of BD= 10 * tan 30° = 5.77 cm

Length of AB= Length of AD + Length of BD = 5.77 +5.77

Length of AB= 11.54 cm

The length of leg of ΔABC is 11.54 cm.

jml angle lkj & angle mj = lm

the answer is b.

64

step-by-step explanation:

c=2πr=2·π·8≈64