Question

In given figure is an equilateral triangle with side $a$ prove that

(i) $Altitude=\frac{a\sqrt{3}}{2}$

(ii) $Area=\frac{\sqrt{3}}{4}{a}^2$

Answer 1

Let ABC be an equilateral triangle the length of whose each side is 'a' units.

Draw $AD\perp BC$. then, D is the mid-point of BC.

$\Rightarrow$ $AB=a,BD=\frac{1}{2}BC=\frac{a}{2}$

Since $△ABD$ is a right-angled triangle at D.

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

$\Rightarrow$ $a^2={AD}^2+{\left(\frac{a}{2}\right)}^2$

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

$\Rightarrow$ $AD=\frac{\sqrt{3a}}{2}$

$\therefore$ $Altitude=\frac{\sqrt{3}}{2}a$ $\left[Henceproved\right]$

Now,

$Areaof△ABC=\left(1/2\right)(Base\times Height)$

$\Rightarrow$ $Areaof△ABC=\frac{1}{2}(BC\times AD)$

$\Rightarrow$ $Areaof△ABC=\frac{1}{2}\times a\times \frac{\sqrt{3}}{2}a=\frac{\sqrt{3}}{4}{a}^2$ $\left[Henceproved\right]$