SOLUTION: Given: ∆ABC –iso. ∆, m∠BAC = 120° AH ⊥ BC , HD ⊥ AC AD = a cm, HD = b cm Find: P∆ADH

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Question 1131761: Given: ∆ABC –iso. ∆, m∠BAC = 120°

AH
⊥
BC
,
HD
⊥
AC

AD = a cm, HD = b cm
Find: P∆ADH
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

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   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%28a%5E2+%2B+b%5E2%29
P ∆ ADH+=+AD+%2B+HD+%2B+AH
P ∆ ADH+=+a+%2B+b+%2B+sqrt%28a%5E2+%2B+b%5E2%29
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%2F60%2F90+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+%2B+b+%2B+2a+=+3a+%2B+b
P ∆ ADH+=3a+%2B+b

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The vertex angle BAC is 120 degrees; the angle bisector creates two angles of 60 degrees each.

Since the triangle is isosceles, the angle bisector AH is perpendicular to side BC.

Then triangle ADH is a 30-60-90 right triangle.

Given AD = a and HD = b, with ADH a 30-60-90 right triangle, we know AH = 2a. So the perimeter of triangle ADH is 2a+a+b = 3a+b.