Question

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

Answer 1

$\textbf{ Hello Mate! }$

In traingle ABC, AD is perpenducular to BC.

Here, BC = 10 cm.

Now, as triangle is equilateral so every side will be 10 cm

And as AD is falling on BC, it divides it into two equal parts [ Altitude drawn from vertex of equal. triangle on opposite aide divides it into two equal parts.

So, BD = CD = 5 cm [ :. 10 / 2 = 5 cm ]

Now, as traingle ABD is right angles at angle D so

h^2 = b^2 + p^2

10^2 = 5^2 + p^2

10^2 - 5^2 = p^2

( 10 + 5 )( 10 - 5 ) = p^2

15 × 5 = p^2

Root 75 = p or AD

8.67 cm = AD

Hence, value of AD is 8.7 cm correct to 1 decimal place.

Have great future ahead!

Answer 2

See through the picture I am posting here..

I hope it will be helpful