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?

Respuesta :

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.

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