Question

Solution⬇⬇ of the question which is present in the downward

Answer 1

In the figure attached with your question

ΔABC is an equilateral triangle with side a

and, AD is an altitude on BC

since , Δ ABC is equilateral and AD is an altitude therefore

AD bisect BC ,, i.e, BD = CD = BC/ 2 = a / 2

now consider Δ ADC in which ∠ ADC = 90°

by pythagorean theorem

AC² = AD² + CD²

so,

AD² = AC² - CD²

AD² = ( a )² - ( a / 2 )²

AD² = a² - ( a² / 4)

$AD^{2} = a^{2} - \frac{a^{2} }{4} \\\\AD^2 = \frac{4a^2-a^2}{4} = \frac{3a^2}{4}$

HENCE,

the value of AD² = 3a² / 4 .