The perimeter of the isosceles ΔABC with the base

1 answer:

5 0

ΔBCD is equilateral, so

|BC| = |CD| = |DA|

and

$P=|BC|+|CD|+|DA|\\\\45=|BC|+|BC|+|BC|\\\\45=3\cdot|BC|\quad|:3\\\\|BC|=45:3\\\\\boxed{|BC|=15\text{ cm}}$

ΔABC is isosceles and

|AB| = |CA| (legs)

$P=|AB|+|BC|+|CA|\\\\P=|AB|+|BC|+|AB|\\\\40=2\cdot|AB|+15\\\\40-15=2\cdot|AB|\\\\25=2\cdot|AB|\quad|:2\\\\\boxed{|AB|=12.5\text{ cm}}$

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