Question

The side of an equilateral triangle is 16cm.Find the length of its altitude​

Answer 1

${\underline{\underline{\huge{\mathfrak{\green{Answer:-}}}}}}$

Hey Mate!!

Area of equilateral ∆ = $\frac{\sqrt{3}}{4}a^{2}$

$= > \frac{ \sqrt{3} }{4} \times 16 \times 16 \\ = > \sqrt{3} \times 4 \times 16 \\ = > 64 \sqrt{3}\:cm^{2}$

Area of ∆ = 1/2 × base × height

=> 64√3 = 1/2 × 16 × h

=> 64√3 = 8h

=> 64√3 ÷ 8 = h

=> h = 8√3 cm

Hope it helps...✌️✌️

Answer 2

${\huge\color{Red}{Answer~is~ 8 \sqrt{3}~cm }}$

Step-by-step explanation:

We know that the altitude of equilateral triangle cuts it's base in two equal parts.

So ∆ABC if altitude is AD and cuts AC into AD and DC in 8cm.

Now, In ∆ABD, angle D = 90°

${{AB}^{2} = {AD}^{2} + {BD}^{2}}$

${16}^{2} = {8}^{2} + {x}^{2} \\ \\ 256 = 64 + {x}^{2} \\ \\ 256 - 64 = {x}^{2} \\ \\ 192 = {x}^{2} \\ \\ x = \sqrt{192} \\ \\ x = \sqrt{64 \times 3} \\ \\ x = 8 \sqrt{3} \: cm$

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