Question

$\purple{\Large{\bf{Hello! \: Math \: Lovers}}}$
$\gray{\sf{Here \: is \: an \: Question \: for \: you \: all}}$
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

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Answer 1

Given:

An equilateral triangle with an altitude.

To Prove:

3 times the square of one of its sides is equal to four times the altitude.

Solution:

Let's name the triangle ABC, with AD as it's altitude, and let's name all the sides "a" for our convenience.

     $\setlength{\unitlength}{1cm}\begin{picture}(20,15)\thicklines\qbezier(0,0)(0,0)(6,0)\qbezier(0,0)(0,0)(3,4)\qbezier(6,0)(6,0)(3,4)\put(3, 0){\line(0,1){4}}\put(2.9, -0.5){\sf D}\put(0, -0.5){\sf B}\put(5.9, -0.5){\sf C}\put(2.9, 4.3){\sf A}\put(0.5, 2.1){\sf a}\put(5.3, 2.1){\sf a}\put(1.3, -0.5){\sf a/2}\put(4.3, -0.5){\sf a/2}\end{picture}$

In an equilateral triangle, the altitude bisects the base into two equal parts, therefore:

BD = DC = a/2

In ΔABD;

∠ADB = 90°, using Pythagoras theorem we get:

$\Longrightarrow \sf AB^2 = AD^2 + DB^2$

Substitute BD = a/2 above.

$\Longrightarrow \sf (a)^2 = AD^2 + \Bigg(\dfrac{a}{2} \Bigg)^2$

$\Longrightarrow \sf a^2 = AD^2 + \dfrac{ \ a^2}{\ 4 \ }$

$\Longrightarrow \sf a^2 - \dfrac{ \ a^2}{\ 4 \ } = AD^2$

Taking LCM we get:

$\Longrightarrow \sf \dfrac{4a^2 - a^2}{4} = AD^2$

$\Longrightarrow \sf \dfrac{3a^2}{4} = AD^2$

$\Longrightarrow \sf 3a^2 = 4AD^2$

Here a is equal to AB, BC and CD, and AD is the altitude. Hence proved.

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

$\huge{\boxed{\rm{\red{Answer}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• An equilateral triangle with an altitude

${\bigstar}\large{\boxed{\sf{\pink{To \: find}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Or \: to \: prove}}}}$

• Three time square of one of its side it is equal to four times the altitude.

${\bigstar}\large{\boxed{\sf{\pink{Solution}}}}$

• Let name the triangle ∆ ABC with AD as it's altitude ( height ) & let's name all the sides ''a'' for our convenience.
• In an equilateral triangle ∆ altitude bisects the base into 2 equal parts therefore, $\large\green{\texttt{BD = DC = a/2}}$

In ∆ABD

<ABD = 90° {Right angle} using phythagoras theorm we get the results -

$\large\green{\texttt{AB² = AD² + DB²}}$

Substitute, BD = a/2 above,

(a)² = AD² + (a/2)²

(a)² = AD² + a² / 4

4a² - a² / 4 = AD²

3a² / 4 = AD²

From taking LCM we get ➜

4a² - a² / 4 = AD²

3a² / 4 = AD²

3a² = 4AD²

Hence a is equal to AB , BC and CD and AD is the altitude.

Hence, the given Question is proved.

@Itzbeautyqueen23

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