Question

AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC$=10$cm, find the length of AD, correct to $1$ place of decimal.

Answer 1

The given Traingle is a Equilateral triangle , hence all Sides are Equal

$AB=BC=AC=10cm$

We have, THEOREM : Any line drawn perpendicular to the opposite side is an equilateral triangle bisects the opposite side.

$\therefore BD=\frac{1}{2}\times BC$

$=\frac{1}{2}\times 10=5cm$

Hence According to the Pythagoras Theorem

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

${10}^2=A{D}^2+5^2$

$A{D}^2=100-25=75$

$AD=5\sqrt{3}$

The length of $AD=5\sqrt{3}=5\left(1.73205081\right)=8.66025405$

Correcting to 1 decimal place,

Hence your answer : The length of $AD=8.6$