Question

ABC is an equilateral triangle of side 2a. Find each of its altitudes​

Answer 1

$\huge\underline\blue {Answer}$

$\large\pink {Given:-}$

In ∆ABC

AB = BC = QC = 2a

$\large\pink {To \:prove:-}$

Length of AD

$\large\pink {Proof:-}$

In ∆ABC,

⇒ AB = BC = QC = 2a

and AD ⊥BC

$⇒ \: BD = \frac{1}{2} \times 2a = a$

In ∆ADB,

⇒ AD²+BD² = AB²

⇒ AD² = AB²-BD²

⇒ AD² = (2a)² - (a)² = 4a² - a²

$⇒ AD = \sqrt{3a}$

Answer 2

Given -

Equilateral triangle ABC with each side 2a

Altitude AD is drawn such that AD||BC

To find -

AD

solution :-

In△ADB and △ADC

AB = Ac ( both are 2a as it is equilateral triangle )

AD = AD (common)

√ADB = √ ADC ( both 90° as AD || BC )

Hence △ ADB = △ADC (by R.H.S congruency)

Hence, BD = DC (CPCT)

BD = DC

BD = DC = $\frac{ 1 }{ 2 }$ BC

BD=DC $\frac{ 2a }{ 2 }$

BD=DC = a

Hence BD = a

Hence in right △ADB

using pythagoras theorem

(Hypotenuse) 2 = (height)2 +(Base)2

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

$(2a) ^ { 2 }$ = $(AD) ^ { 2 }$ + $(a) ^ { 2 }$

$(4a) ^ { 2 }$ = $(AD) ^ { 2 }$ + $(a) ^ { 2 }$

$(4a) ^ { 2 }$ - $(a) ^ { 2 }$ = $(AD) ^ { 2 }$

$(3a) ^ { 2 }$ = $(AD) ^ { 2 }$

$(AD) ^ { 2 }$ = $(3a) ^ { 2 }$

AD = $\sqrt{ 3a }$

AD = a$\sqrt{ 3a }$

Thus, length of altitude AD = a$\sqrt{ 3a }$

similarity,

length of altitude BE = a$\sqrt{ 3a }$

length of altitude CF = a$\sqrt{ 3a }$

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More to know :)

• The Pythagoras theorem that one uses to find the length of a side of a right triangle.

• although first Pythagoras proved the Pythagorean theorem, the idea is linked to the early Babylonians.

• History is that Pythagoras studied with men known as "rope stretchers " who were ment that built Pyramids. The Pythagorean theorem was discovered through the ropes that were used to calculate lengths needed to build pyramids in order to lay their foundations accurately .

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