If ABC is an equilateral triangle of side a, prove that its altitude $= \frac{\sqrt{3}}{2} a$ [3 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark
Proof : 1 Mark

$Δ A B C$ is an equilateral triangle.

We are given that AB = BC = CA = a. AD is the altitude, i.e., $A D ⊥ B C$ .

Now, in right angled triangles ABD and ACD, we have

$A B = A C$ [Given]

and $A D = A D$ [Common side]

$\Rightarrow Δ A B D \sim Δ A C D$ [By RHS congruence]

$\Rightarrow B D = C D$ [ C.P.C.T]

$\Rightarrow B D = \frac{1}{2} B C = \frac{a}{2}$

From right triangle ABD,

${A B}^{2} = {A D}^{2} + {B D}^{2}$

$\Rightarrow a^{2} = {A D}^{2} + {(\frac{a}{2})}^{2}$

$\Rightarrow {A D}^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3}{4} a^{2}$

$\Rightarrow A D = \frac{\sqrt{3} a}{2}$

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Similar questions

If ABC is an equilateral triangle of side a, prove that its altitude = $\frac{\sqrt{3}}{2}$ a

a

\frac{\sqrt{3} a}{2}

A B C

A D

A

B C

3 {A B}^{2} = 4 {A D}^{2}

a

A l t i t u d e = \frac{a \sqrt{3}}{2}

A r e a = \frac{\sqrt{3}}{4} a^{2}

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