Question

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?

Answer 1

20 cm.

Explanation:

The perimeter of triangle

• ACD is
  `P_(ACD)=AC+CD+AD;`
• BCD is
  `P_(BCD)=BC+CD+BD;`
• ABC is
  `P_(ABC)=AB+BC+AC=AD+DB+AC+BC.`

Since the sum of the perimeters of △ACD and △BCD is 20 cm more than the perimeter of △ABC, you have that

`AC+CD+AD+BC+CD+BD=AD+BD+AC+BC+20,\\ \\2CD=20,\\ \\CD=10\ cm.`

Consider right triangle ACD. In this triangle
`\angle A=30^(\circ),` then the hypotenuse AC is twice the leg CD. Hence,
`AC=2\cdot 10=20\ cm.`