Question

triangle ABC is an equilateral triangle with side 12 cm prove that the altitude when squared is 108 cm (ANSWER FAST)​

Answer 1

Answer:

6√3

Step-by-step explanation:

Given: ABC is an equilateral triangle

∴ AB = AC = BC = 12cm And let AD is an altitude on BC.

Therefore, BD = 1/2 x BC = 6 cm

Now, In ∆ADB, using Pythagoras theorem, we have (Perpendicular)2 + (Base)^2 = (Hypotenuse)^2

⇒ (AD)^2 + (BD)^2 = (AB)^2

⇒ (AD)^2 + (6)^2 = (12)^2

⇒ (AD)^2 = 144 – 36

⇒ (AD)^2 = 108

⇒ AD = √108

⇒ AD = 6√3

Hence, the height of an equilateral triangle is 6√3 cm

Answer 2

Answer:

Given: ABC is a equilateral triangle.

Step-by-step explanation:

therefore: AB=AC=BC=12cm

And let AD is an altitude on BC .

therefore ,BDA=1/2×BC=6cm

Now,in ∆ADB, using Pythagoras theorem,we have

(perpendicular)^2+(Base)^2=(hypothuse)^

=(AD)^2+(BD)^2=(AB)^2

=(AB)^2+(6)^2=(12)^2

=(AD)^2=144-36

=(AD)^2=108

=AD=√108

=AD=6√3

hence,the height of an equilateral triangle is 6√3cm