Question

Prove that three times the square of any side of an equilateral triangle is equal to
four times the square of the altitude?
​

Answer 1

Answer:

Step-by-step explanation:

I don't know plz check in google,

Answer 2

$\boxed{\tt{\red{question}}}$

Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.

$\boxed{\tt{\red{answer}}}$

$\pink{ \frak{refer \: to \: attachment \: for \: figure}}$

$\underline{\blue{\frak{given}}}$

∆ ABC is equilateral

AD is perpendicular to BC.

such that,

BD = CD = BC /2 = AB / 2

$\frak{\underline{\blue{to \: prove}}}$

$\tt \: 3 {AB}^{2} = 4 {AD}^{2}$

$\underline{\blue{\frak{proof}}}$

In ∆ ABD

angle ADB = 90°

so,

by pythagoras theorem

$\tt \: {AB}^{2} = {AD}^{2} + {BD}^{2} \\ \\ \tt \: we \: know ,\: \\ \tt \: BD \: = \frac{AB}{2} \\ \tt \: so, \\ \\\tt \: {AB}^{2} = {AD}^{2} + ( { \frac{AB}{2} })^{2} \\ \\\tt \: {AB}^{2} = {AD}^{2} + \frac{ {AB}^{2} }{4} \\ \\ \tt {AB}^{2} - \frac{ {AB}^{2} }{4} = {AD}^{2} \\ \\ \tt \: \frac{4 {AB}^{2} - {AB}^{2} }{4} = {AD}^{2} \\ \\\tt \: \frac{3 {AB}^{2} }{4} = {AD}^{2} \\ \\ \tt \: cross \: multiplying \\ \\ \tt \: 3 {AB}^{2} = 4 {AD}^{2}$

$\blue{\boxed{proved}}$