Me given: ∆abc –iso. ∆, m∠bac = 120° ah ⊥ bc, hd⊥ac ad = a cm, hd = b cm find: p∆adh

3a + b
   This is really an exercise in learning about what information is actually needed and what can be ignored. We don't care that triangle ABC is an isosceles triangles. All that matters is noticing that triangle ADH is a right triangle with side AD having a length of a cm, and side HD having a length of b cm. So you can use the Pythagorean theorem to calculate the length of the hypotenuse AH and from there, calculate the perimeter. So: AH = sqrt(a^2 + b^2) P∆ADH = AD + HD + AH P∆ADH = a + b + sqrt(a^2 + b^2) Now the above is a "correct" answer, but we can take advantage of the extra information that was provided and determine that not only is triangle ADH a right triangle, but that it's a 30/60/90 triangle. With that extra knowledge, we know that AD is half the length of HA. So we can simplify the length of the perimeter to: a + b + 2a = 3a + b

Arhombus is a quadrilateral.