Question

Given: ∆ABC –iso. ∆, m∠BAC = 120°

AH ⊥ BC

HD ⊥ AC

AD = a cm, HD = b cm

Find: P∆ADH

Answer 1

3a+b

Explanation:

Statement-Reasoning Format:

1. CB is base- Given

2. △ABC – isos. △- Given

3. AC=CB- Def. of isos. △

4. AH is an altitude, median, and angle bisector- Def. of alt. med., ∠ bisect. in isos. △

5. m∠BAC= 120°- Given

6. m∠DAH=m∠BAH=m∠BAC/2=120/2=60°-Def of ∠ bisect.

7. 180°-m∠CDH=m∠HDA=180-90=90°-Linear Pair

8. m∠DHA=180- (m∠HDA+m∠HAD)= 180-(90+60)=180-150=30°-Sum of ∠s in a △

9. AD=a cm-Given

10. HA=2AD=2(a)=2a cm- Leg Opposite to 30°

11. HD=b cm- Given

12. P△ADH=a+2a+b= 3a+b cm-Part Whole Postulate

Hope This Helps!

Answer 2

Answer:
`P\triangle ADH = (a+b+√(a^2+b^2) )` unit

Explanation:

Since Here Δ ABC is a isosceles triangle.

Also,
`AH\perp BC` and
`HD\perp AC`

Thus, ∠ ADH = 90°

That is, Δ ADH is the right angle triangle.

In which, AD = a unit and HD = b unit

And, By the definition of Pythagoras theorem,

`AH^2 = HD^2 + AD^2`

⇒
`AH^2 = a^2 +b^2`

⇒
`AH = √(a^2+b^2)`

Since, the perimeter of a triangle = sum of the all sides of the triangle.

Therefore, perimeter of ΔADH = AD + DH + AH

=
`a + b + √(a^2+b^2)` unit