Question

find the altitude of an equilateral triangle when each of its side is 'a' cm.

Answer 1

let draw the perpendicular AD on the base BC in triangle ABC then side AB=a,BD=1/2a then by Pythagoras theorem AB^2-BD^2=AD^2
a^2 - 1/2a^2 = AD^2
a^2- 1/4(a^2)=AD^2
4a^2-a^2/4=AD^2
a^2(4-1)/4=AD^2
√a^2√(4-1)/√4=AD
AD= {a √(4-1)} /2

Answer 2

The altitude of an equilateral triangle is (√3/2)a when each side is a.

We have to find the altitude of an equilateral triangle when each of its sides is 'a' cm.

Concepts :

• An equilateral triangle is a triangle which has equal sides. for example, If we say, ABC is an equilateral triangle then AB = BC = CA
• Altitude of an triangle is nothing but the perpendicular line segment upon the base of the triangle. for example, ABC is a triangle where AD⊥BC then, AC is  altitude on BC.

Let ABC is an equilateral triangle where AB = BC = CA = a and we draw  an altitude AD on BC as shown in figure.

Length of altitude = AD

We know, area of an equilateral triangle is given by,

$A=\frac{\sqrt{3}}{4}a^2\:\:...(1)$

Also we know area of any triangle is given by,

$A=\frac{1}{2}\times\text{height}\times\text{base}$

For ABC equilateral triangle,

height = AD , base = BC = a

$\implies A=\frac{1}{2}\times AD\times a\:\:...(2)$

From equations (1) and (2) we get,

$\frac{\sqrt{3}}{4}a^2=\frac{1}{2}\times AD\times a\\\\\implies AD=\frac{\sqrt{3}}{2}a$

Therefore the altitude of an equilateral triangle is (√3/2)a when each side is a.

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