Question

In an equilateral triangle ABC , AD perpendicular to BC and AD square / BC square = x ??? Find the value of x.

Answer 1

Answer:

Step-by-step explanation:

ABC is an equilateral triangle

Thus, AB = AC = BC = 2a

We know that in an equilateral triangle, the altitude is also the median

Thus, AD is the median

Thus, CD = BC/2 = 2a/2 = a

In triangle ADC, angle D = 90

Therefore, AC$AC^{2} = AD^2 + CD^2\\Thus, AD^2 = AC^2-CD^2\\= (2a)^2-(a)2\\= 4a^2-a2\\=3a^2\\\\Thus, AD=\sqrt{3} a$

Now, $\frac{AD^2}{BC^2}= \frac{3a^2}{4a^2}  \\\\x = \frac{3}{4}$

Answer 2

hope you understand.