Question

2. Prove that the altitude in an equilateral triangle are equal
please give me answer in method​

Answer 1

Answer:

Derivation of Area of an equilateral triangle ;

Let ABC be an equilateral triangle with sides 'a'. Now, draw AD perpendicular to BC.

Here, we have ΔABD = ΔADC.

We will find area of ΔABD using pythagorean theorem, according to which, the square of hypotenuse is equal to the sum of the squares of the other two sides.

Here, we have ;

\begin{gathered}\sf a {}^{2} = h {}^{2} + (\frac{a}{2} ) {}^{2} \\ \\ \sf h {}^{2} = a {}^{2} - \frac{a {}^{2} }{4} \\ \\ \sf h {}^{2} = \frac{3a {}^{2} }{4} \\ \\ \sf h = \frac{ \sqrt{3} }{2} a\end{gathered}a2=h2+(2a)2h2=a2−4a2h2=43a2h=23a

Now, we get the height ;

\begin{gathered}\sf area \: of \: \Delta = \frac{1}{2} \times base \times height \\ \\ \sf area \: of \: \Delta = \frac{1}{2} \times a \times \frac{ \sqrt{3} }{2} a \\ \\ \sf area \: of \: \Delta = \frac{ \sqrt{3} }{4} a {}^{2}\end{gathered}areaofΔ=21×base×heightareaofΔ=21×a×23aareaofΔ=43a2

Hence, area of equilateral triangle is

\sf area \: of \: \Delta = \frac{ \sqrt{3} }{4} a {}^{2}areaofΔ=43a2